Section 9.2 Hypothesis Tests for a Population Mean, Standard Deviation Known 415 Objective 2 Perform hypothesis tests with the P-value method The P-Value Method We will introduce the P-value method with our SAT example. An online coaching program is supposed to increase the mean SAT math score to a value greater than 530. The null and alternate hypotheses are H0 : �� = 530 H1: �� > 530 Now assume that 100 students are randomly chosen to participate in the program, and their sample mean score is ̄x = 562. Suppose that the population standard deviation for SAT math scores is known to be �� = 116. Does this provide strong evidence against the null hypothesis �� = 530? To measure just how strong the evidence against H0 is, we compute a quantity called the P-value. The P-value is the probability that a number drawn from the distribution of the sample mean would be as extreme as or more extreme than our observed value of 562. The more extreme the value, the stronger is the evidence against H0. Because our alternate hypothesis is H1: �� > 530, this is a right-tailed test, so values of ̄x greater than 562 are more extreme than our observed value is. We find the P-value by computing the z-score of our observed sample mean ̄x = 562. We now explain how to do this. Recall that we begin by assuming that H0 is true. We therefore assume that the mean of ̄x is �� = 530. The sample size is large (n = 100), so we know that ̄x is approximately normally distributed. The standard deviation of ̄x is Recall: When the sample size is large (n > 30), the sample mean x̄ is approximately normally distributed with mean √ �� and standard deviation ��∕ n. �� √ n = 116 √ 100 = 11.6 Therefore, the P-value is the probability that ̄x is greater than 562 when �� is assumed to be 530 and the standard deviation is 11.6. The z-score for ̄x is z = ̄x − �� ��∕ √ n = 562 − 530 116∕ √ 100 = 2.76 Explain It Again Using technology: The P-value is the area to the right of x̄ = 562 when the mean is 530 and the standard deviation is 11.6. This area can be found with technology. The following display illustrates the normalcdf command on the TI-84 Plus calculator. The P-value is therefore the area under the normal curve to the right of z = 2.76. Using Table A.2, we see that the area to the left of z = 2.76 is 0.9971. Therefore, the area to the right of z = 2.76 is 1 − 0.9971 = 0.0029 (see Figure 9.4). Therefore, the P-value for this test is 0.0029. 530 ¯ x = 562 z = 2.76 Test statistic H0 value for μ Area = 0.0029 Figure 9.4 If H0 is true, the probability that ̄x takes on a value as extreme as or more extreme than the observed value of 562 is 0.0029. This is the P-value. This P-value tells us that if H0 were true, the probability of observing a value of ̄x as large as 562 is only 0.0029. Therefore, there are only two possibilities: ∙ H0 is false. ∙ H0 is true, and we got an unusual sample, whose mean lies in the most extreme 0.0029 of its distribution. In practice, events in the most extreme 0.0029 of their distributions are very unusual. This means that a P-value as small as 0.0029 is very unlikely to occur if H0 is true. A P-value of 0.0029 is very strong evidence against H0.
navidi_monk_elementary_statistics_2e_ch7-9
To see the actual publication please follow the link above