Hypothesis Testing for Means & Proportions
Lisa Sullivan, PhD
Professor of Biostatistics
Boston University School of Public Health
This is the first of three modules that will addresses the second area of statistical inference, which is hypothesis testing, in which a specific statement or hypothesis is generated about a population parameter, and sample statistics are used to assess the likelihood that the hypothesis is true. The hypothesis is based on available information and the investigator's belief about the population parameters. The process of hypothesis testing involves setting up two competing hypotheses, the null hypothesis and the alternate hypothesis. One selects a random sample (or multiple samples when there are more comparison groups), computes summary statistics and then assesses the likelihood that the sample data support the research or alternative hypothesis. Similar to estimation, the process of hypothesis testing is based on probability theory and the Central Limit Theorem.
This module will focus on hypothesis testing for means and proportions. The next two modules in this series will address analysis of variance and chi-squared tests.
After completing this module, the student will be able to:
Techniques for hypothesis testing .
The techniques for hypothesis testing depend on
In estimation we focused explicitly on techniques for one and two samples and discussed estimation for a specific parameter (e.g., the mean or proportion of a population), for differences (e.g., difference in means, the risk difference) and ratios (e.g., the relative risk and odds ratio). Here we will focus on procedures for one and two samples when the outcome is either continuous (and we focus on means) or dichotomous (and we focus on proportions).
The Centers for Disease Control (CDC) reported on trends in weight, height and body mass index from the 1960's through 2002. 1 The general trend was that Americans were much heavier and slightly taller in 2002 as compared to 1960; both men and women gained approximately 24 pounds, on average, between 1960 and 2002. In 2002, the mean weight for men was reported at 191 pounds. Suppose that an investigator hypothesizes that weights are even higher in 2006 (i.e., that the trend continued over the subsequent 4 years). The research hypothesis is that the mean weight in men in 2006 is more than 191 pounds. The null hypothesis is that there is no change in weight, and therefore the mean weight is still 191 pounds in 2006.
Null Hypothesis | H : μ= 191 (no change) |
Research Hypothesis | H : μ> 191 (investigator's belief) |
In order to test the hypotheses, we select a random sample of American males in 2006 and measure their weights. Suppose we have resources available to recruit n=100 men into our sample. We weigh each participant and compute summary statistics on the sample data. Suppose in the sample we determine the following:
Do the sample data support the null or research hypothesis? The sample mean of 197.1 is numerically higher than 191. However, is this difference more than would be expected by chance? In hypothesis testing, we assume that the null hypothesis holds until proven otherwise. We therefore need to determine the likelihood of observing a sample mean of 197.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true or under the null hypothesis). We can compute this probability using the Central Limit Theorem. Specifically,
(Notice that we use the sample standard deviation in computing the Z score. This is generally an appropriate substitution as long as the sample size is large, n > 30. Thus, there is less than a 1% probability of observing a sample mean as large as 197.1 when the true population mean is 191. Do you think that the null hypothesis is likely true? Based on how unlikely it is to observe a sample mean of 197.1 under the null hypothesis (i.e., <1% probability), we might infer, from our data, that the null hypothesis is probably not true.
Suppose that the sample data had turned out differently. Suppose that we instead observed the following in 2006:
How likely it is to observe a sample mean of 192.1 or higher when the true population mean is 191 (i.e., if the null hypothesis is true)? We can again compute this probability using the Central Limit Theorem. Specifically,
There is a 33.4% probability of observing a sample mean as large as 192.1 when the true population mean is 191. Do you think that the null hypothesis is likely true?
Neither of the sample means that we obtained allows us to know with certainty whether the null hypothesis is true or not. However, our computations suggest that, if the null hypothesis were true, the probability of observing a sample mean >197.1 is less than 1%. In contrast, if the null hypothesis were true, the probability of observing a sample mean >192.1 is about 33%. We can't know whether the null hypothesis is true, but the sample that provided a mean value of 197.1 provides much stronger evidence in favor of rejecting the null hypothesis, than the sample that provided a mean value of 192.1. Note that this does not mean that a sample mean of 192.1 indicates that the null hypothesis is true; it just doesn't provide compelling evidence to reject it.
In essence, hypothesis testing is a procedure to compute a probability that reflects the strength of the evidence (based on a given sample) for rejecting the null hypothesis. In hypothesis testing, we determine a threshold or cut-off point (called the critical value) to decide when to believe the null hypothesis and when to believe the research hypothesis. It is important to note that it is possible to observe any sample mean when the true population mean is true (in this example equal to 191), but some sample means are very unlikely. Based on the two samples above it would seem reasonable to believe the research hypothesis when x̄ = 197.1, but to believe the null hypothesis when x̄ =192.1. What we need is a threshold value such that if x̄ is above that threshold then we believe that H 1 is true and if x̄ is below that threshold then we believe that H 0 is true. The difficulty in determining a threshold for x̄ is that it depends on the scale of measurement. In this example, the threshold, sometimes called the critical value, might be 195 (i.e., if the sample mean is 195 or more then we believe that H 1 is true and if the sample mean is less than 195 then we believe that H 0 is true). Suppose we are interested in assessing an increase in blood pressure over time, the critical value will be different because blood pressures are measured in millimeters of mercury (mmHg) as opposed to in pounds. In the following we will explain how the critical value is determined and how we handle the issue of scale.
First, to address the issue of scale in determining the critical value, we convert our sample data (in particular the sample mean) into a Z score. We know from the module on probability that the center of the Z distribution is zero and extreme values are those that exceed 2 or fall below -2. Z scores above 2 and below -2 represent approximately 5% of all Z values. If the observed sample mean is close to the mean specified in H 0 (here m =191), then Z will be close to zero. If the observed sample mean is much larger than the mean specified in H 0 , then Z will be large.
In hypothesis testing, we select a critical value from the Z distribution. This is done by first determining what is called the level of significance, denoted α ("alpha"). What we are doing here is drawing a line at extreme values. The level of significance is the probability that we reject the null hypothesis (in favor of the alternative) when it is actually true and is also called the Type I error rate.
α = Level of significance = P(Type I error) = P(Reject H 0 | H 0 is true).
Because α is a probability, it ranges between 0 and 1. The most commonly used value in the medical literature for α is 0.05, or 5%. Thus, if an investigator selects α=0.05, then they are allowing a 5% probability of incorrectly rejecting the null hypothesis in favor of the alternative when the null is in fact true. Depending on the circumstances, one might choose to use a level of significance of 1% or 10%. For example, if an investigator wanted to reject the null only if there were even stronger evidence than that ensured with α=0.05, they could choose a =0.01as their level of significance. The typical values for α are 0.01, 0.05 and 0.10, with α=0.05 the most commonly used value.
Suppose in our weight study we select α=0.05. We need to determine the value of Z that holds 5% of the values above it (see below).
The critical value of Z for α =0.05 is Z = 1.645 (i.e., 5% of the distribution is above Z=1.645). With this value we can set up what is called our decision rule for the test. The rule is to reject H 0 if the Z score is 1.645 or more.
With the first sample we have
Because 2.38 > 1.645, we reject the null hypothesis. (The same conclusion can be drawn by comparing the 0.0087 probability of observing a sample mean as extreme as 197.1 to the level of significance of 0.05. If the observed probability is smaller than the level of significance we reject H 0 ). Because the Z score exceeds the critical value, we conclude that the mean weight for men in 2006 is more than 191 pounds, the value reported in 2002. If we observed the second sample (i.e., sample mean =192.1), we would not be able to reject the null hypothesis because the Z score is 0.43 which is not in the rejection region (i.e., the region in the tail end of the curve above 1.645). With the second sample we do not have sufficient evidence (because we set our level of significance at 5%) to conclude that weights have increased. Again, the same conclusion can be reached by comparing probabilities. The probability of observing a sample mean as extreme as 192.1 is 33.4% which is not below our 5% level of significance.
The procedure for hypothesis testing is based on the ideas described above. Specifically, we set up competing hypotheses, select a random sample from the population of interest and compute summary statistics. We then determine whether the sample data supports the null or alternative hypotheses. The procedure can be broken down into the following five steps.
H 0 : Null hypothesis (no change, no difference);
H 1 : Research hypothesis (investigator's belief); α =0.05
Upper-tailed, Lower-tailed, Two-tailed Tests The research or alternative hypothesis can take one of three forms. An investigator might believe that the parameter has increased, decreased or changed. For example, an investigator might hypothesize: : μ > μ , where μ is the comparator or null value (e.g., μ =191 in our example about weight in men in 2006) and an increase is hypothesized - this type of test is called an ; : μ < μ , where a decrease is hypothesized and this is called a ; or : μ ≠ μ where a difference is hypothesized and this is called a .The exact form of the research hypothesis depends on the investigator's belief about the parameter of interest and whether it has possibly increased, decreased or is different from the null value. The research hypothesis is set up by the investigator before any data are collected.
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The test statistic is a single number that summarizes the sample information. An example of a test statistic is the Z statistic computed as follows:
When the sample size is small, we will use t statistics (just as we did when constructing confidence intervals for small samples). As we present each scenario, alternative test statistics are provided along with conditions for their appropriate use.
The decision rule is a statement that tells under what circumstances to reject the null hypothesis. The decision rule is based on specific values of the test statistic (e.g., reject H 0 if Z > 1.645). The decision rule for a specific test depends on 3 factors: the research or alternative hypothesis, the test statistic and the level of significance. Each is discussed below.
The following figures illustrate the rejection regions defined by the decision rule for upper-, lower- and two-tailed Z tests with α=0.05. Notice that the rejection regions are in the upper, lower and both tails of the curves, respectively. The decision rules are written below each figure.
Rejection Region for Upper-Tailed Z Test (H : μ > μ ) with α=0.05 The decision rule is: Reject H if Z 1.645. |
Rejection Region for Lower-Tailed Z Test (H 1 : μ < μ 0 ) with α =0.05 The decision rule is: Reject H 0 if Z < 1.645.
Rejection Region for Two-Tailed Z Test (H 1 : μ ≠ μ 0 ) with α =0.05 The decision rule is: Reject H 0 if Z < -1.960 or if Z > 1.960.
The complete table of critical values of Z for upper, lower and two-tailed tests can be found in the table of Z values to the right in "Other Resources." Critical values of t for upper, lower and two-tailed tests can be found in the table of t values in "Other Resources."
Here we compute the test statistic by substituting the observed sample data into the test statistic identified in Step 2.
The final conclusion is made by comparing the test statistic (which is a summary of the information observed in the sample) to the decision rule. The final conclusion will be either to reject the null hypothesis (because the sample data are very unlikely if the null hypothesis is true) or not to reject the null hypothesis (because the sample data are not very unlikely). If the null hypothesis is rejected, then an exact significance level is computed to describe the likelihood of observing the sample data assuming that the null hypothesis is true. The exact level of significance is called the p-value and it will be less than the chosen level of significance if we reject H 0 . Statistical computing packages provide exact p-values as part of their standard output for hypothesis tests. In fact, when using a statistical computing package, the steps outlined about can be abbreviated. The hypotheses (step 1) should always be set up in advance of any analysis and the significance criterion should also be determined (e.g., α =0.05). Statistical computing packages will produce the test statistic (usually reporting the test statistic as t) and a p-value. The investigator can then determine statistical significance using the following: If p < α then reject H 0 .
H 0 : μ = 191 H 1 : μ > 191 α =0.05 The research hypothesis is that weights have increased, and therefore an upper tailed test is used.
Because the sample size is large (n > 30) the appropriate test statistic is
In this example, we are performing an upper tailed test (H 1 : μ> 191), with a Z test statistic and selected α =0.05. Reject H 0 if Z > 1.645. We now substitute the sample data into the formula for the test statistic identified in Step 2. We reject H 0 because 2.38 > 1.645. We have statistically significant evidence at a =0.05, to show that the mean weight in men in 2006 is more than 191 pounds. Because we rejected the null hypothesis, we now approximate the p-value which is the likelihood of observing the sample data if the null hypothesis is true. An alternative definition of the p-value is the smallest level of significance where we can still reject H 0 . In this example, we observed Z=2.38 and for α=0.05, the critical value was 1.645. Because 2.38 exceeded 1.645 we rejected H 0 . In our conclusion we reported a statistically significant increase in mean weight at a 5% level of significance. Using the table of critical values for upper tailed tests, we can approximate the p-value. If we select α=0.025, the critical value is 1.96, and we still reject H 0 because 2.38 > 1.960. If we select α=0.010 the critical value is 2.326, and we still reject H 0 because 2.38 > 2.326. However, if we select α=0.005, the critical value is 2.576, and we cannot reject H 0 because 2.38 < 2.576. Therefore, the smallest α where we still reject H 0 is 0.010. This is the p-value. A statistical computing package would produce a more precise p-value which would be in between 0.005 and 0.010. Here we are approximating the p-value and would report p < 0.010. Type I and Type II ErrorsIn all tests of hypothesis, there are two types of errors that can be committed. The first is called a Type I error and refers to the situation where we incorrectly reject H 0 when in fact it is true. This is also called a false positive result (as we incorrectly conclude that the research hypothesis is true when in fact it is not). When we run a test of hypothesis and decide to reject H 0 (e.g., because the test statistic exceeds the critical value in an upper tailed test) then either we make a correct decision because the research hypothesis is true or we commit a Type I error. The different conclusions are summarized in the table below. Note that we will never know whether the null hypothesis is really true or false (i.e., we will never know which row of the following table reflects reality). Table - Conclusions in Test of Hypothesis
In the first step of the hypothesis test, we select a level of significance, α, and α= P(Type I error). Because we purposely select a small value for α, we control the probability of committing a Type I error. For example, if we select α=0.05, and our test tells us to reject H 0 , then there is a 5% probability that we commit a Type I error. Most investigators are very comfortable with this and are confident when rejecting H 0 that the research hypothesis is true (as it is the more likely scenario when we reject H 0 ). When we run a test of hypothesis and decide not to reject H 0 (e.g., because the test statistic is below the critical value in an upper tailed test) then either we make a correct decision because the null hypothesis is true or we commit a Type II error. Beta (β) represents the probability of a Type II error and is defined as follows: β=P(Type II error) = P(Do not Reject H 0 | H 0 is false). Unfortunately, we cannot choose β to be small (e.g., 0.05) to control the probability of committing a Type II error because β depends on several factors including the sample size, α, and the research hypothesis. When we do not reject H 0 , it may be very likely that we are committing a Type II error (i.e., failing to reject H 0 when in fact it is false). Therefore, when tests are run and the null hypothesis is not rejected we often make a weak concluding statement allowing for the possibility that we might be committing a Type II error. If we do not reject H 0 , we conclude that we do not have significant evidence to show that H 1 is true. We do not conclude that H 0 is true. The most common reason for a Type II error is a small sample size. Tests with One Sample, Continuous OutcomeHypothesis testing applications with a continuous outcome variable in a single population are performed according to the five-step procedure outlined above. A key component is setting up the null and research hypotheses. The objective is to compare the mean in a single population to known mean (μ 0 ). The known value is generally derived from another study or report, for example a study in a similar, but not identical, population or a study performed some years ago. The latter is called a historical control. It is important in setting up the hypotheses in a one sample test that the mean specified in the null hypothesis is a fair and reasonable comparator. This will be discussed in the examples that follow. Test Statistics for Testing H 0 : μ= μ 0
Note that statistical computing packages will use the t statistic exclusively and make the necessary adjustments for comparing the test statistic to appropriate values from probability tables to produce a p-value. The National Center for Health Statistics (NCHS) published a report in 2005 entitled Health, United States, containing extensive information on major trends in the health of Americans. Data are provided for the US population as a whole and for specific ages, sexes and races. The NCHS report indicated that in 2002 Americans paid an average of $3,302 per year on health care and prescription drugs. An investigator hypothesizes that in 2005 expenditures have decreased primarily due to the availability of generic drugs. To test the hypothesis, a sample of 100 Americans are selected and their expenditures on health care and prescription drugs in 2005 are measured. The sample data are summarized as follows: n=100, x̄ =$3,190 and s=$890. Is there statistical evidence of a reduction in expenditures on health care and prescription drugs in 2005? Is the sample mean of $3,190 evidence of a true reduction in the mean or is it within chance fluctuation? We will run the test using the five-step approach.
H 0 : μ = 3,302 H 1 : μ < 3,302 α =0.05 The research hypothesis is that expenditures have decreased, and therefore a lower-tailed test is used. This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645.
We do not reject H 0 because -1.26 > -1.645. We do not have statistically significant evidence at α=0.05 to show that the mean expenditures on health care and prescription drugs are lower in 2005 than the mean of $3,302 reported in 2002. Recall that when we fail to reject H 0 in a test of hypothesis that either the null hypothesis is true (here the mean expenditures in 2005 are the same as those in 2002 and equal to $3,302) or we committed a Type II error (i.e., we failed to reject H 0 when in fact it is false). In summarizing this test, we conclude that we do not have sufficient evidence to reject H 0 . We do not conclude that H 0 is true, because there may be a moderate to high probability that we committed a Type II error. It is possible that the sample size is not large enough to detect a difference in mean expenditures. The NCHS reported that the mean total cholesterol level in 2002 for all adults was 203. Total cholesterol levels in participants who attended the seventh examination of the Offspring in the Framingham Heart Study are summarized as follows: n=3,310, x̄ =200.3, and s=36.8. Is there statistical evidence of a difference in mean cholesterol levels in the Framingham Offspring? Here we want to assess whether the sample mean of 200.3 in the Framingham sample is statistically significantly different from 203 (i.e., beyond what we would expect by chance). We will run the test using the five-step approach. H 0 : μ= 203 H 1 : μ≠ 203 α=0.05 The research hypothesis is that cholesterol levels are different in the Framingham Offspring, and therefore a two-tailed test is used.
This is a two-tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.960 or is Z > 1.960. We reject H 0 because -4.22 ≤ -1. .960. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level in the Framingham Offspring is different from the national average of 203 reported in 2002. Because we reject H 0 , we also approximate a p-value. Using the two-sided significance levels, p < 0.0001. Statistical Significance versus Clinical (Practical) SignificanceThis example raises an important concept of statistical versus clinical or practical significance. From a statistical standpoint, the total cholesterol levels in the Framingham sample are highly statistically significantly different from the national average with p < 0.0001 (i.e., there is less than a 0.01% chance that we are incorrectly rejecting the null hypothesis). However, the sample mean in the Framingham Offspring study is 200.3, less than 3 units different from the national mean of 203. The reason that the data are so highly statistically significant is due to the very large sample size. It is always important to assess both statistical and clinical significance of data. This is particularly relevant when the sample size is large. Is a 3 unit difference in total cholesterol a meaningful difference? Consider again the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. Suppose a new drug is proposed to lower total cholesterol. A study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients are enrolled in the study and asked to take the new drug for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows: n=15, x̄ =195.9 and s=28.7. Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new drug for 6 weeks? We will run the test using the five-step approach. H 0 : μ= 203 H 1 : μ< 203 α=0.05
Because the sample size is small (n<30) the appropriate test statistic is This is a lower tailed test, using a t statistic and a 5% level of significance. In order to determine the critical value of t, we need degrees of freedom, df, defined as df=n-1. In this example df=15-1=14. The critical value for a lower tailed test with df=14 and a =0.05 is -2.145 and the decision rule is as follows: Reject H 0 if t < -2.145. We do not reject H 0 because -0.96 > -2.145. We do not have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower than the national mean in patients taking the new drug for 6 weeks. Again, because we failed to reject the null hypothesis we make a weaker concluding statement allowing for the possibility that we may have committed a Type II error (i.e., failed to reject H 0 when in fact the drug is efficacious). This example raises an important issue in terms of study design. In this example we assume in the null hypothesis that the mean cholesterol level is 203. This is taken to be the mean cholesterol level in patients without treatment. Is this an appropriate comparator? Alternative and potentially more efficient study designs to evaluate the effect of the new drug could involve two treatment groups, where one group receives the new drug and the other does not, or we could measure each patient's baseline or pre-treatment cholesterol level and then assess changes from baseline to 6 weeks post-treatment. These designs are also discussed here. Video - Comparing a Sample Mean to Known Population Mean (8:20) Link to transcript of the video Tests with One Sample, Dichotomous OutcomeHypothesis testing applications with a dichotomous outcome variable in a single population are also performed according to the five-step procedure. Similar to tests for means, a key component is setting up the null and research hypotheses. The objective is to compare the proportion of successes in a single population to a known proportion (p 0 ). That known proportion is generally derived from another study or report and is sometimes called a historical control. It is important in setting up the hypotheses in a one sample test that the proportion specified in the null hypothesis is a fair and reasonable comparator. In one sample tests for a dichotomous outcome, we set up our hypotheses against an appropriate comparator. We select a sample and compute descriptive statistics on the sample data. Specifically, we compute the sample size (n) and the sample proportion which is computed by taking the ratio of the number of successes to the sample size, We then determine the appropriate test statistic (Step 2) for the hypothesis test. The formula for the test statistic is given below. Test Statistic for Testing H 0 : p = p 0 if min(np 0 , n(1-p 0 )) > 5 The formula above is appropriate for large samples, defined when the smaller of np 0 and n(1-p 0 ) is at least 5. This is similar, but not identical, to the condition required for appropriate use of the confidence interval formula for a population proportion, i.e., Here we use the proportion specified in the null hypothesis as the true proportion of successes rather than the sample proportion. If we fail to satisfy the condition, then alternative procedures, called exact methods must be used to test the hypothesis about the population proportion. Example: The NCHS report indicated that in 2002 the prevalence of cigarette smoking among American adults was 21.1%. Data on prevalent smoking in n=3,536 participants who attended the seventh examination of the Offspring in the Framingham Heart Study indicated that 482/3,536 = 13.6% of the respondents were currently smoking at the time of the exam. Suppose we want to assess whether the prevalence of smoking is lower in the Framingham Offspring sample given the focus on cardiovascular health in that community. Is there evidence of a statistically lower prevalence of smoking in the Framingham Offspring study as compared to the prevalence among all Americans? H 0 : p = 0.211 H 1 : p < 0.211 α=0.05 We must first check that the sample size is adequate. Specifically, we need to check min(np 0 , n(1-p 0 )) = min( 3,536(0.211), 3,536(1-0.211))=min(746, 2790)=746. The sample size is more than adequate so the following formula can be used: This is a lower tailed test, using a Z statistic and a 5% level of significance. Reject H 0 if Z < -1.645. We reject H 0 because -10.93 < -1.645. We have statistically significant evidence at α=0.05 to show that the prevalence of smoking in the Framingham Offspring is lower than the prevalence nationally (21.1%). Here, p < 0.0001. The NCHS report indicated that in 2002, 75% of children aged 2 to 17 saw a dentist in the past year. An investigator wants to assess whether use of dental services is similar in children living in the city of Boston. A sample of 125 children aged 2 to 17 living in Boston are surveyed and 64 reported seeing a dentist over the past 12 months. Is there a significant difference in use of dental services between children living in Boston and the national data? Calculate this on your own before checking the answer. Video - Hypothesis Test for One Sample and a Dichotomous Outcome (3:55) Tests with Two Independent Samples, Continuous OutcomeThere are many applications where it is of interest to compare two independent groups with respect to their mean scores on a continuous outcome. Here we compare means between groups, but rather than generating an estimate of the difference, we will test whether the observed difference (increase, decrease or difference) is statistically significant or not. Remember, that hypothesis testing gives an assessment of statistical significance, whereas estimation gives an estimate of effect and both are important. Here we discuss the comparison of means when the two comparison groups are independent or physically separate. The two groups might be determined by a particular attribute (e.g., sex, diagnosis of cardiovascular disease) or might be set up by the investigator (e.g., participants assigned to receive an experimental treatment or placebo). The first step in the analysis involves computing descriptive statistics on each of the two samples. Specifically, we compute the sample size, mean and standard deviation in each sample and we denote these summary statistics as follows: for sample 1: for sample 2: The designation of sample 1 and sample 2 is arbitrary. In a clinical trial setting the convention is to call the treatment group 1 and the control group 2. However, when comparing men and women, for example, either group can be 1 or 2. In the two independent samples application with a continuous outcome, the parameter of interest in the test of hypothesis is the difference in population means, μ 1 -μ 2 . The null hypothesis is always that there is no difference between groups with respect to means, i.e., The null hypothesis can also be written as follows: H 0 : μ 1 = μ 2 . In the research hypothesis, an investigator can hypothesize that the first mean is larger than the second (H 1 : μ 1 > μ 2 ), that the first mean is smaller than the second (H 1 : μ 1 < μ 2 ), or that the means are different (H 1 : μ 1 ≠ μ 2 ). The three different alternatives represent upper-, lower-, and two-tailed tests, respectively. The following test statistics are used to test these hypotheses. Test Statistics for Testing H 0 : μ 1 = μ 2
NOTE: The formulas above assume equal variability in the two populations (i.e., the population variances are equal, or s 1 2 = s 2 2 ). This means that the outcome is equally variable in each of the comparison populations. For analysis, we have samples from each of the comparison populations. If the sample variances are similar, then the assumption about variability in the populations is probably reasonable. As a guideline, if the ratio of the sample variances, s 1 2 /s 2 2 is between 0.5 and 2 (i.e., if one variance is no more than double the other), then the formulas above are appropriate. If the ratio of the sample variances is greater than 2 or less than 0.5 then alternative formulas must be used to account for the heterogeneity in variances. The test statistics include Sp, which is the pooled estimate of the common standard deviation (again assuming that the variances in the populations are similar) computed as the weighted average of the standard deviations in the samples as follows: Because we are assuming equal variances between groups, we pool the information on variability (sample variances) to generate an estimate of the variability in the population. Note: Because Sp is a weighted average of the standard deviations in the sample, Sp will always be in between s 1 and s 2 .) Data measured on n=3,539 participants who attended the seventh examination of the Offspring in the Framingham Heart Study are shown below.
Suppose we now wish to assess whether there is a statistically significant difference in mean systolic blood pressures between men and women using a 5% level of significance. H 0 : μ 1 = μ 2 H 1 : μ 1 ≠ μ 2 α=0.05 Because both samples are large ( > 30), we can use the Z test statistic as opposed to t. Note that statistical computing packages use t throughout. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The guideline suggests investigating the ratio of the sample variances, s 1 2 /s 2 2 . Suppose we call the men group 1 and the women group 2. Again, this is arbitrary; it only needs to be noted when interpreting the results. The ratio of the sample variances is 17.5 2 /20.1 2 = 0.76, which falls between 0.5 and 2 suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is We now substitute the sample data into the formula for the test statistic identified in Step 2. Before substituting, we will first compute Sp, the pooled estimate of the common standard deviation. Notice that the pooled estimate of the common standard deviation, Sp, falls in between the standard deviations in the comparison groups (i.e., 17.5 and 20.1). Sp is slightly closer in value to the standard deviation in the women (20.1) as there were slightly more women in the sample. Recall, Sp is a weight average of the standard deviations in the comparison groups, weighted by the respective sample sizes. Now the test statistic: We reject H 0 because 2.66 > 1.960. We have statistically significant evidence at α=0.05 to show that there is a difference in mean systolic blood pressures between men and women. The p-value is p < 0.010. Here again we find that there is a statistically significant difference in mean systolic blood pressures between men and women at p < 0.010. Notice that there is a very small difference in the sample means (128.2-126.5 = 1.7 units), but this difference is beyond what would be expected by chance. Is this a clinically meaningful difference? The large sample size in this example is driving the statistical significance. A 95% confidence interval for the difference in mean systolic blood pressures is: 1.7 + 1.26 or (0.44, 2.96). The confidence interval provides an assessment of the magnitude of the difference between means whereas the test of hypothesis and p-value provide an assessment of the statistical significance of the difference. Above we performed a study to evaluate a new drug designed to lower total cholesterol. The study involved one sample of patients, each patient took the new drug for 6 weeks and had their cholesterol measured. As a means of evaluating the efficacy of the new drug, the mean total cholesterol following 6 weeks of treatment was compared to the NCHS-reported mean total cholesterol level in 2002 for all adults of 203. At the end of the example, we discussed the appropriateness of the fixed comparator as well as an alternative study design to evaluate the effect of the new drug involving two treatment groups, where one group receives the new drug and the other does not. Here, we revisit the example with a concurrent or parallel control group, which is very typical in randomized controlled trials or clinical trials (refer to the EP713 module on Clinical Trials). A new drug is proposed to lower total cholesterol. A randomized controlled trial is designed to evaluate the efficacy of the medication in lowering cholesterol. Thirty participants are enrolled in the trial and are randomly assigned to receive either the new drug or a placebo. The participants do not know which treatment they are assigned. Each participant is asked to take the assigned treatment for 6 weeks. At the end of 6 weeks, each patient's total cholesterol level is measured and the sample statistics are as follows.
Is there statistical evidence of a reduction in mean total cholesterol in patients taking the new drug for 6 weeks as compared to participants taking placebo? We will run the test using the five-step approach. H 0 : μ 1 = μ 2 H 1 : μ 1 < μ 2 α=0.05 Because both samples are small (< 30), we use the t test statistic. Before implementing the formula, we first check whether the assumption of equality of population variances is reasonable. The ratio of the sample variances, s 1 2 /s 2 2 =28.7 2 /30.3 2 = 0.90, which falls between 0.5 and 2, suggesting that the assumption of equality of population variances is reasonable. The appropriate test statistic is: This is a lower-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table (in More Resources to the right). In order to determine the critical value of t we need degrees of freedom, df, defined as df=n 1 +n 2 -2 = 15+15-2=28. The critical value for a lower tailed test with df=28 and α=0.05 is -1.701 and the decision rule is: Reject H 0 if t < -1.701. Now the test statistic, We reject H 0 because -2.92 < -1.701. We have statistically significant evidence at α=0.05 to show that the mean total cholesterol level is lower in patients taking the new drug for 6 weeks as compared to patients taking placebo, p < 0.005. The clinical trial in this example finds a statistically significant reduction in total cholesterol, whereas in the previous example where we had a historical control (as opposed to a parallel control group) we did not demonstrate efficacy of the new drug. Notice that the mean total cholesterol level in patients taking placebo is 217.4 which is very different from the mean cholesterol reported among all Americans in 2002 of 203 and used as the comparator in the prior example. The historical control value may not have been the most appropriate comparator as cholesterol levels have been increasing over time. In the next section, we present another design that can be used to assess the efficacy of the new drug. Video - Comparison of Two Independent Samples With a Continuous Outcome (8:02) Tests with Matched Samples, Continuous OutcomeIn the previous section we compared two groups with respect to their mean scores on a continuous outcome. An alternative study design is to compare matched or paired samples. The two comparison groups are said to be dependent, and the data can arise from a single sample of participants where each participant is measured twice (possibly before and after an intervention) or from two samples that are matched on specific characteristics (e.g., siblings). When the samples are dependent, we focus on difference scores in each participant or between members of a pair and the test of hypothesis is based on the mean difference, μ d . The null hypothesis again reflects "no difference" and is stated as H 0 : μ d =0 . Note that there are some instances where it is of interest to test whether there is a difference of a particular magnitude (e.g., μ d =5) but in most instances the null hypothesis reflects no difference (i.e., μ d =0). The appropriate formula for the test of hypothesis depends on the sample size. The formulas are shown below and are identical to those we presented for estimating the mean of a single sample presented (e.g., when comparing against an external or historical control), except here we focus on difference scores. Test Statistics for Testing H 0 : μ d =0 A new drug is proposed to lower total cholesterol and a study is designed to evaluate the efficacy of the drug in lowering cholesterol. Fifteen patients agree to participate in the study and each is asked to take the new drug for 6 weeks. However, before starting the treatment, each patient's total cholesterol level is measured. The initial measurement is a pre-treatment or baseline value. After taking the drug for 6 weeks, each patient's total cholesterol level is measured again and the data are shown below. The rightmost column contains difference scores for each patient, computed by subtracting the 6 week cholesterol level from the baseline level. The differences represent the reduction in total cholesterol over 4 weeks. (The differences could have been computed by subtracting the baseline total cholesterol level from the level measured at 6 weeks. The way in which the differences are computed does not affect the outcome of the analysis only the interpretation.)
Because the differences are computed by subtracting the cholesterols measured at 6 weeks from the baseline values, positive differences indicate reductions and negative differences indicate increases (e.g., participant 12 increases by 2 units over 6 weeks). The goal here is to test whether there is a statistically significant reduction in cholesterol. Because of the way in which we computed the differences, we want to look for an increase in the mean difference (i.e., a positive reduction). In order to conduct the test, we need to summarize the differences. In this sample, we have The calculations are shown below.
Is there statistical evidence of a reduction in mean total cholesterol in patients after using the new medication for 6 weeks? We will run the test using the five-step approach. H 0 : μ d = 0 H 1 : μ d > 0 α=0.05 NOTE: If we had computed differences by subtracting the baseline level from the level measured at 6 weeks then negative differences would have reflected reductions and the research hypothesis would have been H 1 : μ d < 0.
This is an upper-tailed test, using a t statistic and a 5% level of significance. The appropriate critical value can be found in the t Table at the right, with df=15-1=14. The critical value for an upper-tailed test with df=14 and α=0.05 is 2.145 and the decision rule is Reject H 0 if t > 2.145. We now substitute the sample data into the formula for the test statistic identified in Step 2. We reject H 0 because 4.61 > 2.145. We have statistically significant evidence at α=0.05 to show that there is a reduction in cholesterol levels over 6 weeks. Here we illustrate the use of a matched design to test the efficacy of a new drug to lower total cholesterol. We also considered a parallel design (randomized clinical trial) and a study using a historical comparator. It is extremely important to design studies that are best suited to detect a meaningful difference when one exists. There are often several alternatives and investigators work with biostatisticians to determine the best design for each application. It is worth noting that the matched design used here can be problematic in that observed differences may only reflect a "placebo" effect. All participants took the assigned medication, but is the observed reduction attributable to the medication or a result of these participation in a study. Video - Hypothesis Testing With a Matched Sample and a Continuous Outcome (3:11) Tests with Two Independent Samples, Dichotomous OutcomeThere are several approaches that can be used to test hypotheses concerning two independent proportions. Here we present one approach - the chi-square test of independence is an alternative, equivalent, and perhaps more popular approach to the same analysis. Hypothesis testing with the chi-square test is addressed in the third module in this series: BS704_HypothesisTesting-ChiSquare. In tests of hypothesis comparing proportions between two independent groups, one test is performed and results can be interpreted to apply to a risk difference, relative risk or odds ratio. As a reminder, the risk difference is computed by taking the difference in proportions between comparison groups, the risk ratio is computed by taking the ratio of proportions, and the odds ratio is computed by taking the ratio of the odds of success in the comparison groups. Because the null values for the risk difference, the risk ratio and the odds ratio are different, the hypotheses in tests of hypothesis look slightly different depending on which measure is used. When performing tests of hypothesis for the risk difference, relative risk or odds ratio, the convention is to label the exposed or treated group 1 and the unexposed or control group 2. For example, suppose a study is designed to assess whether there is a significant difference in proportions in two independent comparison groups. The test of interest is as follows: H 0 : p 1 = p 2 versus H 1 : p 1 ≠ p 2 . The following are the hypothesis for testing for a difference in proportions using the risk difference, the risk ratio and the odds ratio. First, the hypotheses above are equivalent to the following:
Suppose a test is performed to test H 0 : RD = 0 versus H 1 : RD ≠ 0 and the test rejects H 0 at α=0.05. Based on this test we can conclude that there is significant evidence, α=0.05, of a difference in proportions, significant evidence that the risk difference is not zero, significant evidence that the risk ratio and odds ratio are not one. The risk difference is analogous to the difference in means when the outcome is continuous. Here the parameter of interest is the difference in proportions in the population, RD = p 1 -p 2 and the null value for the risk difference is zero. In a test of hypothesis for the risk difference, the null hypothesis is always H 0 : RD = 0. This is equivalent to H 0 : RR = 1 and H 0 : OR = 1. In the research hypothesis, an investigator can hypothesize that the first proportion is larger than the second (H 1 : p 1 > p 2 , which is equivalent to H 1 : RD > 0, H 1 : RR > 1 and H 1 : OR > 1), that the first proportion is smaller than the second (H 1 : p 1 < p 2 , which is equivalent to H 1 : RD < 0, H 1 : RR < 1 and H 1 : OR < 1), or that the proportions are different (H 1 : p 1 ≠ p 2 , which is equivalent to H 1 : RD ≠ 0, H 1 : RR ≠ 1 and H 1 : OR ≠ 1). The three different alternatives represent upper-, lower- and two-tailed tests, respectively. The formula for the test of hypothesis for the difference in proportions is given below. Test Statistics for Testing H 0 : p 1 = p
The formula above is appropriate for large samples, defined as at least 5 successes (np > 5) and at least 5 failures (n(1-p > 5)) in each of the two samples. If there are fewer than 5 successes or failures in either comparison group, then alternative procedures, called exact methods must be used to estimate the difference in population proportions. The following table summarizes data from n=3,799 participants who attended the fifth examination of the Offspring in the Framingham Heart Study. The outcome of interest is prevalent CVD and we want to test whether the prevalence of CVD is significantly higher in smokers as compared to non-smokers.
The prevalence of CVD (or proportion of participants with prevalent CVD) among non-smokers is 298/3,055 = 0.0975 and the prevalence of CVD among current smokers is 81/744 = 0.1089. Here smoking status defines the comparison groups and we will call the current smokers group 1 (exposed) and the non-smokers (unexposed) group 2. The test of hypothesis is conducted below using the five step approach. H 0 : p 1 = p 2 H 1 : p 1 ≠ p 2 α=0.05
We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group. In this example, we have more than enough successes (cases of prevalent CVD) and failures (persons free of CVD) in each comparison group. The sample size is more than adequate so the following formula can be used: Reject H 0 if Z < -1.960 or if Z > 1.960. We now substitute the sample data into the formula for the test statistic identified in Step 2. We first compute the overall proportion of successes: We now substitute to compute the test statistic.
We do not reject H 0 because -1.960 < 0.927 < 1.960. We do not have statistically significant evidence at α=0.05 to show that there is a difference in prevalent CVD between smokers and non-smokers. A 95% confidence interval for the difference in prevalent CVD (or risk difference) between smokers and non-smokers as 0.0114 + 0.0247, or between -0.0133 and 0.0361. Because the 95% confidence interval for the risk difference includes zero we again conclude that there is no statistically significant difference in prevalent CVD between smokers and non-smokers. Smoking has been shown over and over to be a risk factor for cardiovascular disease. What might explain the fact that we did not observe a statistically significant difference using data from the Framingham Heart Study? HINT: Here we consider prevalent CVD, would the results have been different if we considered incident CVD? A randomized trial is designed to evaluate the effectiveness of a newly developed pain reliever designed to reduce pain in patients following joint replacement surgery. The trial compares the new pain reliever to the pain reliever currently in use (called the standard of care). A total of 100 patients undergoing joint replacement surgery agreed to participate in the trial. Patients were randomly assigned to receive either the new pain reliever or the standard pain reliever following surgery and were blind to the treatment assignment. Before receiving the assigned treatment, patients were asked to rate their pain on a scale of 0-10 with higher scores indicative of more pain. Each patient was then given the assigned treatment and after 30 minutes was again asked to rate their pain on the same scale. The primary outcome was a reduction in pain of 3 or more scale points (defined by clinicians as a clinically meaningful reduction). The following data were observed in the trial.
We now test whether there is a statistically significant difference in the proportions of patients reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) using the five step approach. H 0 : p 1 = p 2 H 1 : p 1 ≠ p 2 α=0.05 Here the new or experimental pain reliever is group 1 and the standard pain reliever is group 2. We must first check that the sample size is adequate. Specifically, we need to ensure that we have at least 5 successes and 5 failures in each comparison group, i.e., In this example, we have min(50(0.46), 50(1-0.46), 50(0.22), 50(1-0.22)) = min(23, 27, 11, 39) = 11. The sample size is adequate so the following formula can be used We reject H 0 because 2.526 > 1960. We have statistically significant evidence at a =0.05 to show that there is a difference in the proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever. A 95% confidence interval for the difference in proportions of patients on the new pain reliever reporting a meaningful reduction (i.e., a reduction of 3 or more scale points) as compared to patients on the standard pain reliever is 0.24 + 0.18 or between 0.06 and 0.42. Because the 95% confidence interval does not include zero we concluded that there was a statistically significant difference in proportions which is consistent with the test of hypothesis result. Again, the procedures discussed here apply to applications where there are two independent comparison groups and a dichotomous outcome. There are other applications in which it is of interest to compare a dichotomous outcome in matched or paired samples. For example, in a clinical trial we might wish to test the effectiveness of a new antibiotic eye drop for the treatment of bacterial conjunctivitis. Participants use the new antibiotic eye drop in one eye and a comparator (placebo or active control treatment) in the other. The success of the treatment (yes/no) is recorded for each participant for each eye. Because the two assessments (success or failure) are paired, we cannot use the procedures discussed here. The appropriate test is called McNemar's test (sometimes called McNemar's test for dependent proportions). Vide0 - Hypothesis Testing With Two Independent Samples and a Dichotomous Outcome (2:55) Here we presented hypothesis testing techniques for means and proportions in one and two sample situations. Tests of hypothesis involve several steps, including specifying the null and alternative or research hypothesis, selecting and computing an appropriate test statistic, setting up a decision rule and drawing a conclusion. There are many details to consider in hypothesis testing. The first is to determine the appropriate test. We discussed Z and t tests here for different applications. The appropriate test depends on the distribution of the outcome variable (continuous or dichotomous), the number of comparison groups (one, two) and whether the comparison groups are independent or dependent. The following table summarizes the different tests of hypothesis discussed here.
Once the type of test is determined, the details of the test must be specified. Specifically, the null and alternative hypotheses must be clearly stated. The null hypothesis always reflects the "no change" or "no difference" situation. The alternative or research hypothesis reflects the investigator's belief. The investigator might hypothesize that a parameter (e.g., a mean, proportion, difference in means or proportions) will increase, will decrease or will be different under specific conditions (sometimes the conditions are different experimental conditions and other times the conditions are simply different groups of participants). Once the hypotheses are specified, data are collected and summarized. The appropriate test is then conducted according to the five step approach. If the test leads to rejection of the null hypothesis, an approximate p-value is computed to summarize the significance of the findings. When tests of hypothesis are conducted using statistical computing packages, exact p-values are computed. Because the statistical tables in this textbook are limited, we can only approximate p-values. If the test fails to reject the null hypothesis, then a weaker concluding statement is made for the following reason. In hypothesis testing, there are two types of errors that can be committed. A Type I error occurs when a test incorrectly rejects the null hypothesis. This is referred to as a false positive result, and the probability that this occurs is equal to the level of significance, α. The investigator chooses the level of significance in Step 1, and purposely chooses a small value such as α=0.05 to control the probability of committing a Type I error. A Type II error occurs when a test fails to reject the null hypothesis when in fact it is false. The probability that this occurs is equal to β. Unfortunately, the investigator cannot specify β at the outset because it depends on several factors including the sample size (smaller samples have higher b), the level of significance (β decreases as a increases), and the difference in the parameter under the null and alternative hypothesis. We noted in several examples in this chapter, the relationship between confidence intervals and tests of hypothesis. The approaches are different, yet related. It is possible to draw a conclusion about statistical significance by examining a confidence interval. For example, if a 95% confidence interval does not contain the null value (e.g., zero when analyzing a mean difference or risk difference, one when analyzing relative risks or odds ratios), then one can conclude that a two-sided test of hypothesis would reject the null at α=0.05. It is important to note that the correspondence between a confidence interval and test of hypothesis relates to a two-sided test and that the confidence level corresponds to a specific level of significance (e.g., 95% to α=0.05, 90% to α=0.10 and so on). The exact significance of the test, the p-value, can only be determined using the hypothesis testing approach and the p-value provides an assessment of the strength of the evidence and not an estimate of the effect. Answers to Selected ProblemsDental services problem - bottom of page 5.
α=0.05
First, determine whether the sample size is adequate. Therefore the sample size is adequate, and we can use the following formula:
Reject H0 if Z is less than or equal to -1.96 or if Z is greater than or equal to 1.96.
We reject the null hypothesis because -6.15<-1.96. Therefore there is a statistically significant difference in the proportion of children in Boston using dental services compated to the national proportion. Hypothesis tests about the meanby Marco Taboga , PhD This lecture explains how to conduct hypothesis tests about the mean of a normal distribution. We tackle two different cases: when we know the variance of the distribution, then we use a z-statistic to conduct the test; when the variance is unknown, then we use the t-statistic. In each case we derive the power and the size of the test. We conclude with two solved exercises on size and power. Table of contents Known variance: the z-testThe null hypothesis, the test statistic, the critical region, the decision, the power function, the size of the test, how to choose the critical value, unknown variance: the t-test, how to choose the critical values, solved exercises. The assumptions are the same we made in the lecture on confidence intervals for the mean . A test of hypothesis based on it is called z-test . Otherwise, it is not rejected. We explain how to do this in the page on critical values . This case is similar to the previous one. The only difference is that we now relax the assumption that the variance of the distribution is known. The test of hypothesis based on it is called t-test . Otherwise, we do not reject it. The page on critical values explains how this equation is solved. Below you can find some exercises with explained solutions. Suppose that a statistician observes 100 independent realizations of a normal random variable. The mean and the variance of the random variable, which the statistician does not know, are equal to 1 and 4 respectively. Find the probability that the statistician will reject the null hypothesis that the mean is equal to zero if: she runs a t-test based on the 100 observed realizations; A statistician observes 100 independent realizations of a normal random variable. She performs a t-test of the null hypothesis that the mean of the variable is equal to zero. How to citePlease cite as: Taboga, Marco (2021). "Hypothesis tests about the mean", Lectures on probability theory and mathematical statistics. Kindle Direct Publishing. Online appendix. https://www.statlect.com/fundamentals-of-statistics/hypothesis-testing-mean. Most of the learning materials found on this website are now available in a traditional textbook format.
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Hypothesis Testing | A Step-by-Step Guide with Easy ExamplesPublished on November 8, 2019 by Rebecca Bevans . Revised on June 22, 2023. Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics . It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories. There are 5 main steps in hypothesis testing:
Though the specific details might vary, the procedure you will use when testing a hypothesis will always follow some version of these steps. Table of contentsStep 1: state your null and alternate hypothesis, step 2: collect data, step 3: perform a statistical test, step 4: decide whether to reject or fail to reject your null hypothesis, step 5: present your findings, other interesting articles, frequently asked questions about hypothesis testing. After developing your initial research hypothesis (the prediction that you want to investigate), it is important to restate it as a null (H o ) and alternate (H a ) hypothesis so that you can test it mathematically. The alternate hypothesis is usually your initial hypothesis that predicts a relationship between variables. The null hypothesis is a prediction of no relationship between the variables you are interested in.
Here's why students love Scribbr's proofreading servicesDiscover proofreading & editing For a statistical test to be valid , it is important to perform sampling and collect data in a way that is designed to test your hypothesis. If your data are not representative, then you cannot make statistical inferences about the population you are interested in. There are a variety of statistical tests available, but they are all based on the comparison of within-group variance (how spread out the data is within a category) versus between-group variance (how different the categories are from one another). If the between-group variance is large enough that there is little or no overlap between groups, then your statistical test will reflect that by showing a low p -value . This means it is unlikely that the differences between these groups came about by chance. Alternatively, if there is high within-group variance and low between-group variance, then your statistical test will reflect that with a high p -value. This means it is likely that any difference you measure between groups is due to chance. Your choice of statistical test will be based on the type of variables and the level of measurement of your collected data .
Based on the outcome of your statistical test, you will have to decide whether to reject or fail to reject your null hypothesis. In most cases you will use the p -value generated by your statistical test to guide your decision. And in most cases, your predetermined level of significance for rejecting the null hypothesis will be 0.05 – that is, when there is a less than 5% chance that you would see these results if the null hypothesis were true. In some cases, researchers choose a more conservative level of significance, such as 0.01 (1%). This minimizes the risk of incorrectly rejecting the null hypothesis ( Type I error ). Receive feedback on language, structure, and formattingProfessional editors proofread and edit your paper by focusing on:
See an example The results of hypothesis testing will be presented in the results and discussion sections of your research paper , dissertation or thesis . In the results section you should give a brief summary of the data and a summary of the results of your statistical test (for example, the estimated difference between group means and associated p -value). In the discussion , you can discuss whether your initial hypothesis was supported by your results or not. In the formal language of hypothesis testing, we talk about rejecting or failing to reject the null hypothesis. You will probably be asked to do this in your statistics assignments. However, when presenting research results in academic papers we rarely talk this way. Instead, we go back to our alternate hypothesis (in this case, the hypothesis that men are on average taller than women) and state whether the result of our test did or did not support the alternate hypothesis. If your null hypothesis was rejected, this result is interpreted as “supported the alternate hypothesis.” These are superficial differences; you can see that they mean the same thing. You might notice that we don’t say that we reject or fail to reject the alternate hypothesis . This is because hypothesis testing is not designed to prove or disprove anything. It is only designed to test whether a pattern we measure could have arisen spuriously, or by chance. If we reject the null hypothesis based on our research (i.e., we find that it is unlikely that the pattern arose by chance), then we can say our test lends support to our hypothesis . But if the pattern does not pass our decision rule, meaning that it could have arisen by chance, then we say the test is inconsistent with our hypothesis . If you want to know more about statistics , methodology , or research bias , make sure to check out some of our other articles with explanations and examples.
Methodology
Research bias
Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is used by scientists to test specific predictions, called hypotheses , by calculating how likely it is that a pattern or relationship between variables could have arisen by chance. A hypothesis states your predictions about what your research will find. It is a tentative answer to your research question that has not yet been tested. For some research projects, you might have to write several hypotheses that address different aspects of your research question. A hypothesis is not just a guess — it should be based on existing theories and knowledge. It also has to be testable, which means you can support or refute it through scientific research methods (such as experiments, observations and statistical analysis of data). Null and alternative hypotheses are used in statistical hypothesis testing . The null hypothesis of a test always predicts no effect or no relationship between variables, while the alternative hypothesis states your research prediction of an effect or relationship. Cite this Scribbr articleIf you want to cite this source, you can copy and paste the citation or click the “Cite this Scribbr article” button to automatically add the citation to our free Citation Generator. Bevans, R. (2023, June 22). Hypothesis Testing | A Step-by-Step Guide with Easy Examples. Scribbr. Retrieved August 30, 2024, from https://www.scribbr.com/statistics/hypothesis-testing/ Is this article helpful?Rebecca BevansOther students also liked, choosing the right statistical test | types & examples, understanding p values | definition and examples, what is your plagiarism score. User PreferencesContent preview. Arcu felis bibendum ut tristique et egestas quis:
Keyboard Shortcuts5.3 - hypothesis testing for one-sample mean. In the previous section, we learned how to perform a hypothesis test for one proportion. The concepts of hypothesis testing remain constant for any hypothesis test. In these next few sections, we will present the hypothesis test for one mean. We start with our knowledge of the sampling distribution of the sample mean. Hypothesis Test for One-Sample Mean SectionRecall that under certain conditions, the sampling distribution of the sample mean, \(\bar{x} \), is approximately normal with mean, \(\mu \), standard error \(\dfrac{\sigma}{\sqrt{n}} \), and estimated standard error \(\dfrac{s}{\sqrt{n}} \). \(H_0\colon \mu=\mu_0\) Conditions:
Test Statistic: If at least one of conditions are satisfied, then... \( t=\dfrac{\bar{x}-\mu_0}{\frac{s}{\sqrt{n}}} \) will follow a t-distribution with \(n-1 \) degrees of freedom. Notice when working with continuous data we are going to use a t statistic as opposed to the z statistic. This is due to the fact that the sample size impacts the sampling distribution and needs to be taken into account. We do this by recognizing “degrees of freedom”. We will not go into too much detail about degrees of freedom in this course. Let’s look at an example. Example 5-1 SectionThis depends on the standard deviation of \(\bar{x} \) . \begin{align} t^*&=\dfrac{\bar{x}-\mu}{\frac{s}{\sqrt{n}}}\\&=\dfrac{8.3-8.5}{\frac{1.2}{\sqrt{61}}}\\&=-1.3 \end{align} Thus, we are asking if \(-1.3\) is very far away from zero, since that corresponds to the case when \(\bar{x}\) is equal to \(\mu_0 \). If it is far away, then it is unlikely that the null hypothesis is true and one rejects it. Otherwise, one cannot reject the null hypothesis. Statistics TutorialDescriptive statistics, inferential statistics, stat reference, statistics - hypothesis testing a mean. A population mean is an average of value a population. Hypothesis tests are used to check a claim about the size of that population mean. Hypothesis Testing a MeanThe following steps are used for a hypothesis test:
For example:
And we want to check the claim: "The average age of Nobel Prize winners when they received the prize is more than 55" By taking a sample of 30 randomly selected Nobel Prize winners we could find that: The mean age in the sample (\(\bar{x}\)) is 62.1 The standard deviation of age in the sample (\(s\)) is 13.46 From this sample data we check the claim with the steps below. 1. Checking the ConditionsThe conditions for calculating a confidence interval for a proportion are:
A moderately large sample size, like 30, is typically large enough. In the example, the sample size was 30 and it was randomly selected, so the conditions are fulfilled. Note: Checking if the data is normally distributed can be done with specialized statistical tests. 2. Defining the ClaimsWe need to define a null hypothesis (\(H_{0}\)) and an alternative hypothesis (\(H_{1}\)) based on the claim we are checking. The claim was: In this case, the parameter is the mean age of Nobel Prize winners when they received the prize (\(\mu\)). The null and alternative hypothesis are then: Null hypothesis : The average age was 55. Alternative hypothesis : The average age was more than 55. Which can be expressed with symbols as: \(H_{0}\): \(\mu = 55 \) \(H_{1}\): \(\mu > 55 \) This is a ' right tailed' test, because the alternative hypothesis claims that the proportion is more than in the null hypothesis. If the data supports the alternative hypothesis, we reject the null hypothesis and accept the alternative hypothesis. Advertisement 3. Deciding the Significance LevelThe significance level (\(\alpha\)) is the uncertainty we accept when rejecting the null hypothesis in a hypothesis test. The significance level is a percentage probability of accidentally making the wrong conclusion. Typical significance levels are:
A lower significance level means that the evidence in the data needs to be stronger to reject the null hypothesis. There is no "correct" significance level - it only states the uncertainty of the conclusion. Note: A 5% significance level means that when we reject a null hypothesis: We expect to reject a true null hypothesis 5 out of 100 times. 4. Calculating the Test StatisticThe test statistic is used to decide the outcome of the hypothesis test. The test statistic is a standardized value calculated from the sample. The formula for the test statistic (TS) of a population mean is: \(\displaystyle \frac{\bar{x} - \mu}{s} \cdot \sqrt{n} \) \(\bar{x}-\mu\) is the difference between the sample mean (\(\bar{x}\)) and the claimed population mean (\(\mu\)). \(s\) is the sample standard deviation . \(n\) is the sample size. In our example: The claimed (\(H_{0}\)) population mean (\(\mu\)) was \( 55 \) The sample mean (\(\bar{x}\)) was \(62.1\) The sample standard deviation (\(s\)) was \(13.46\) The sample size (\(n\)) was \(30\) So the test statistic (TS) is then: \(\displaystyle \frac{62.1-55}{13.46} \cdot \sqrt{30} = \frac{7.1}{13.46} \cdot \sqrt{30} \approx 0.528 \cdot 5.477 = \underline{2.889}\) You can also calculate the test statistic using programming language functions: With Python use the scipy and math libraries to calculate the test statistic. With R use built-in math and statistics functions to calculate the test statistic. 5. ConcludingThere are two main approaches for making the conclusion of a hypothesis test:
Note: The two approaches are only different in how they present the conclusion. The Critical Value ApproachFor the critical value approach we need to find the critical value (CV) of the significance level (\(\alpha\)). For a population mean test, the critical value (CV) is a T-value from a student's t-distribution . This critical T-value (CV) defines the rejection region for the test. The rejection region is an area of probability in the tails of the standard normal distribution. Because the claim is that the population mean is more than 55, the rejection region is in the right tail: The student's t-distribution is adjusted for the uncertainty from smaller samples. This adjustment is called degrees of freedom (df), which is the sample size \((n) - 1\) In this case the degrees of freedom (df) is: \(30 - 1 = \underline{29} \) Choosing a significance level (\(\alpha\)) of 0.01, or 1%, we can find the critical T-value from a T-table , or with a programming language function: With Python use the Scipy Stats library t.ppf() function find the T-Value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df). With R use the built-in qt() function to find the t-value for an \(\alpha\) = 0.01 at 29 degrees of freedom (df). Using either method we can find that the critical T-Value is \(\approx \underline{2.462}\) For a right tailed test we need to check if the test statistic (TS) is bigger than the critical value (CV). If the test statistic is bigger than the critical value, the test statistic is in the rejection region . When the test statistic is in the rejection region, we reject the null hypothesis (\(H_{0}\)). Here, the test statistic (TS) was \(\approx \underline{2.889}\) and the critical value was \(\approx \underline{2.462}\) Here is an illustration of this test in a graph: Since the test statistic was bigger than the critical value we reject the null hypothesis. This means that the sample data supports the alternative hypothesis. And we can summarize the conclusion stating: The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 1% significance level . The P-Value ApproachFor the P-value approach we need to find the P-value of the test statistic (TS). If the P-value is smaller than the significance level (\(\alpha\)), we reject the null hypothesis (\(H_{0}\)). The test statistic was found to be \( \approx \underline{2.889} \) For a population proportion test, the test statistic is a T-Value from a student's t-distribution . Because this is a right tailed test, we need to find the P-value of a t-value bigger than 2.889. The student's t-distribution is adjusted according to degrees of freedom (df), which is the sample size \((30) - 1 = \underline{29}\) We can find the P-value using a T-table , or with a programming language function: With Python use the Scipy Stats library t.cdf() function find the P-value of a T-value bigger than 2.889 at 29 degrees of freedom (df): With R use the built-in pt() function find the P-value of a T-Value bigger than 2.889 at 29 degrees of freedom (df): Using either method we can find that the P-value is \(\approx \underline{0.0036}\) This tells us that the significance level (\(\alpha\)) would need to be bigger than 0.0036, or 0.36%, to reject the null hypothesis. This P-value is smaller than any of the common significance levels (10%, 5%, 1%). So the null hypothesis is rejected at all of these significance levels. The sample data supports the claim that "The average age of Nobel Prize winners when they received the prize is more than 55" at a 10%, 5%, or 1% significance level . Note: An outcome of an hypothesis test that rejects the null hypothesis with a p-value of 0.36% means: For this p-value, we only expect to reject a true null hypothesis 36 out of 10000 times. Calculating a P-Value for a Hypothesis Test with ProgrammingMany programming languages can calculate the P-value to decide outcome of a hypothesis test. Using software and programming to calculate statistics is more common for bigger sets of data, as calculating manually becomes difficult. The P-value calculated here will tell us the lowest possible significance level where the null-hypothesis can be rejected. With Python use the scipy and math libraries to calculate the P-value for a right tailed hypothesis test for a mean. Here, the sample size is 30, the sample mean is 62.1, the sample standard deviation is 13.46, and the test is for a mean bigger than 55. With R use built-in math and statistics functions find the P-value for a right tailed hypothesis test for a mean. Left-Tailed and Two-Tailed TestsThis was an example of a right tailed test, where the alternative hypothesis claimed that parameter is bigger than the null hypothesis claim. You can check out an equivalent step-by-step guide for other types here:
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Type II Error
The first step in hypothesis testing is to calculate the test statistic. The formula for the test statistic depends on whether the population standard deviation (σ) is known or unknown. If σ is known, our hypothesis test is known as a z test and we use the z distribution. If σ is unknown, our hypothesis test is known as a t test and we use the t distribution. Use of the t distribution relies on the degrees of freedom, which is equal to the sample size minus one. Furthermore, if the population standard deviation σ is unknown, the sample standard deviation s is used instead. To switch from σ known to σ unknown, click on $\boxed{\sigma}$ and select $\boxed{s}$ in the Hypothesis Testing Calculator.
Next, the test statistic is used to conduct the test using either the p-value approach or critical value approach. The particular steps taken in each approach largely depend on the form of the hypothesis test: lower tail, upper tail or two-tailed. The form can easily be identified by looking at the alternative hypothesis (H a ). If there is a less than sign in the alternative hypothesis then it is a lower tail test, greater than sign is an upper tail test and inequality is a two-tailed test. To switch from a lower tail test to an upper tail or two-tailed test, click on $\boxed{\geq}$ and select $\boxed{\leq}$ or $\boxed{=}$, respectively.
In the p-value approach, the test statistic is used to calculate a p-value. If the test is a lower tail test, the p-value is the probability of getting a value for the test statistic at least as small as the value from the sample. If the test is an upper tail test, the p-value is the probability of getting a value for the test statistic at least as large as the value from the sample. In a two-tailed test, the p-value is the probability of getting a value for the test statistic at least as unlikely as the value from the sample. To test the hypothesis in the p-value approach, compare the p-value to the level of significance. If the p-value is less than or equal to the level of signifance, reject the null hypothesis. If the p-value is greater than the level of significance, do not reject the null hypothesis. This method remains unchanged regardless of whether it's a lower tail, upper tail or two-tailed test. To change the level of significance, click on $\boxed{.05}$. Note that if the test statistic is given, you can calculate the p-value from the test statistic by clicking on the switch symbol twice. In the critical value approach, the level of significance ($\alpha$) is used to calculate the critical value. In a lower tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the lower tail of the sampling distribution of the test statistic. In an upper tail test, the critical value is the value of the test statistic providing an area of $\alpha$ in the upper tail of the sampling distribution of the test statistic. In a two-tailed test, the critical values are the values of the test statistic providing areas of $\alpha / 2$ in the lower and upper tail of the sampling distribution of the test statistic. To test the hypothesis in the critical value approach, compare the critical value to the test statistic. Unlike the p-value approach, the method we use to decide whether to reject the null hypothesis depends on the form of the hypothesis test. In a lower tail test, if the test statistic is less than or equal to the critical value, reject the null hypothesis. In an upper tail test, if the test statistic is greater than or equal to the critical value, reject the null hypothesis. In a two-tailed test, if the test statistic is less than or equal the lower critical value or greater than or equal to the upper critical value, reject the null hypothesis.
When conducting a hypothesis test, there is always a chance that you come to the wrong conclusion. There are two types of errors you can make: Type I Error and Type II Error. A Type I Error is committed if you reject the null hypothesis when the null hypothesis is true. Ideally, we'd like to accept the null hypothesis when the null hypothesis is true. A Type II Error is committed if you accept the null hypothesis when the alternative hypothesis is true. Ideally, we'd like to reject the null hypothesis when the alternative hypothesis is true.
Hypothesis testing is closely related to the statistical area of confidence intervals. If the hypothesized value of the population mean is outside of the confidence interval, we can reject the null hypothesis. Confidence intervals can be found using the Confidence Interval Calculator . The calculator on this page does hypothesis tests for one population mean. Sometimes we're interest in hypothesis tests about two population means. These can be solved using the Two Population Calculator . The probability of a Type II Error can be calculated by clicking on the link at the bottom of the page. Hypothesis Test for a MeanUse the calculator below to analyze the results of a hypothesis test for a mean. Enter your null hypothesis's mean, sample mean, sample standard deviation, sample size, test type, and significance level to find your results. You will find a description of how to conduct a hypothesis test of a mean below the calculator. Define the t-test
Sample Mean Under the Null DistributionConducting single mean hypothesis tests. A hypothesis test of a sample mean can help you make inferences about the population from which you drew it. It is a tool to determine what is probably true about an event or phenomena. Testing a MeanFor the results of a hypothesis test of a mean to be valid, you should follow these steps: Check Your ConditionsState your hypothesis, determine your analysis plan, analyze your sample, interpret your results. To use the testing procedure described below, you should check the following conditions:
You must state a null hypothesis and an alternative hypothesis to conduct a hypothesis test for a mean. The null hypothesis, is a skeptical claim that you would like to test. It is defined by the null hypothesis's mean, which is often labeled μ 0 . The alternative hypothesis represents an alternative claim to the null hypothesis. Your null hypothesis and alternative hypothesis should be stated in one of three mutually exclusive ways listed in the table below.
Before conducting a hypothesis test, you must determine a reasonable significance level, α, or the probability of rejecting the null hypothesis assuming it is true. The lower your significance level, the more confident you can be of the conclusion of your hypothesis test. Common significance levels are 10%, 5%, and 1%. To evaluate your hypothesis test at the significance level that you set, consider if you are conducting a one or two tail test:
The graphical results section of the calculator above shades rejection regions blue. After checking your conditions, stating your hypothesis, determining your significance level, and collecting your sample, you are ready to analyze your hypothesis. Sample means follow the Normal Distribution with the following parameters:
In a hypothesis test for a mean, we calculate the probability that we would observe the sample mean, x̄, assuming the null hypothesis is true, also known as the p-value . If the p-value is less than the significance level, then we can reject the null hypothesis. You can determine a precise p-value using the calculator above, but we can find an estimate of the p-value manually by calculating the t-score, or t-statistic, as follows: t = (x̄ - μ 0 ) / SE The t-score is a test statistic that tells us how far our observation is from the null hypothesis's mean under the null distribution. Using any t-score table, we can look up the probability of observing the results under the null distribution. You will need to look up the t-score for the type of test you are conducting, i.e. one or two tail. A hypothesis test for a mean is sometimes known as a t-test because of the use of a t-score in analyzing results. If you find the probability is below the significance level, we reject the null hypothesis. The conclusion of a hypothesis test for a mean is always either:
If you reject the null hypothesis, you cannot say that your sample mean is the true population mean. If you do not reject the null hypothesis, you cannot say that the null hypothesis is true. A hypothesis test is simply a way to look at a sample and conclude if it provides sufficient evidence to reject the null hypothesis. Example: Hypothesis Test for a MeanLet’s say that you manage a clothing store with a historical average transaction amount of $53.24. You believe the best way to improve your business is to increase your average sale amount. So, you have been working hard on training your sales staff on how sell more items to customers to increase sales. To test if your sales training has increased your average sale amount, you decide to run a hypothesis test for a mean with a sample of 50 transactions to see if your average sale amount has increased.
In this example, you found that you cannot reject the claim that your current transaction amount is less than or equal to your historical transaction amount of $53.24. Your results do not guarantee that your transaction amount is $53.24 or below, but they do indicate that your sales training has likely not had the effect that you wanted on your store's average sale amount. Module 10: Inference for MeansHypothesis test for a population mean (1 of 5), learning outcomes.
IntroductionIn Inference for Means , our focus is on inference when the variable is quantitative, so the parameters and statistics are means. In “Estimating a Population Mean,” we learned how to use a sample mean to calculate a confidence interval. The confidence interval estimates a population mean. In “Hypothesis Test for a Population Mean,” we learn to use a sample mean to test a hypothesis about a population mean. We did hypothesis tests in earlier modules. In Inference for One Proportion , each claim involved a single population proportion. In Inference for Two Proportions , the claim was a statement about a treatment effect or a difference in population proportions. In “Hypothesis Test for a Population Mean,” the claims are statements about a population mean. But we will see that the steps and the logic of the hypothesis test are the same. Before we get into the details, let’s practice identifying research questions and studies that involve a population mean. Cell Phone DataCell phones and cell phone plans can be very expensive, so consumers must think carefully when choosing a cell phone and service. This decision is as much about choosing the right cellular company as it is about choosing the right phone. Many people use the data/Internet capabilities of a phone as much as, if not more than, they use voice capability. The data service of a cell company is therefore an important factor in this decision. In the following example, a student named Melanie from Los Angeles applies what she learned in her statistics class to help her make a decision about buying a data plan for her smartphone. Melanie read an advertisement from the Cell Phone Giants (CPG, for short, and yes, we’re using a fictitious company name) that she thinks is too good to be true. The CPG ad states that customers in Los Angeles get average data download speeds of 4 Mbps. With this speed, the ad claims, it takes, on average, only 12 seconds to download a typical 3-minute song from iTunes. Only 12 seconds on average to download a 3-minute song from iTunes! Melanie has her doubts about this claim, so she gathers data to test it. She asks a friend who uses the CPG plan to download a song, and it takes 13 seconds to download a 3-minute song using the CPG network. Melanie decides to gather more evidence. She uses her friend’s phone and times the download of the same 3-minute song from various locations in Los Angeles. She gets a mean download time of 13.5 seconds for her sample of downloads. What can Melanie conclude? Her sample has a mean download time that is greater than 12 seconds. Isn’t this evidence that the CPG claim is wrong? Why is a hypothesis test necessary? Isn’t the conclusion clear? Let’s review the reason Melanie needs to do a hypothesis test before she can reach a conclusion. Why should Melanie do a hypothesis test? Melanie’s data (with a mean of 13.5 seconds) suggest that the average download time overall is greater than the 12 seconds claimed by the manufacturer. But wait. We know that samples will vary. If the CPG claim is correct, we don’t expect all samples to have a mean download time exactly equal to 12 seconds. There will be variability in the sample means. But if the overall average download time is 12 seconds, how much variability in sample means do we expect to see? We need to determine if the difference Melanie observed can be explained by chance. We have to judge Melanie’s data against random samples that come from a population with a mean of 12. For this reason, we must do a simulation or use a mathematical model to examine the sampling distribution of sample means. Based on the sampling distribution, we ask, Is it likely that the samples will have mean download times that are greater than 13.5 seconds if the overall mean is 12 seconds? This probability (the P-value) determines whether Melanie’s data provides convincing evidence against the CPG claim. Now let’s do the hypothesis test. Step 1: Determine the hypotheses. As always, hypotheses come from the research question. The null hypothesis is a hypothesis that the population mean equals a specific value. The alternative hypothesis reflects our claim. The alternative hypothesis says the population mean is “greater than” or “less than” or “not equal to” the value we assume is true in the null hypothesis. Melanie’s hypotheses:
We can write the hypotheses in terms of µ. When we do so, we should always define µ. Here μ = the average number of seconds it takes to download Melanie’s song on the CPG network in Los Angeles.
Step 2: Collect the data. To conduct a hypothesis test, Melanie knows she has to use a t-model of the sampling distribution. She thinks ahead to the conditions required, which helps her collect a useful sample. Recall the conditions for use of a t-model.
Melanie collects a random sample of 45 downloads by using her friend’s phone to download her song from iTunes according to the randomly selected days, times, and locations. Melanie’s sample of size 45 downloads has an average download time of 13.5 seconds. The standard deviation for the sample is 3.2 seconds. Now Melanie needs to determine how unlikely this data is if CPG’s claim is actually true. Step 3: Assess the evidence. Assuming the average download time for Melanie’s song is really 12 seconds, what is the probability that 45 random downloads of this song will have a mean of 13.5 seconds or more? This is a question about sampling variability. Melanie must determine the standard error. She knows the standard error of random sample means is [latex]\sigma \text{}/\sqrt{n}[/latex]. Since she has no way of knowing the population standard deviation, σ, Melanie uses the sample standard deviation, s = 3.2, as an approximation. Therefore, Melanie approximates the standard error of all sample means ( n = 45) to be [latex]s\text{}/\sqrt{n}\text{}=\text{}3.2\text{}/\sqrt{45}\text{}=\text{}0.48[/latex] Now she can assess how far away her sample is from the claimed mean in terms of standard errors. That is, she can compute the t-score of her sample mean. [latex]T\text{}=\text{}\frac{\mathrm{statistic}-\mathrm{parameter}}{\mathrm{standard}\text{}\mathrm{error}}\text{}=\text{}\frac{\stackrel{¯}{x}-μ}{s\text{}/\sqrt{n}}\text{}=\text{}\frac{13.5-12}{0.48}\text{}=\text{}3.14[/latex] The sample mean for Melanie’s random sample is approximately 3.14 standard errors above the overall mean of 12. We know from previous experience that a sample mean this far above µ is very unlikely. With a t-score this large, the P-value is very small. We use a simulation of the t-model for 44 degrees of freedom to verify this. We want the probability that the sample mean is greater than 13.5. This corresponds to the probability that T is greater than 3.14. The P-value is 0.0015. Step 4: State a conclusion. Here the logic is the same as for other hypothesis tests. We use the P-value to make a decision. The P-value helps us determine if the difference we see between the data and the hypothesized value of µ is statistically significant or due to chance. One of two outcomes can occur:
Melanie’s data is very unlikely if µ = 12. The probability is essentially zero (P-value = 0.0015). This means we will rarely see sample means greater than 13.5 if µ = 12. So we reject the null and accept the alternative hypothesis. In other words, this sample provides strong evidence that CPG has overstated the speed of its data download capability. The following activities give you an opportunity to practice parts of the hypothesis testing process for a population mean. Later you will have the opportunity to practice the hypothesis test from start to finish. For the following scenarios, give the null and alternative hypotheses and state in words what µ represents in your hypotheses. A good definition of µ describes both the variable and the population. In the previous example, Melanie did not state a significance level for her test. If she had, the logic is the same as we used for hypothesis tests in Modules 8 and 9. To come to a conclusion about H 0 , we compare the P-value to the significance level α.
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Teach yourself statistics Hypothesis Test: Difference Between Paired MeansThis lesson explains how to conduct a hypothesis test for the difference between paired means . The test procedure, called the matched-pairs t-test , is appropriate when the following conditions are met:
This approach consists of four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. State the HypothesesEvery hypothesis test requires the analyst to state a null hypothesis and an alternative hypothesis . The hypotheses are stated in such a way that they are mutually exclusive. That is, if one is true, the other must be false; and vice versa. The hypotheses concern a new variable d, which is based on the difference between paired values from two data sets. d = x 1 - x 2 where x 1 is the value of variable x in the first data set, and x 2 is the value of the variable from the second data set that is paired with x 1 . The table below shows three sets of null and alternative hypotheses. Each makes a statement about how the true difference in population values μ d is related to some hypothesized value D. (In the table, the symbol ≠ means " not equal to ".)
The first set of hypotheses (Set 1) is an example of a two-tailed test , since an extreme value on either side of the sampling distribution would cause a researcher to reject the null hypothesis. The other two sets of hypotheses (Sets 2 and 3) are one-tailed tests , since an extreme value on only one side of the sampling distribution would cause a researcher to reject the null hypothesis. Formulate an Analysis PlanThe analysis plan describes how to use sample data to accept or reject the null hypothesis. It should specify the following elements.
Analyze Sample DataUsing sample data, find the standard deviation, standard error, degrees of freedom, test statistic, and the P-value associated with the test statistic. s d = sqrt [ (Σ(d i - d ) 2 / (n - 1) ] SE = s d * sqrt{ ( 1/n ) * [ (N - n) / ( N - 1 ) ] } SE = s d / sqrt( n )
t = [ ( x 1 - x 2 ) - D ] / SE = ( d - D) / SE
Interpret ResultsIf the sample findings are unlikely, given the null hypothesis, the researcher rejects the null hypothesis. Typically, this involves comparing the P-value to the significance level , and rejecting the null hypothesis when the P-value is less than the significance level. Test Your UnderstandingForty-four sixth graders were randomly selected from a school district. Then, they were divided into 22 matched pairs, each pair having equal IQ's. One member of each pair was randomly selected to receive special training. Then, all of the students were given an IQ test. Test results are summarized below.
Σ(d - d ) 2 = 270 Do these results provide evidence that the special training helped or hurt student performance? Use an 0.05 level of significance. Assume that the mean differences are approximately normally distributed. The solution to this problem takes four steps: (1) state the hypotheses, (2) formulate an analysis plan, (3) analyze sample data, and (4) interpret results. We work through those steps below: Null hypothesis: μ d = 0 Alternative hypothesis: μ d ≠ 0
s = sqrt [ (Σ(d i - d ) 2 / (n - 1) ] s = sqrt[ 270/(22-1) ] = sqrt(12.857) = 3.586 SE = s / sqrt(n) = 3.586 / [ sqrt(22) ] SE = 3.586/4.69 = 0.765 DF = n - 1 = 22 -1 = 21 t = [ ( x 1 - x 2 ) - D ] / SE t = ( d - D)/ SE = (1 - 0)/0.765 = 1.307 where d i is the observed difference for pair i , d is mean difference between sample pairs, D is the hypothesized mean difference between population pairs, and n is the number of pairs. Since we have a two-tailed test , the P-value is the probability that the t statistic having 21 degrees of freedom is less than -1.307 or greater than 1.307. We use the t Distribution Calculator to find P(t < -1.307) is about 0.103.
Note: If you use this approach on an exam, you may also want to mention why this approach is appropriate. Specifically, the approach is appropriate because the sampling method was simple random sampling, the samples consisted of paired data, and the mean differences were normally distributed. In addition, we used the approximation formula to compute the standard error, since the sample size was small relative to the population size. |
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How to conduct a hypothesis test for a mean value, using a one-sample t-test. The test procedure is illustrated with examples for one- and two-tailed tests.
8.4.3 Hypothesis Testing for the Mean Here, we would like to discuss some common hypothesis testing problems. We assume that we have a random sample X1, X2 ,..., Xn from a distribution and our goal is to make inference about the mean of the distribution μ. We consider three hypothesis testing problems. The first one is a test to decide between the following hypotheses: H0: μ = μ0, H1: μ ...
The calculator returns the alternative hypothesis (check and make sure you selected the correct sign), the test statistic, p-value, sample mean, and sample size.
We finish our discussion of the hypothesis test for a population mean with a review of the meaning of the P-value, along with a review of type I and type II errors.
Learn how to test hypotheses for means and proportions with this online module from Boston University School of Public Health.
Learn how to conduct hypothesis tests for means with examples and explanations. Compare different methods and assumptions for testing population means.
How to conduct a hypothesis test to determine whether the difference between two mean scores is significant. Includes examples for one- and two-tailed tests.
The test statistic To construct a test statistic, we use the sample mean The test statistic, called z-statistic , is A test of hypothesis based on it is called z-test . We prove below that has a normal distribution with zero mean and unit variance.
Hypothesis testing is a formal procedure for investigating our ideas about the world using statistics. It is most often used by scientists to test specific predictions, called hypotheses, that arise from theories.
5.3 - Hypothesis Testing for One-Sample Mean In the previous section, we learned how to perform a hypothesis test for one proportion. The concepts of hypothesis testing remain constant for any hypothesis test. In these next few sections, we will present the hypothesis test for one mean. We start with our knowledge of the sampling distribution of the sample mean.
The test statistic is used to decide the outcome of the hypothesis test. The test statistic is a standardized value calculated from the sample. The formula for the test statistic (TS) of a population mean is: x ¯ − μ s ⋅ n. x ¯ − μ is the difference between the sample mean ( x ¯) and the claimed population mean ( μ ).
Hypothesis Testing Calculator. The first step in hypothesis testing is to calculate the test statistic. The formula for the test statistic depends on whether the population standard deviation (σ) is known or unknown. If σ is known, our hypothesis test is known as a z test and we use the z distribution. If σ is unknown, our hypothesis test is ...
Steps for performing Hypothesis Test of a Single Population Mean Step 1: State your hypotheses about the population mean. Step 2: Summarize the data. State a significance level. State and check conditions required for the procedure
8.3: Hypothesis Testing of Single Mean is shared under a license and was authored, remixed, and/or curated by LibreTexts. Previous hypotheses testing for population means was described in the case of large samples. The statistical validity of the tests was insured by the Central Limit Theorem, with essentially no ….
Learn how to use t-tests and hypothesis testing to compare sample means and draw statistical inferences. A simple and intuitive explanation with examples.
A hypothesis test consists of five steps: 1. State the hypotheses. State the null and alternative hypotheses. These two hypotheses need to be mutually exclusive, so if one is true then the other must be false. 2. Determine a significance level to use for the hypothesis. Decide on a significance level.
In a hypothesis test for a mean, we calculate the probability that we would observe the sample mean, x̄, assuming the null hypothesis is true, also known as the p-value. If the p-value is less than the significance level, then we can reject the null hypothesis. You can determine a precise p-value using the calculator above, but we can find an ...
In "Hypothesis Test for a Population Mean," we learn to use a sample mean to test a hypothesis about a population mean. We did hypothesis tests in earlier modules. In Inference for One Proportion, each claim involved a single population proportion. In Inference for Two Proportions, the claim was a statement about a treatment effect or a ...
A statistical hypothesis test is a method of statistical inference used to decide whether the data sufficiently supports a particular hypothesis. A statistical hypothesis test typically involves a calculation of a test statistic. Then a decision is made, either by comparing the test statistic to a critical value or equivalently by evaluating a ...
How to conduct a hypothesis test for the difference between paired means. Includes step-by-step example of the test procedure, a matched-pairs t-test.
Learn about the procedure for conducting a test of hypotheses for a population mean with a sample size of at least 30.
Learn how to conduct a two sample hypothesis test for the difference in means and use the two sample t-test calculator to find the results of a test.
NE POPULATION MEAN WORKSHEET MTH 1210, SPRING 2019We are responsible for 2 types of hypothesis tests that produce inferences about the un-known population mean, , each of which has. 3 possible alternative hypothesis (6 cases total). As is the case with con dence intervals, our inferences are made with. a prescribed certainty called the con ...