Question

Question

Using the information from above, with p′=0.143, q′=0.857, and n=350, what is the 95% confidence interval for the proportion of the population who play games on their phones?

z0.10	z0.05	z0.025	z0.01	z0.005
1.282	1.645	1.960	2.326	2.576

Use the table of common z-scores above.

• Round the final answer to three decimal places.

Provide your answer below:

Answer 1

Solution :

Given that,

n = 350

Point estimate = sample proportion = $\hat p$ = 0.143

1 - $\hat p$ = 0.857

At 95% confidence level the t is ,

$\alpha$ = 1 - 95% = 1 - 0.95 = 0.05

$\alpha$ / 2 = 0.05 / 2 = 0.025

Z_$\alpha$/2 = Z_0.025 = 1.960

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

= 1.960 * ($\sqrt$((0.143*0.857) / 350)

= 0.037

A 95% confidence interval for population proportion p is ,

$\hat p$ - E < p < $\hat p$ + E

0.143 - 0.037 < p < 0.143 + 0.037

0.106 < p < 0.180

The 95% confidence interval for the population proportion p is : ( 0.106 , 0.180 )