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9.1.17 Question Help Construct a 99% confidence interval of the population proportion using the given information. x = 80, n

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Answer #1

Solution :

Given that,

n = 200

x = 80

Point estimate = sample proportion = \hat p = x / n = 80/200=0.4

1 - \hat p = 0.6

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 = Z0.005 = 2.576 ( Using z table )

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

= 2.576 (\sqrt((0.4*0.6) /200 )

= 0.089

A 99% confidence interval for population proportion p is ,

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

0.4-0.089 < p < 0.4+0.089

0.311< p < 0.489

lower bound 0.311

upper bound 0.489

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