Here, Sample mean= r = 4.39 Sample S.D=5= 2.29 Sample size = n = 80 Degrees of freedom = 79 Significance Level = Q = 0.05 Step 1: H : 4 = 4.5 H: 4 + 4.5 (Two tailed test) Step 2: The test statistic is, t=* = -0.430 Step 3: P-value = P(1>-0.430) = 0.6684 Step 4: Conclusion: Since the P-value is greater than the significance level, we fail to reject H. There is not sufficient evidence to reject the claim that population of student course evaluations has a mean equal to 4.50.