Question
In: Statistics and Probability
In a survey of 633 males ages? 18-64, 394 say they have gone to the dentist...
In a survey of 633 males ages? 18-64, 394 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.
Solutions
Expert Solution
Sample proportion 
= 394 / 633 = 0.622
a)
90% confidence interval for p is
- Z * Sqrt(
( 1 -
) / n) < p <
+ Z * Sqrt(
( 1 -
) / n)
0.622 - 1.645 * sqrt(0.622 * 0.378 / 633) < p < 0.622 + 1.645 * sqrt(0.622 * 0.378 / 633)
0.590 < p < 0.654
90% CI for p is (0.590 , 0.654)
Interpretation - We are 90% confident that population proportion of the males who says
they have gone to the dentist in the past year is between 0.590 and 0.654
b)
95% confidence interval for p is
- Z * Sqrt(
( 1 -
) / n) < p <
+ Z * Sqrt(
( 1 -
) / n)
0.622 - 1.96 * sqrt(0.622 * 0.378 / 633) < p < 0.622 + 1.96 * sqrt(0.622 * 0.378 / 633)
0.584 < p < 0.660
95% CI for p is ( 0.584 , 0.660)
Interpretation - We are 95% confident that population proportion of the males who says
they have gone to the dentist in the past year is between 0.584 and 0.660.
c)
Width of 95% confidence interval is wider than 90% confidence interval.
orchestra answered 2 years agoRelated Solutions
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