Question

5. Construct a confidence interval of the population proportion at the given level of confidence. x 220, n 800, 90 % confidence a) The 90% confidence interval is ( ). (Round to three decimal places) b) Make a statement in a complete sentence about your finding: CE ntel sampe apv

number 5 A & B. please explain.

Answer 1

a)

Solution :

Given that,

n = 800

x = 220

Point estimate = sample proportion = $\hat p$ = x / n = 0.275

1 - $\hat p$ = 1 - 0.275 = 0.725

At 90% confidence level

$\alpha$ = 1 - 90%  

$\alpha$ = 1 - 0.90 =0.10

$\alpha$/2 = 0.05

Z$\alpha$/2 = Z_{0.05 = 1.645} ( Using z table )

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

= 1.645 ($\sqrt$( 0.275( 0.725 ) / 800 )

= 0.026

A 90% confidence interval for population proportion p is ,

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

0.275 - 0.026 < p < 0.275 + 0.026

0.249 < p < 0.301

(0.249,0.301)

b)

A 90% confidence interval for the population proportion p (0.249,0.301) is fall between lower bound and upper bound