In a survey of 638 males ages 18-64, 394 say they have gone to the dentist in th
ID: 3224964 • Letter: I
Question
In a survey of 638 males ages 18-64, 394 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? The 90% confidence interval The 95% confidence intervalExplanation / Answer
Answer to the question:
pcap = 394/638 = .618
1-pcap = 1-.618 = .382
CI is given by : pcap +/- Z*sqrt(pcap*pcap'/n)
The 90% CI for the population proportion p is
.618 +/- 1.645*Sqrt(.618*.382/638)
=.586 to .650
The 95% CI for the population proportion p is
.618 +/- 1.96*Sqrt(.618*.382/638)
=.580 to .656
The option B is the right interpretation of the confidence level.
> A CI should always be about the population proportion
> CI says pop. proportion is isnside these limits. Hence, B remains.
The 95% confidence level is obsviously wider
Why? The second term is wider ( Margin of error is wider)
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