410 Chapter 9 Hypothesis Testing Recall: The z-score tells us how many standard deviations x̄ is from ��. The sample mean, ̄x = 562, differs somewhat from the null hypothesis value for the population mean, �� = 530. The key idea behind a hypothesis test is to measure how large this difference is. If the sample differs from H0 only slightly, then H0 may well be true, because slight differences can easily be due to chance. However, if the difference is larger, it is less likely to be due to chance, and H0 is less likely to be true. We must now determine how strong the disagreement is between the sample mean ̄x = 562 and the null hypothesis �� = 530. We do this by calculating the value of a test statistic. In this example, the test statistic is the z-score of the sample mean ̄x. We now show how to compute the z-score. Recall: When the sample size is large (n > 30), the sample mean x̄ is approximately normally distributed with mean √ �� and standard deviation ��∕ n. Recall that in a hypothesis test, we begin by assuming that H0 is true. We therefore assume that the mean of ̄x is �� = 530. Because the sample size is large (n = 100), we know that ̄x is approximately normally distributed. Suppose the population standard deviation is known to be �� = 116. The standard deviation of ̄x is �� √ n = 116 √ 100 = 11.6 The z-score for ̄x is z = ̄x − �� √ n ��∕ = 562 − 530 116∕ √ 100 = 2.76 We have found that the value of the test statistic is z = 2.76. Does this present strong evidence against H0? Figure 9.1 presents the distribution of the sample mean under the assumption that H0 is true. The value ̄x = 562 that we observed has a z-score of 2.76, which means that our observed mean is 2.76 standard deviations away from the assumed mean of 530. Visually, we can see from Figure 9.1 that our observed value ̄x = 562 is pretty far out in the tail of the distribution—far from the null hypothesis value �� = 530. Intuitively, therefore, it appears that the evidence against H0 is fairly strong. 530 ¯x = 562 z = 2.76 H0 value for μ Test statistic Figure 9.1 If H0 is true, then the value of ̄x is in the tail of the distribution, and far from the null hypothesis mean �� = 530. Visually, it appears that the evidence against H0 is fairly strong. The critical value method is based on the idea that we should reject H0 if the value of the test statistic is unusual when we assume H0 to be true. In this method, we choose a critical value, which forms a boundary between values that are considered unusual and values that are not. The region that contains the unusual values is called the critical region. If the value of the test statistic is in the critical region, we reject H0. The critical value we choose depends on how small we believe a probability should be for an event to be considered unusual. Let’s say that an event with a probability of 0.05 or less is unusual. Figure 9.2 illustrates a critical value of 1.645 and a critical region consisting of z-scores greater than or equal to 1.645. The probability that a z-score is in the critical region is 0.05, so the critical region contains the z-scores that are considered unusual. We have observed a z-score of 2.76, which is in the critical region. Therefore, we reject H0. We conclude that the mean SAT math score for students completing the online coaching program is greater than 530.
navidi_monk_elementary_statistics_2e_ch7-9
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