F-Test for Variances, Then Two-Sample t-Test: 04-BS-2, December 2014, Q7

Question 7: F-Test for Variances, Then Two-Sample t-Test (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. Maker A: $n_A=10$, $m_A=29.5$ s, $s_A=0.7$ s. Maker B: $n_B=9$, $m_B=30.3$ s, $s_B=0.5$ s.

Find. (a) test H₀: $\sigma_A^2=\sigma_B^2$ at α=0.05; (b) test H₀: $\mu_A=\mu_B$ at α=0.05.

Approach. Use an F-test on the variance ratio first (this decides whether the t-test in (b) should pool the two sample variances), then a two-sample t-test with the pooled variance since the F-test finds no significant variance difference.

(a) F-test for equality of variances. Placing the larger sample variance on top, $$F=\dfrac{s_A^2}{s_B^2}=\dfrac{0.7^2}{0.5^2}=\boxed{1.96},\qquad df_1=n_A-1=9,\ df_2=n_B-1=8.$$ The two-tailed critical value is $F_{0.025,9,8}=4.357$. Since $1.96<4.357$, fail to reject H₀: the variability of Maker A and Maker B's activation times is NOT significantly different at α=0.05 — this justifies pooling the variances in part (b).
(b) Pooled two-sample t-test. Since (a) supports equal variances, $$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{9(0.49)+8(0.25)}{17}=0.3771,\qquad s_p=0.6141.$$ $$t=\dfrac{m_A-m_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{29.5-30.3}{0.6141\sqrt{1/10+1/9}}=\boxed{-2.836},\qquad df=17.$$ The critical value is $t_{0.025,17}=2.110$. Since $|t|=2.836>2.110$, reject H₀: the mean activation time of Maker A IS significantly different (lower) than Maker B's at α=0.05.

Final results — Question 7
Part	Quantity	Result
(a)	F-stat vs F-crit; conclusion	1.96 vs 4.357; fail to reject H₀
(b)	pooled t-stat vs t-crit; conclusion	−2.836 vs 2.110; reject H₀

Topic: Sequential F-test for variance equality followed by the appropriate (pooled or Welch) two-sample t-test for means.

Key relations: $F=s_1^2/s_2^2$ (larger over smaller) compared to $F_{\alpha/2,\,df_1,df_2}$; pooled $t=(\bar x_A-\bar x_B)/\big(s_p\sqrt{1/n_A+1/n_B}\big)$, $df=n_A+n_B-2$, with $s_p^2$ a weighted average of the two sample variances.

Why this works: the pooled t-test assumes a single common population variance; the F-test is the standard preliminary check of that assumption — only once it is not rejected is it valid to combine $s_A^2$ and $s_B^2$ into one pooled estimate rather than using Welch's unequal-variance t-test.

Common pitfall: skipping the F-test and pooling variances regardless, or placing the smaller variance in the F-ratio's numerator (the ratio must be ≥1 by convention when comparing to a one-sided table of critical values doubled for a two-tailed test).

Source: Walpole, Myers, Myers & Ye, Ch. 10 (Two-sample inference, F- and t-tests).

Question 7 of 8: F-Test for Variances, Then Two-Sample t-Test

Question 7: F-Test for Variances, Then Two-Sample t-Test (20 marks)