if x-bar = 57.2 and s = 9.6, compute the test statistic

Question

To test H0: mean = 60 vs. H1: mean < 60, a simple random sample of size n = 20 is obtained from a population that is known to be normally distributed.

Answer 1

The t-test statistic t=(60-57.2)/(9.6/√20)=1.30 approx. (The sample standard deviation has to be adjusted to 9.6/√n where n is the sample size.) When using the t-tables the number of degrees of freedom = n-1 = 19 in this case. Since we are only considering mean < 60, it is a one-tail test. (A 2-tail test would apply if we were considering mean ≠ 60, that is, mean <> 60.) Then the significance level or confidence level needs to be known to decide whether to accept or reject H₀. If the significance level is 0.05 (95% confidence level), tables give a critical value of 1.79, compared to the calculated 1.30 test value. So it would appear that we do not reject H₀.