Question

Suppose a professor wants to estimate the proportion of UM students who have a satisfying experience with the e-learning approach adopted by the school in the current semester. What is the minimum sample size that he should use if he wants the estimate to be accurate within ±0.06 with a 90% confidence?

Select one:

a. 752

b. 267

c. 188

d. 456

Answer #1

Solution :

Given that,

= 0.5

1 - = 0.5

margin of error = E = 0.06

At 90% confidence level the z is ,

= 1 - 90% = 1 - 0.90 = 0.10

/ 2 = 0.10 / 2 = 0.05

Z_{/2}
= Z_{0.05} = 1.645

sample size = n = (Z_{
/ 2} / E )^{2} *
* (1 -
)

= (1.645 / 0.06)^{2} * 0.5 * 0.5

= 187.92

**sample size = 188**

**Option c is correct.**

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