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.)
Solutions
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 agoRelated Solutions
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