Chapter 9: Q.11.46 (page 460)
A sample size that will ensure a margin of error of at most the one specified.
The required sample size is 6, 766
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Find and interpret 95 % confidence interval for the proportion of all US adults who never clothes-shop online.
Determining Sample Size The sample size needed to estimate the difference between two population proportions to within a margin of error E with a confidence level of 1 - a can be found by using the following expression:
\({\bf{E = }}{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}}\sqrt {\frac{{{{\bf{p}}_{\bf{1}}}{{\bf{q}}_{\bf{1}}}}}{{{{\bf{n}}_{\bf{1}}}}}{\bf{ + }}\frac{{{{\bf{p}}_{\bf{2}}}{{\bf{q}}_{\bf{2}}}}}{{{{\bf{n}}_{\bf{2}}}}}} \)
Replace \({{\bf{n}}_{\bf{1}}}\;{\bf{and}}\;{{\bf{n}}_{\bf{2}}}\) by n in the preceding formula (assuming that both samples have the same size) and replace each of \({{\bf{p}}_{\bf{1}}}{\bf{,}}{{\bf{q}}_{\bf{1}}}{\bf{,}}{{\bf{p}}_{\bf{2}}}\;{\bf{and}}\;{{\bf{q}}_{\bf{2}}}\)by 0.5 (because their values are not known). Solving for n results in this expression:
\({\bf{n = }}\frac{{{\bf{z}}_{\frac{{\bf{\alpha }}}{{\bf{2}}}}^{\bf{2}}}}{{{\bf{2}}{{\bf{E}}^{\bf{2}}}}}\)
Use this expression to find the size of each sample if you want to estimate the difference between the proportions of men and women who own smartphones. Assume that you want 95% confidence that your error is no more than 0.03.
Testing Claims About Proportions. In Exercises 7–22, test the given claim. Identify the null hypothesis, alternative hypothesis, test statistic, P-value or critical value(s), then state the conclusion about the null hypothesis, as well as the final conclusion that addresses the original claim.
Cell Phones and Handedness A study was conducted to investigate the association between cell phone use and hemispheric brain dominance. Among 216 subjects who prefer to use their left ear for cell phones, 166 were right-handed. Among 452 subjects who prefer to use their right ear for cell phones, 436 were right-handed (based on data from “Hemi- spheric Dominance and Cell Phone Use,” by Seidman et al., JAMA Otolaryngology—Head & Neck Surgery, Vol. 139, No. 5). We want to use a 0.01 significance level to test the claim that the rate of right-handedness for those who prefer to use their left ear for cell phones is less than the rate of right-handedness for those who prefer to use their right ear for cell phones. (Try not to get too confused here.)
a. Test the claim using a hypothesis test.
b. Test the claim by constructing an appropriate confidence interval.
Equivalence of Hypothesis Test and Confidence Interval Two different simple random samples are drawn from two different populations. The first sample consists of 20 people with 10 having a common attribute. The second sample consists of 2000 people with 1404 of them having the same common attribute. Compare the results from a hypothesis test of \({p_1} = {p_2}\) (with a 0.05 significance level) and a 95% confidence interval estimate of \({p_1} - {p_2}\).
In Exercises 5–20, assume that the two samples are independent simple random samples selected from normally distributed populations, and do not assume that the population standard deviations are equal. (Note: Answers in Appendix D include technology answers based on Formula 9-1 along with “Table” answers based on Table A-3 with df equal to the smaller of\({n_1} - 1\)and\({n_2} - 1\).)
IQ and Lead Exposure Data Set 7 “IQ and Lead” in Appendix B lists full IQ scores for a random sample of subjects with low lead levels in their blood and another random sample of subjects with high lead levels in their blood. The statistics are summarized below.
a. Use a 0.05 significance level to test the claim that the mean IQ score of people with low blood lead levels is higher than the mean IQ score of people with high blood lead levels.
b. Construct a confidence interval appropriate for the hypothesis test in part (a).
c. Does exposure to lead appear to have an effect on IQ scores?
Low Blood Lead Level: n = 78, \(\bar x\) = 92.88462, s = 15.34451
High Blood Lead Level: n = 21,\(\bar x\)= 86.90476, s = 8.988352
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