Question

In a recent survey conducted by a professor of UM, 200 students were asked whether or not they have a satisfying experience with the e-learning approach adopted by the school in the current semester. Among the 200 students interviewed, 121 said they have a satisfying experience. What is the 99% confidence interval for the proportion of all UM students who have a satisfying experience with the e-learning approach?

Answer 1

Solution :

Given that,

Point estimate = sample proportion = = x / n = 121 / 200 = 0.605

1 - = 1 - 0.605 = 0.395

Z/2 = Z0.005 = 2.576

Margin of error = E = Z / 2 * (( * (1 - )) / n).    = 2.576 * (((0.605 * 0.395) / 200)

E = 0.089

A 99% confidence interval for population proportion p is ,

- E < p < + E

0.605 - 0.089 < p < 0.605 + 0.089

0.516 < p < 0.694

The 95% confidence interval for the population proportion p is : (0.516 , 0.694)