Question

#7, #8, #10

Using and Interpreting Concepts Hypothesis Testing Using Rejection Regions In Exercises 7-1 (a) identify the claim and state Ho and Ha, (b) find the critical value(s) and identi the rejection region(s), (c) find the standardized test statistic z, (d) decide v to reject or fail to reject the null hypothesis, and (e) interpret the decistion in context of the original claim. 7, Vaccination Requirement A medical researcher says that less than 80% of U.S. adults think that healthy children should be required to be vaccinated. In a random sample of 200 US. adults, 82% think that healthy children should be required to be vaccinated. At α 0.05, is there enough evidence to support the researcher's claim? (Adapted from Pew Research Center) 8. Internal Revenue Service Audits A research center claims that at least 27% of US. adults think that the IRS will audit their taxes. In a random sample of 1000 U.S. adults in a recent year, 23% say they are concerned that the IRS will audit their taxes. At α 0.01, is there enough evidence to reject the center's claim? (Source: Rasmussen Reports) 9. Student Employment An eduction researcher claims that at most 3% of working college students are employed as teachers or teaching assistants. In a random sample of 200 working college students, 4% are employed as teachers or teaching assistants. At α-00, is there enough evidence to reject the researcher's claim? (Adapted from Sallie Mae 10. working Students An education researcher claims that 57% of college students work year-round. In a random sample of 300 college students, 171 say they work year-round. At a 0.10, is there enough evidence to support the researcher's claim? (Adapted from Sallie Mac) 11. Zika Virus A researcher claims that 85% percent of Americans think they

Answer 1

7. We have to test, H0: p = 0.8 against H1: p < 0.8

The test-statistic is given by, Z = $\frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$ , where, $\hat{p}$ = 0.82, p = 0.8, n = 200

Hence, Z = 0.7071.

Thus, p-value = P(Z < 0.7071) = $\Phi$(0.7071) =0.7602.

[$\Phi$(.) is the cdf of N(0,1)].

Since, p-value > level of significance = 0.05, we fail to reject H0.

8. We have to test, H0: p = 0.27 against H1: p > 0.27

The test-statistic is given by, Z = $\frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$ , $\hat{p}$ = 0.23, p = 0.27, n = 1000

Hence, Z = -3.0670

Thus, p-value = P(Z > -3.0670) = 0.9989.

Since, p-value > level of significance = 0.01, we fail to reject H0.

10. We have to test, H0: p = 0.57 against H1: p $\neq$​​​​​​​ 0.57.

The test-statistic is given by, Z = $\frac{\hat{p} - p}{\sqrt{\frac{p(1-p)}{n}}}$ , where, $\hat{p}$ = 171/300 = 0.57, p = 0.57, n = 300.

Hence, Z = 0.

Thus, p-value = 2 (Z < 0) = 2 * $\Phi$(0) = 2 * 0.5 = 1.

Since, p-value > level of significance = 0.1, we fail to reject H0.