Question

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 199 with 52 successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places.

C.I. =

Homework Answers

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 52 / 199 = 0.261

1 - = 1 - 0.261 = 0.739

Z/2 = 2.326

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

= 2.326 * (((0.261 * 0.739) /199 )

= 0.072

A 98% confidence interval for population proportion p is ,

- E < p < + E

0.261 - 0.072 < p < 0.261 + 0.072

0.189 < p < 0.333

C.I = 0.189 , 0.333

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