Question

In a survey of 615 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.

Answer 1

Solution :

n = 615

x = 390

= x / n = 390 / 615 = 0.634

1 - = 1 - 0.634 = 0.366

a ) At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_/2 = Z_0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.634 * 0.366) / 615 )

= 0.032

A 90 % confidence interval for population proportion p is ,

- E < P < + E

0.634 - 0.032 < p < 0.634 + 0.032

0.602 < p < 0.666

The 90% confidence interval for the population proportion p is : ( 0.602 , 0.666 )

b ) At 95% confidence level the z is ,

  = 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z_/2 = Z_0.025 = 1.960

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.960* (((0.634 * 0.366) / 615 )

= 0.038

A 95 % confidence interval for population proportion p is ,

- E < P < + E

0.634 - 0.038 < p < 0.634 + 0.038

0.596 < p < 0.672

The 95% confidence interval for the population proportion p is : ( 0.596 , 0.672 )