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
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 = Z0.05 = 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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