Two alloys, equal spreads: pooled t-test

Question

Two alloys are tested for hardness on separate specimens with comparable spreads, so equal variances are assumed. Alloy X: 8 specimens, mean 60, sample standard deviation 4. Alloy Y: 8 specimens, mean 64, sample standard deviation 5.

Run a pooled two-sample t-test at the 5% level (two-sided): do the mean hardnesses differ? Report the t-statistic.

Accepted Answer

Equal variances assumed, so pool. Hypotheses: H0: \muX = \muY versus H1: \muX 
eq \muY. Pooled variance: sp^2 = \frac{7(16) + 7(25)}{14} = \frac{112 + 175}{14} = \frac{287}{14} = 20.5. So sp = 4.528. Standard error of the difference: 	ext{SE} = 4.528\sqrt{\frac{1}{8} + \frac{1}{8}} = 4.528\sqrt{0.25} = 4.528(0.5) = 2.264. Test statistic: t = \frac{60 - 64}{2.264} = \frac{-4}{2.264} = -1.77. Degrees of freedom: 8 + 8 - 2 = 14; the two-sided 5% critical value is t{0.025,\,14} = 2.145. Since -1.77 = 1.77 < 2.145, we fail to reject H0. The p-value is about 0.10. Conclusion: the 4-unit gap is no…

Two alloys, equal spreads: pooled t-test

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