Question

In: Statistics and Probability

n a survey of 602 males ages 18-64, 390 say they have gone to the dentist...

n a survey of 602 males ages 18-64, 390 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 ( nothing, nothing). (Round to three decimal places as needed.) The 95% confidence interval for the population proportion p is ( nothing, nothing). (Round to three decimal places as needed.)

Expert Solution

Sample proportion = 390 / 602 = 0.648

90% confidence interval for p is

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

0.648 - 1.645 * sqrt( 0.648 * 0.352 / 602) < p < 0.648 + 1.645 * sqrt( 0.648 * 0.352 / 602)

0.616 < p < 0.680

The 90% CI is ( 0.616 , 0.680 )

95% confidence interval for p is

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

0.648 - 1.96 * sqrt( 0.648 * 0.352 / 602) < p < 0.648 + 1.96 * sqrt( 0.648 * 0.352 / 602)

0.610 < p < 0.686

The 95% CI is ( 0.610 , 0.686 )

orchestra answered 2 years ago

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n a survey of 602 males ages 18-64, 390 say they have gone to the dentist...

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Expert Solution

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n a survey of 602 males ages​ 18-64, 390 say they have gone to the dentist...

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