Question

construct a 95% confidence interval for the mean number of defects per screen for all screens produced by this manufacturer. a) what is the lower limit on the interval? give your answer to three decimal places. 2.077 incorrect: your answer is incorrect. b) what is the upper limit on the interval? give your answer to three decimal places. 2.203 incorrect: your answer is incorrect.

Answer 1

The 99% confidence interval for the population proportion p is 0.833< p < 0.863.

Given that,

x = 3059

n = 3604

Point estimate = sample proportion = x / n =3059/3604=0.848

$1 - \hat p$  = $1-0.848 =0.152$

At 99% confidence level the z is ,

$\alpha$ = 1 - 99% = 1 - 0.99 = 0.01

$\alpha$/ 2 = 0.01 / 2 = 0.005

$Z\alpha$/2 = $Z*0.005$ = 2.576  (  Using z table   )

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

                              = $2.576* (\sqrt((0.848*0.152) /3604 )$

                               =0.015

A 99% confidence interval for population proportion p is ,

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

0.848-0.015< p <0.848+ 0.015

0.833< p < 0.863

Hence, the 99% confidence interval for the population proportion p is 0.833< p < 0.863.

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