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Assume that a sample is used to estimate a population proportion p. Find the 98% confidence...

Assume that a sample is used to estimate a population proportion p. Find the 98% confidence interval for a sample of size 176 with 114 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

Homework Answers

Answer #1

Given that,

Point estimate = sample proportion = = x / n = 114 / 176 = 0.648

Z/2 = 2.326

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

= 2.326 * (((0.648 * 0.352) / 176)

= 0.084

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.648 - 0.084 < p < 0.648 + 0.084

0.564 < p < 0.731

(0.564 , 0.731)

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