Question

In: Statistics and Probability

In a survey of 633 males ages? 18-64, 394 say they have gone to the dentist...

In a survey of 633 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.

Expert Solution

Sample proportion = 394 / 633 = 0.622

a)

90% confidence interval for p is

- Z * Sqrt( ( 1 - ) / n) < p < + Z * Sqrt( ( 1 - ) / n)

0.622 - 1.645 * sqrt(0.622 * 0.378 / 633) < p < 0.622 + 1.645 * sqrt(0.622 * 0.378 / 633)

0.590 < p < 0.654

90% CI for p is (0.590 , 0.654)

Interpretation - We are 90% confident that population proportion of the males who says

they have gone to the dentist in the past year is between 0.590 and 0.654

b)

95% confidence interval for p is

- Z * Sqrt( ( 1 - ) / n) < p < + Z * Sqrt( ( 1 - ) / n)

0.622 - 1.96 * sqrt(0.622 * 0.378 / 633) < p < 0.622 + 1.96 * sqrt(0.622 * 0.378 / 633)

0.584 < p < 0.660

95% CI for p is ( 0.584 , 0.660)

Interpretation - We are 95% confident that population proportion of the males who says

they have gone to the dentist in the past year is between 0.584 and 0.660.

c)

Width of 95% confidence interval is wider than 90% confidence interval.

orchestra answered 2 years ago

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In a survey of 633 males ages? 18-64, 394 say they have gone to the dentist...

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