04-BS-2, May 2015, Q7 | Nivaar Exam Prep

Question 7 (20 marks)

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. Make A: $n_A=11$, $\bar x_A=84$, $s_A=0.25$. Make B: $n_B=10$, $\bar x_B=76$, $s_B=0.35$.

Find. (a) Test $H_0:\sigma_A^2=\sigma_B^2$ at $\alpha=0.05$. (b) Test $H_0:\mu_A=\mu_B$ at $\alpha=0.05$ (using the result of (a) to choose the correct $t$-test form).

Approach. An $F$-test compares the two variances first; if they are not significantly different, the pooled-variance two-sample $t$-test is the correct form for comparing the means.

(a) $F$-test for equality of variances. Assumption: both populations are approximately Normal and the two samples are independent. Put the larger sample variance on top: $$F=\frac{s_B^2}{s_A^2}=\frac{0.35^2}{0.25^2}=\boxed{1.96}\qquad (df_1=n_B-1=9,\ df_2=n_A-1=10)$$ Critical values $F_{0.025,9,10}=3.78$ and $F_{0.975,9,10}=0.259$. Since $0.259<1.96<3.78$, fail to reject $H_0$: the variabilities of the two makes are not significantly different.
(b) Pooled-variance two-sample $t$-test. Since (a) supports equal variances, pool them: $$s_p^2=\frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\frac{10(0.0625)+9(0.1225)}{19}=\boxed{0.0909}$$ $$t=\frac{\bar x_A-\bar x_B}{\sqrt{s_p^2\left(\frac1{n_A}+\frac1{n_B}\right)}}=\frac{84-76}{\sqrt{0.0909(1/11+1/10)}}=\boxed{60.72}\qquad df=19$$ Critical value $t_{0.025,19}=2.093$. Since $60.72\gg2.093$, reject $H_0$: the mean ratings of the two makes ARE significantly different (Make A rates markedly higher, and the very small within-make standard deviations, 0.25 and 0.35, leave essentially no overlap around an 8-unit mean gap).

Question 7 — final results
Quantity	Value
(a) $F$, critical range	1.96; (0.259, 3.78) — fail to reject
(b) pooled $s_p^2$	0.0909
(b) $t$, conclusion	60.72; reject $H_0$, means differ

Topic: Two-sample inference: $F$-test for variances, pooled $t$-test for means.

Key relations: $F=s_1^2/s_2^2\sim F_{n_1-1,n_2-1}$; pooled variance $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}$; $t=\dfrac{\bar x_A-\bar x_B}{\sqrt{s_p^2(1/n_A+1/n_B)}}$, df $=n_A+n_B-2$.

Why this works: The $F$-test is the natural ratio-of-variances statistic under independent Normal samples; pooling is valid (and more powerful than Welch's unequal-variance form) precisely when the $F$-test fails to detect a variance difference, since pooling assumes a single common $\sigma^2$.

Common pitfall: Running the pooled $t$-test without first checking equal variances (the correct workflow is always $F$-test first); putting the smaller variance on top of the $F$ ratio (convention is larger over smaller, or track which critical value corresponds to the ratio actually computed).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, Ch. 9 (Two-Sample Tests).

Question 7 of 8

Question 7 (20 marks)