Question

The standard IQ test is designed so that the mean is 100 and the standard deviation is 15 for the population of all adults. We wish to find the sample size necessary to estimate the mean IQ score of all people who have successfully passed a college statistics course. We want to create a confidence interval that is no wider than 8 IQ points. The standard deviation for this sub-population is certainly less than 15 as it should be a less variable population. Therefore by using σ σ = 15 we will obtain a conservative sample size, meaning it will be sufficient large enough. How large a sample should we utilize for a 93% confidence interval? (hint: the z-score is 1.81191 )

Answer 1

Solution :

Given that,

standard deviation = = 15

margin of error = E = 8

At 93% confidence level the z is ,

= 1 - 93% = 1 - 0.93 = 0.07

/ 2 = 0.07 / 2 = 0.035

Z_/2 = Z_0.035 = 1.812

Sample size = n = ((Z_/2 * ) / E)²

= ((1.812 * 15) / 8)²

= 11.54 = 12

Sample size = 12