Question

Suppose that a random sample of 144 graduate-admissions personnel was asked what role scores on standardized tests play in consideration of a candidate for graduate school. Of these sample members, 79 answered "very important." Find a 95% confidence interval for the population proportion of graduate admissions personnel with this view.

Answer #1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 79 / 144 = 0.549

1 - = 1 - 0.549 = 0.451

Z/2 = 1.96

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

= 1.96 * (((0.549 * 0.451) / 144)

Margin of error = E = 0.081

A 95% confidence interval for population proportion p is ,

- E < p < + E

0.549 - 0.081 < p < 0.549 + 0.081

0.468 < p < 0.630

The 95% confidence interval for the population proportion p is :
**0.468 , 0.630**

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