Case Study #4 – Chapter 11
First Hypothesis Test
Ho: or Ho:
H1: or Ho:
Step 2: α = 0.05
Step 3: For 2-Sample TTest, enter:
Step 4: Since the smaller n is 57, we use df = 56 and half of α = 0.025 to find the critical values: ± t 0.025 for 56 df = ± 2.004. Therefore, the critical values and rejection regions are:
Reject H0ifor .
Step 4: The P-value would be 2 x P(t < -3.099) since this is a two-tailed test. Using DISTR tcdf(left value, right value, df), we get P-value = 2 x DISTR tcdf(-100000000, -3.099, 56) = 2 x 0.0015 = 0.0030
Note: the P-value on the calculator for the 2-Sample TTest (0.0024) is a little bit different than the one calculated above. This is because it uses a different value for degrees of freedom (122.31 instead of 56). I prefer the P-value using degrees of freedom n – 1 for the smaller sample size (df 56).
Step 5: Using the Classical Approach, we reject H0 since the test statistic, , lies in the rejection region for α = 0.05. Using the P-value Approach, we can reach the same conclusion: reject H0 because the p-value = 0.0030 is less than the significance level α = 0.05 (and less than the significance level α = 0.01).
Step 6: Conclusion in words: There is sufficient evidence at the α = 0.05 (and at the α = 0.01) level of significance to conclude that there is a significant difference in the mean age of the subjects in the control group versus the ESRD group.