Chapters 18-19 Summative Assessment

Question

Allison wants to estimate the proportion of U.S. teens who think that a middle part is the best hairstyle. She chooses a random sample of 500 teenagers and finds that 64.8% believe a middle part is the best hairstyle. 

1. Find a 90% confidence interval for the proportion of all U.S. teens who believe a middle part is the best hairstyle. Give an interpretation of the confidence interval and an interpretation of the confidence level. 

1. An advocate for Ms. Timlin’s generation claims that only 60% of teenagers believe a middle part is best. Is this supported by Allison’s confidence interval? 

1. If Allison wanted to be 99% confident instead, what would happen to the confidence interval?

1. If Allison chose a sample size of 1000 instead, what would happen to the confidence interval?

Answer 1

CI = phat +/- z-critical*sqrt(phat*(1-phat)/n), where phat = 0.648 and n = 500

z-critical for 90% CI uses z-score at 95% probability = 1.645

CI = point estimate +/- margin of error =

CI = 0.648 +/- 1.645 * sqrt(0.648*0.352/500) = 0.63, 0.67

since 60% is not inside CI her claim is not supported by the CI at 90% confidence level (0.10 level of significance)

increasing the confidence level would increase the size of the confidence interval because the z-critical value because z-critical would be 2.58

doubling the sample size to 1000 would decrease the size the of the confidence interval because dividing by a larger number in computing the margin of error.