Question

In: Statistics and Probability

In a survey of 636 males ages 18-64, 399 say they have gone to the dentist...

In a survey of 636 males ages 18-64, 399 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.

Expert Solution

Sample proportion = 399 / 636 = 0.627

90% confidence interval for p is

- Z/2 * Sqrt( ( 1 - ) / n) < p < + Z/2 * Sqrt( ( 1 - ) / n)

0.627 - 1.645 * sqrt( 0.627* 0.373 / 636) < p < 0.627 + 1.645 * sqrt( 0.627* 0.373 / 636)

0.595 < p < 0.659

90% CI is ( 0.595 , 0.659)

Interpretation - We are 90% confident that the true proportion is between 0.595 and 0.659.

95% confidence interval for p is

- Z/2 * Sqrt( ( 1 - ) / n) < p < + Z/2 * Sqrt( ( 1 - ) / n)

0.627 - 1.96 * sqrt( 0.627* 0.373 / 636) < p < 0.627 + 1.96 * sqrt( 0.627* 0.373 / 636)

0.599 < p < 0.665

95% CI is ( 0.589 , 0.665)

Interpretation - We are 95% confident that the true proportion is between 0.589 and 0.665.

The 95% confidence interval is wider than 90% confidence interval.

orchestra answered 2 years ago

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