Question

In: Statistics and Probability

In a survey of 609609 males ages​ 18-64, 397397 say they have gone to the dentist...

In a survey of

609609

males ages​ 18-64,

397397

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.)

Solutions

Expert Solution

Sample proportion = 397 / 609 = 0.652

90% confidence interval for p is

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

0.652 - 1.645 * sqrt( 0.652 * ( 1 - 0.652) / 609) < p < 0.652 + 1.645 * sqrt( 0.652 * ( 1 - 0.652) / 609)

0.620 < p < 0.684

90% CI is ( 0.620 , 0.684 )

Interpretation - We are 90% confident that population proportion of males who say they have

gone to the dentist in the past year is between 0.620 and 0.684

95% confidence interval for p is

- z * sqrt( ( 1 - ) / n) < p < + z * sqrt( ( 1 - ​​​​​​​) / n)

0.652 - 1.96 * sqrt( 0.652 * ( 1 - 0.652) / 609) < p < 0.652 + 1.96 * sqrt( 0.652 * ( 1 - 0.652) / 609)

0.614 < p < 0.690

95% CI is ( 0.614 , 0.690 )

Interpretation - We are 95% confident that population proportion of males who say they have

gone to the dentist in the past year is between 0.614 and 0.690

Comparison = 95% confidence interval is wider than 90% confidence interval.


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