Refer to the computer attitude scores (CAS) of students given in Table D.4 of the Data Bank. A computer analysis produces the output

(a) What is the conclusion to testing the equality of mean computer attitude scores at level a = .05?
(b) Find a 95% confidence interval for the mean attitude score for males minus the mean attitude score for females.
(c) Test the null hypothesis that, on average, males score.1 lower than females against a two-sided alternative. Take α = .05.

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Two-sample T for CAS sex 15 20 Mean StDev 2.643 0.554 2.9450.390 Difference = mu M-mu F Estimate for difference: 0.302 9018 CI for the difference 0.032, 0.571) T-Test of difference 0 (vs not =): T-value = 1.89 P-Value = 0.067 DF # 33 Both use Pooled StDev = 0.467