9-2 Two Means: Independent Samples 459 of degrees of freedom is 11, which is the smaller of 11 and 14. In Table A-3 with df = 11 and a = 0.05 in the left tail, we get a critical value of t = -1.796. (Table A-3 gives us t = 1.796, but we must make it negative because this is a left-tailed test.) Technology can be used to find the more accurate critical value of t = -1.727 (as shown in the preceding Statdisk display). Figure 9-2 shows the more accurate critical value of t = -1.727. Figure 9-2 shows that the test statistic does not fall within the critical region, so we fail to reject the null hypothesis, as we did in Example 1. t 5 0 a5 0.05 Test Statistic: t 521.679 t 521.727 FIGURE 9-2 Hypothesis Test of Means from Two Independent Populations Confidence Intervals Confidence Interval Estimate of Difference Between 1988 Heights and 2012 Heights EXAMPLE 2 Repeat Example 1 by constructing a confidence interval estimate of the difference between the mean height in 1988 and the mean height in 2012. SOLUTION First, we must be careful to use the appropriate confidence level. The test in Example 1 is left-tailed with a 0.05 significance level, so we should construct the confidence interval using a confidence level of 90%. (See Table 8-1 on page 376, which shows that for a significance level of 0.05 in a one-tailed test, we should use a 90% confidence interval.) REQUIREMENT CHECK Because we are using the same data from Example 1, the same requirement check applies here, so the requirements are satisfied. Technology Technology can be used to find this 90% confidence interval estimate of the difference between the two populations: -77.9 mm 6 m1 - m2 6 1.1 mm (as shown in the Statdisk display included with Example 1). Manual Calculations For manual calculations, we must first find the value of the margin of error E. In Table A-3 with df = 11 and a = 0.05 in one tail, we get the critical value of t = 1.796. (Technology can be used to find the more accurate critical value of t = 1.727.) We use the critical value of t = 1.796 and the sample statistics n1 = 12, s1 = 66.6012 mm, n2 = 15, and s2 = 47.86618 mm to evaluate the margin of error as follows. E = ta>2Bs2 1 n1 + s2 2 n2 = 1.796B66.60122 12 + 47.866182 15 = 41.049035 1 us n dSuper Bowls Students were invited to a Super Bowl game, and half of them were given large 4-liter snack bowls while the other half were given smaller 2-liter bowls. Those using the large bowls consumed 56% more than those using the smaller bowls. (See “Super Bowls: Serving Bowl Size and Food Consumption,” by Wansink and Cheney, Journal of the American Medical Association, Vol. 293, No. 14.) A separate study showed that there is “a significant increase in fatal motor vehicle crashes during the hours following the Super Bowl telecast in the United States.” Researchers analyzed 20,377 deaths on 27 Super Bowl Sundays and 54 other Sundays used as controls. They found a 41% increase in fatalities after Super Bowl games. (See “Do Fatal Crashes Increase Following a Super Bowl Telecast?” by Redelmeier and Stewart, Chance, Vol. 18, No. 1.) continued
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