In a survey of 655 males ages 18-64,395 say they have gone to the dentist in the
ID: 3237422 • Letter: I
Question
In a survey of 655 males ages 18-64,395 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 ( , ).
(Round to three decimal places as needed.)
The 95% confidence interval for the population proportion p is ( , ).
(Round to three decimal places as needed.)
Interpret your results of both confidence intervals.
A. With the given confidence, it can be said that the population proportion of males ages 18-64 who say they have gone to the dentist in the past year is not between the endpoints of the given confidence interval.
B. With the given confidence, it can be said that the population proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.
C. With the given confidence, it can be said that the sample proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.
Which interval is wider?
A. The 90% confidence interval
B. The 95% confidence interval
Explanation / Answer
Solution:
option B. With the given confidence, it can be said that the population proportion of males ages 18-64 who say they have gone to the dentist in the past year is between the endpoints of the given confidence interval.
Explanation:
proportion(p) = 395 / 655 = 0.6031
standard error(se) = sqrt( 0.6031 * (1-0.6031) / 655 ) = 0.0191
90% confidence interval(CI)
alpha() = 1 - (90/100) = 0.10
critical probability(p*) = 1 - (0.10/2) = 0.95
degrees of freedom(df) = 655 -1 = 654
critical value(use t statistic calculator) = 1.647
Margin of Error(ME) =
= 0.6031 ± 1.647*sqrt[0.6031 * (1-0.6031) / 655]
= (0.5716,0.6346)
Thus, a 95% C.I. is:
= 0.6031 ± 1.964*sqrt[0.6031 * (1-0.6031) / 655]
= (0.5655, 0.64065)
We are trying to estimate the population proportion, answer is B
95% CI is wider, answer is b
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