Question

Assume that a sample is used to estimate a population proportion
p. Find the 99% confidence interval for a sample of size 356 with
320 successes. Enter your answer as an open-interval (i.e.,
parentheses) using decimals (not percents) accurate to three
decimal places.

99% 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,

n = 356

x = 320

Point estimate = sample proportion = = x / n = 320/356=0.899

1 - = 1-0.899=0.101

At 99% confidence level the z is ,

= 1 - 99% = 1 - 0.99 = 0.01

/ 2 = 0.01 / 2 = 0.005

Z/2 = Z0.005 = 2.576 ( Using z table )

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

= 2.576 (((0.899*0.101) / 356)

= 0.041

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.899-0.041< p < 0.899+0.041

0.858< p < 0.940

(0.858 , 0.940)

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