Test for significance by using the confidence interval and hypothesis testing methods

Question

An IQ test has a mean of 100 and standard deviation 10. A Group of 20 college students is given the IQ test and their mean score is 106. Is their evidence at the 05 level of significance that the students mean score differs from the mean score of the IQ test.

Answer 1

Ho: mean = 100

Ha: mean != 100

Critical value for two-sided test at 0.05 level of significance is 1.96.

will compute z-statistic for this test:

z = (sample mean - population mean)/(population standard deviation/sqrt(sample size)) = (106-100)/(10/sqrt(20)) = 2.68

will reject Ho if |z| > 1.96, which it is.

Confidence interval (95 % corresponding to level of significance = 0.05):

sample mean +/- z-critical * standard deviation/sqrt(sample size)

106 +/- 1.96 * 10/sqrt(20)

101.62, 110.38

CI does not contain population mean, so 100 is not a plausible value for the population mean at the 95% confidence level.

This matches the conclusion of the hypothesis test.