Two-Sample Tests — Switch Activation Time: 04-BS-2, December 2018, Q7

Question 7: Two-Sample Tests — Switch Activation Time (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. Two independent samples of switch activation times; Make A: n_A=10, mean 19.5 s, sd 0.8 s (one substandard unit already discarded, so Make B has n_B=9). Make B: n_B=9, mean 20.2 s, sd 0.6 s.

Find. (a) Test H₀: σ_A²=σ_B² at α=0.05. (b) Test H₀: μ_A=μ_B at α=0.05.

Approach. (a) an F-test on the ratio of sample variances, assuming both populations are normally distributed and the two samples are independent. (b) if (a) supports equal variances, a pooled two-sample t-test with df=n_A+n_B−2.

(a) F-test for equal variances. Assumptions: both activation-time populations are approximately Normal, and the Make-A and Make-B samples are independent (a reasonable assumption since HST tested the two makes separately). Placing the larger variance on top: $$F=\dfrac{s_A^2}{s_B^2}=\dfrac{0.8^2}{0.6^2}=\dfrac{0.64}{0.36}=1.778$$ with $\nu_1=n_A-1=9$, $\nu_2=n_B-1=8$. For a two-tailed test at $\alpha=0.05$, the critical value is $F_{0.025,9,8}=4.36$ (Appendix 5). Since $1.778\lt 4.36$: $$\boxed{\text{Fail to reject }H_0\text{ -- the variability of Make A and Make B is not significantly different}}$$
(b) Pooled two-sample t-test. Part (a) supports equal population variances, so pool them: $$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{9(0.64)+8(0.36)}{17}=\dfrac{8.64}{17}=0.5082$$ Then $$t=\dfrac{m_A-m_B}{\sqrt{s_p^2\left(\frac{1}{n_A}+\frac{1}{n_B}\right)}}=\dfrac{19.5-20.2}{\sqrt{0.5082\left(\frac{1}{10}+\frac{1}{9}\right)}}=\dfrac{-0.7}{0.3276}=-2.137$$ with df$=n_A+n_B-2=17$; critical value $t_{0.025,17}=2.110$ (two-tailed, Appendix 2). Since $|t|=2.137\gt 2.110$: $$\boxed{\text{Reject }H_0\text{ -- the mean activation time of Make A is significantly different from Make B}}$$ (a narrow margin: the conclusion would flip to “fail to reject” at α=0.02 or smaller.)

Question 7 — final results
Part	Quantity	Result
(a)	F (df 9,8) vs F_0.025,9,8=4.36	1.778 → fail to reject
(b)	pooled t (df=17) vs t_0.025,17=2.110	−2.137 → reject

Topic: Two-sample F-test for variances and pooled two-sample t-test for means.

Key relations: $F=s_1^2/s_2^2$ (larger over smaller), df=(n₁−1, n₂−1); pooled variance $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}$; $t=\dfrac{\bar x_A-\bar x_B}{s_p\sqrt{1/n_A+1/n_B}}$.

Why this works: The F-test is checked first because the correct form of the two-sample t-test depends on whether the population variances can be assumed equal; pooling is only valid once that assumption is supported, since it combines both samples’ variance information into one shared estimate with more degrees of freedom, which sharpens the t-test.

Common pitfall: Skipping the variance-equality check and always pooling (or always using Welch’s unequal-variance t); putting the smaller variance on top of the F-ratio (must always be larger/smaller to compare against an upper-tail critical value).

Source: Walpole et al., Probability & Statistics for Engineers and Scientists, Ch. 10 (Two-Sample Tests).

Question 7 of 8: Two-Sample Tests — Switch Activation Time

Question 7: Two-Sample Tests — Switch Activation Time (20 marks)