Question

A random sample of 160 observations results in 104 successes. [You may find it useful to reference the z table.]

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) Point estimate = sample proportion = $\hat p$ = x / n = 104 / 160 = 0.65

1 - $\hat p$ = 1 - 0.65 = 0.35

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.96 ($\sqrt$((0.65 * 0.35) / 160)

= 0.074

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E  

= 0.65  ± 0.074

= ( 0.576, 0.724 )

b) x = 160 - 104 = 56

Point estimate = sample proportion = $\hat p$ = x / n = 56 / 160 = 0.35

1 - $\hat p$ = 1 - 0.35 = 0.65

Z$\alpha$/2 = Z_0.025 = 1.96

Margin of error = E = Z_{$\alpha$ / 2} * (($\hat p$ * (1 - $\hat p$ ))$\sqrt$ / n)

= 1.96 ($\sqrt$((0.35 * 0.65) / 160)

= 0.074

A 95% confidence interval for population proportion p is ,

$\hat p$ ± E  

= 0.35  ± 0.074

= ( 0.276, 0.424 )