Question

Assume that a sample is used to estimate a population proportion
*p*. Find the 99.5% confidence interval for a sample of size
248 with 196 successes. Enter your answer as an
**open-interval** (*i.e.*, parentheses) using
decimals (not percents) accurate to three decimal places.

99.5% C.I. =

Answer should be obtained without any preliminary rounding.
However, the critical value may be rounded to 3 decimal places.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 196 / 248 = 0.790

1 - = 1 - 0.790 = 0.21

Z/2 = 2.807

Margin of error = E = Z / 2 * (( * (1 - )) / n)

= 2.807 * (((0.790 * 0.21) / 248)

Margin of error = E = 0.073

A 99.5% confidence interval for population proportion p is ,

- E < p < + E

0.790 - 0.073 < p < 0.790 + 0.073

0.717 < p < 0.863

99.5% C.I. = **(0.717 , 0.863)**

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