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.
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
Get Answers For Free
Most questions answered within 1 hours.