Consider the Minitab output below.

(a) Fill in the missing values.
(b) Can the null hypothesis be rejected at the 0.05 level? Why?
(c) Use the output and the t-table to find a 99% CI on the difference in means.
(d) Suppose that the alternative hypothesis was H1: 1 = 2 versus H1: 1 > 2. What is the P-value? What conclusions would you draw?

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Two-Sample T-Test and CI Sauple N Mea StDev SE Mean 0.45 0.56 1.75 15 50.20 15 51.99 2.15 Difference= mu (1)-mu (2) Estimate for difference: ? 95% CI ror a1rrerence: (-3.246, -0.314) T-Test of difference 0 [vs not=)· T-Value-_2.49 D-Value = 0.019 DF Both use Pooled StDev1.9602 ?