Question

Assume that a sample is used to estimate a population proportion p. Find the 80% confidence interval for a sample of size 276 with 43.8% successes. Enter your answer as an open-interval (i.e., parentheses) using decimals (not percents) accurate to three decimal places. 80% 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 = = 0.438

1 - = 1 - 0.438 = 0.562

Z_{/2}
= 1.282

Margin of error = E = Z_{
/ 2} *
((
* (1 -
)) / n)

= 1.282 * (((0.438 * 0.562) / 276)

Margin of error = E = 0.038

A 80% confidence interval for population proportion p is ,

- E < p < + E

0.438 - 0.038 < p < 0.438 + 0.038

0.400 < p < 0.476

**80% C. I.** = **0.400 , 0.476**

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