90% = 95% = Compare widths In a survey of 603 males ages 18-64, 390 say they hav
ID: 3372117 • Letter: 9
Question
90% = 95% = Compare widths In a survey of 603 males ages 18-64, 390 say they have gone to the dentist in the past year Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals. The 90% confidence interval for the population proportion p is CD Round to three decimal places as needed.) Enter your answer in the edit fields and then click Check Answer parts Clear Al inin doExercise(29)Explanation / Answer
solution :
Given that x = 390 , n = 603
=> population proportion p = x/n = 390/603 = 0.647
=> q = 1-p = 0.353
=> for 90% confidence interval , Z = 1.645
=> The 90% confidence interval for the proportion is P +/- Z*sqrt(p*q/n)
=> 0.647 +/- 1.645*sqrt(0.647*0.353/603)
=> (0.615 , 0.679)
=> for 95% confidence interval , Z = 1.96
=> The 95% confidence interval for the proportion is P +/- Z*sqrt(p*q/n)
=> 0.647 +/- 1.96*sqrt(0.647*0.353/603)
=> (0.609 , 0.685)
=> The 95% confidence interval is wider than a 90% confidence interval.
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