Solutions to Practice Problems
Multiple-Choice
The correct answer is (d).
for the one-sided alternative. The calculator answer is P = 0.112. Had the alternative been two-
sided, the correct answer would have been (b).
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 P -value.
The correct answer is (b). In AP Statistics, we consider confidence intervals to be two-sided. A
two-sided α-level 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.
6 (c) is not one of the required steps. You can state a significance level that you will later compare the
P- value with, but it is not required. You can simply argue the strength of the evidence against the null
hypothesis based on the P- value alone—small values of P provide evidence against the null.
- (c) is the most correct response. (a) is incorrect because t- procedures do not work well with, for
example, small samples that come from non-normal populations. (b) is false since the numerical
value of t is, like z , affected by outliers. t- procedures 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 t-
distribution 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 t -test is a one-sample 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 two-proportion z -test. Note that, although it is not one of the choices, a chi-square test for
homogeneity of proportions could also be used, since we are interested in whether the proportions of