Science, Tech, Math › Math An Example of a Hypothesis Test Share Flipboard Email Print Here the test statistic falls within the critical region. C.K.Taylor Math Statistics Inferential Statistics Statistics Tutorials Formulas Probability & Games Descriptive Statistics Applications Of Statistics Math Tutorials Geometry Arithmetic Pre Algebra & Algebra Exponential Decay Functions Worksheets By Grade Resources View More By Courtney Taylor Professor of Mathematics Ph.D., Mathematics, Purdue University M.S., Mathematics, Purdue University B.A., Mathematics, Physics, and Chemistry, Anderson University Courtney K. Taylor, Ph.D., is a professor of mathematics at Anderson University and the author of "An Introduction to Abstract Algebra." our editorial process Courtney Taylor Updated April 13, 2018 Mathematics and statistics are not for spectators. To truly understand what is going on, we should read through and work through several examples. If we know about the ideas behind hypothesis testing and see an overview of the method, then the next step is to see an example. The following shows a worked out example of a hypothesis test. In looking at this example, we consider two different versions of the same problem. We examine both traditional methods of a test of significance and also the p-value method. A Statement of the Problem Suppose that a doctor claims that those who are 17 years old have an average body temperature that is higher than the commonly accepted average human temperature of 98.6 degrees Fahrenheit. A simple random statistical sample of 25 people, each of age 17, is selected. The average temperature of the sample is found to be 98.9 degrees. Further, suppose that we know that the population standard deviation of everyone who is 17 years old is 0.6 degrees. The Null and Alternative Hypotheses The claim being investigated is that the average body temperature of everyone who is 17 years old is greater than 98.6 degrees This corresponds to the statement x > 98.6. The negation of this is that the population average is not greater than 98.6 degrees. In other words, the average temperature is less than or equal to 98.6 degrees. In symbols, this is x ≤ 98.6. One of these statements must become the null hypothesis, and the other should be the alternative hypothesis. The null hypothesis contains equality. So for the above, the null hypothesis H0 : x = 98.6. It is common practice to only state the null hypothesis in terms of an equals sign, and not a greater than or equal to or less than or equal to. The statement that does not contain equality is the alternative hypothesis, or H1 : x >98.6. One or Two Tails? The statement of our problem will determine which kind of test to use. If the alternative hypothesis contains a "not equals to" sign, then we have a two-tailed test. In the other two cases, when the alternative hypothesis contains a strict inequality, we use a one-tailed test. This is our situation, so we use a one-tailed test. Choice of a Significance Level Here we choose the value of alpha, our significance level. It is typical to let alpha be 0.05 or 0.01. For this example we will use a 5% level, meaning that alpha will be equal to 0.05. Choice of Test Statistic and Distribution Now we need to determine which distribution to use. The sample is from a population that is normally distributed as the bell curve, so we can use the standard normal distribution. A table of z-scores will be necessary. The test statistic is found by the formula for the mean of a sample, rather than the standard deviation we use the standard error of the sample mean. Here n=25, which has a square root of 5, so the standard error is 0.6/5 = 0.12. Our test statistic is z = (98.9-98.6)/.12 = 2.5 Accepting and Rejecting At a 5% significance level, the critical value for a one-tailed test is found from the table of z-scores to be 1.645. This is illustrated in the diagram above. Since the test statistic does fall within the critical region, we reject the null hypothesis. The p-Value Method There is a slight variation if we conduct our test using p-values. Here we see that a z-score of 2.5 has a p-value of 0.0062. Since this is less than the significance level of 0.05, we reject the null hypothesis. Conclusion We conclude by stating the results of our hypothesis test. The statistical evidence shows that either a rare event has occurred, or that the average temperature of those who are 17 years old is, in fact, greater than 98.6 degrees.