By Duane C. Hinders — McGrawHill Professional
Updated on Apr 25, 2014
Solutions
for the onesided alternative. The calculator answer is P = 0.112. Had the alternative been twosided, the correct answer would have been (b).
 The correct answer is (d).
 The correct answer is (c). If the sample size conditions are met, it is not necessary that the samples be drawn from a normal population.
 The correct answer is (a). Often you will see the null written as H_{0}: p = 0.7 rather than H_{0}: p = 0.7. Either is correct.
 The correct answer is (e). If we can assume that the variances are equal, then df = n_{1} + n_{2} – 2 = 12 + 15 – 2 = 25. A conservative estimate for the number of degrees of freedom is df = min {n_{1} – 1, n_{2} – 1} = min {12 – 1, 25 – 1} = 11. For a given test statistic, the greater the number of degrees of freedom, the lower the Pvalue.
 The correct answer is (b). In this course, we consider confidence intervals to be twosided. A twosided αlevel significance test will reject the null hypothesis whenever the hypothesized value of the parameter is not contained in the C = 1 – α level confidence interval.
 (c) is not one of the required steps. You can state a significance level that you will later compare the Pvalue with, but it is not required. You can simply argue the strength of the evidence against the null hypothesis based on the Pvalue alone—small values of P provide evidence against the null.
 (c) is the most correct response. (a) is incorrect because tprocedures do not work well with, for example, small samples that come from nonnormal populations. (b) is false since the numerical value of t is, like z, affected by outliers. tprocedures are generally OK to use for samples of size 40 or larger, but this is not what is meant by robust (so (d) is incorrect), (e) is not correct since the tdistribution is more variable than the standard normal. It becomes closer to z as sample size increases but is "as accurate" only in the limiting case.
 The correct answer is (d). The alternative hypothesis, H_{0} : μ = μ_{0}, would require a negative value of z to be evidence against the null. Because the given value is positive, we conclude that the finding is in the wrong direction to support the alternative and, hence, is not going to be significant at any level.
 The correct answer is (b). Because the data are paired, the appropriate ttest is a onesample test for the mean of the difference scores. In this case, df = n – 1 = 8 – 1 = 7.
 The correct answer is (a). The problem states that the teachers will record for comparison the number of students in each class who score above 80%. Since the enrollments differ in the two classes, we need to compare the proportion of students who score above 80% in each class. Thus the appropriate test is a twoproportion ztest. Note that, although it is not one of the choices, a chisquare test for homogeneity of proportions could also be used since we are interested in whether the proportions of those getting above 80% are the same across the two populations.
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From 5 Steps to a 5 AP Statistics. Copyright © 2010 by The McGrawHill Companies. All Rights Reserved.
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