Question

In: Statistics and Probability

In a survey of 600 males ages 18-64, 392 say they have gone to the dentist...

In a survey of 600 males ages 18-64, 392 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.)

Expert Solution

sample proportion = 392 / 600 = 0.653

90% confidence interval for p is

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

0.653 - 1.645 * sqrt( 0.653 * 0.347 / 600) < p < 0.653 + 1.645 * sqrt( 0.653 * 0.347 / 600)

0.621 < p < 0.685

90% CI is ( 0.621 , 0.685 )

We are 90% confident that population proportion is between 0.621 and 0.685

95% confidence interval for p is

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

0.653 - 1.96 * sqrt( 0.653 * 0.347 / 600) < p < 0.653 + 1.96 * sqrt( 0.653 * 0.347 / 600)

0.615 < p < 0.691

90% CI is ( 0.615 , 0.691 )

We are 90% confident that population proportion is between 0.615 and 0.691

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

orchestra answered 2 years ago

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In a survey of 600 males ages 18-64, 392 say they have gone to the dentist...

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