Question 7 of 8

Question 7 (20 marks: (a) 10, (b) 10)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Two independent samples of bulb lifetimes: A ($n_A=11$, $\bar{x}_A=11{,}860$, $s_A=370$) and B ($n_B=10$, $\bar{x}_B=11{,}980$, $s_B=250$).

Find. (a) Test $H_0:\sigma_A=\sigma_B$. (b) Test $H_0:\mu_A=\mu_B$, using the F-test result from (a) to choose the correct t-procedure.

Approach. Compare the two variances with an F-test first (assuming both populations are Normal, independent samples) — the outcome decides whether the means are compared with a pooled-variance or a Welch (unequal-variance) t-test.

1. (a) F-test for equal variances. Put the larger sample variance on top: $$F = \frac{s_A^2}{s_B^2} = \frac{370^2}{250^2} = \frac{136{,}900}{62{,}500} = 2.190.$$ Critical value $F_{0.025,10,9}=3.964$ (df$_A$=10 numerator, df$_B$=9 denominator). Since $2.190 < 3.964$, we fail to reject H₀ — the two population variances are not significantly different, so a pooled-variance t-test is the right procedure for part (b).
2. (b) Pooled t-test for equal means. Pooled variance: $$s_p^2 = \frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2} = \frac{10(370^2)+9(250^2)}{19} = 101{,}657.9, \qquad s_p = 318.8.$$ $$t = \frac{\bar{x}_A-\bar{x}_B}{s_p\sqrt{1/n_A+1/n_B}} = \frac{11{,}860-11{,}980}{318.8\sqrt{1/11+1/10}} = \frac{-120}{139.3} = \boxed{-0.861}.$$ With df $=19$, $t_{0.025,19}=2.093$, and $|-0.861|<2.093$, we fail to reject H₀ — the mean lifetimes of the two manufacturers' bulbs are not significantly different.