Question

5. Construct a confidence interval of the population proportion at the given level of confidence. x 220, n 800, 90 % confiden

number 5 A & B. please explain.

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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 = Z0.05 = 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

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