Question

In Exercises 19–22, let p be the population proportion for the situation. (a) Find point estimates of p and q, (b) construct 90% and 95% confidence intervals for p, and (c) interpret the results of part (b) and compare the widths of the confidence intervals

. In a survey of 2223 U.S. adults, 1334 say an occupation as an athlete is prestigious. (Adapted

Answer 1

Solution :

Given that,

n = 2223

x = 1334

(a)

Point estimate = sample proportion = = x / n = 1334 / 2223 = 0.600

1 - = 1 - 0.600 = 0.400

(b)

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 * (((0.600 * 0.400) / 2223)

= 0.017

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.600 - 0.017 < p < 0.600 + 0.017

0.583 < p < 0.617

(0.583 , 0.617)

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 = 1.96

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

= 1.96 * (((0.600 * 0.400) / 2223)

= 0.020

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.600 - 0.020 < p < 0.600 + 0.020

0.580 < p < 0.620

(0.580 , 0.620)

(c)

Width of the confidence interval at 95% is more than 90% .

If we increase the confidence level then width of the confidence interval is also increases .