Question

A random sample of 140 observations results in 119 successes.

a. Construct the a 95% confidence interval for the population proportion of successes. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)  

b. Construct the a 95% confidence interval for the population proportion of failures. (Round intermediate calculations to at least 4 decimal places. Round "z" value and final answers to 3 decimal places.)

Answer 1

Solution :

Given that,

(a)

n = 140

x = 119

Point estimate = sample proportion = = x / n = 119 / 140 = 0.850

1 - = 1 - 0.850 = 0.150

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 * (((0.850 * 0.150) / 140)

= 0.059

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.850 - 0.059 < p < 0.850 + 0.059

0.791 < p < 0.909

The a 95% confidence interval for the population proportion of successes : (0.791 , 0.909)

(b)

n = 140

x = 21

Point estimate = sample proportion = = x / n = 21 / 140 = 0.150

1 - = 1 - 0.150 = 00.850

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 * (((0.150 * 0.850) / 140)

= 0.059

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.150 - 0.059 < p < 0.150 + 0.059

0.091 < p < 0.209

The a 95% confidence interval for the population proportion of failures : (0.091 , 0.209)