472 CHAPTER 9 Inferences from Two Samples We will follow the same method of hypothesis testing that we used for testing a claim about a mean (see Figure 8-1 on page 376), but we use differences instead of the original raw sample data. Step1: The claim that the measured weights tend to be higher than the reported weights can be expressed as md 7 0 lb. Step2: If the original claim is not true, we have md … 0 lb. Step3: The null hypothesis must express equality and the alternative hypothesis cannot include equality, so we have H0:md = 0 lb H1:md 7 0 lb1original claim2 Step4: The significance level is a = 0.05. Step5: Using Technology Steps 5 and 6 can be done with technology. Shown here is the Excel (XLSTAT) display for this hypothesis test, and it shows the test statistic is t = 0.778 (“observed value”) and the P-value is 0.231. Excel (XLSTAT) Manual Calculation If not using technology, we use the Student t distribution with the differences. Step6: Manual Calculations From Table 9-2 we get these “measured – reported” differences: 2.6, 1.3, 4.8, 0.5, 9.9, 0.3, -8.6, 0.6. We use this list of differences to find these unrounded sample statistics: n = 8, d = 1.425 lb, sd = 5.181216 lb. Using these sample statistics and the assumption from the null hypothesis that md = 0 lb, we can now find the value of the test statistic as shown below. t = d - md sd2 n = 1.425 - 0 5.181216 2 8 = 0.778 P-Value Method Because we are using a t distribution, we refer to Table A-3 for the row with df = 7 and we see that the test statistic t = 0.778 corresponds to an “Area in One Tail” that is greater than 0.10, so P@value 7 0.10. (The preceding XLSTAT display shows that P@value = 0.231.) Critical Value Method Refer to Table A-3 to find the critical value of t = 1.895 as follows: Use the column for 0.05 (Area in One Tail), and use the row with degrees of freedom of n - 1 = 7. We get the critical value of t = 1.895. See Figure 9-3(b). Step7: If we use the P-value method, we fail to reject H0 because the P-value of 0.231 is greater than the significance level of 0.05. If we use the critical value method, we fail to reject H0 because the test statistic of t = 0.778 does not fall in the critical region. Twins in Twinsburg During the first weekend in August of each year, Twinsburg, Ohio, celebrates its annual “Twins Days in Twinsburg” festival. Thousands of twins from around the world have attended this festival in the past. Scientists saw the festival as an opportunity to study identical twins. Because they have the same basic genetic structure, identical twins are ideal for studying the different effects of heredity and environment on a variety of traits, such as male baldness, heart disease, and deafness— traits that were recently studied at one Twinsburg festival. A study of twins showed that myopia (near-sightedness) is strongly affected by hereditary factors, not by environmental factors such as watching television, surfing the Internet, or playing computer or video games. D w A ye O its “T
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