04-BS-2, May 2017, 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. Process A: $n_A=10$, $\bar x_A=26.5$, $s_A=0.50$ MV/m. Process B: $n_B=12$, $\bar x_B=27.1$, $s_B=0.65$ MV/m. Both populations assumed normal with independent random samples.

Find. (a) two-sided test $H_0:\sigma_A^2=\sigma_B^2$; (b) two-sided test $H_0:\mu_A=\mu_B$ (using the outcome of (a) to decide pooled vs. unequal-variance $t$).

Approach. F-test for equality of two variances first; if not rejected, pool the variances and run a two-sample $t$-test with $n_A+n_B-2$ degrees of freedom.

(a) F-test for equal variances. $F=\dfrac{s_B^2}{s_A^2}=\dfrac{0.65^2}{0.50^2}=\dfrac{0.4225}{0.25}=1.69$, with numerator $df_1=n_B-1=11$, denominator $df_2=n_A-1=9$. Critical value (two-sided $\alpha=0.05$, so upper $2.5\%$ point) $F_{0.025,11,9}=3.91$. Since $1.69<3.91$, $\boxed{\text{fail to reject }H_0\text{: the two variances are not significantly different}}$ — assuming both processes yield independent, normally distributed dielectric strengths, we may pool them.
(b) Pooled two-sample t-test. $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{9(0.25)+11(0.4225)}{20}=0.3450$, so $s_p=0.5874$. $t=\dfrac{\bar x_A-\bar x_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{26.5-27.1}{0.5874\sqrt{1/10+1/12}}=\dfrac{-0.6}{0.2514}=-2.386$, with $df=n_A+n_B-2=20$. Critical value $t_{0.025,20}=2.086$. Since $|t|=2.386>2.086$, $\boxed{\text{reject }H_0}$: the mean dielectric strengths ARE significantly different (process B tests higher).

Question 7 — final results
Part	Quantity	Value
(a)	F-stat; F_0.025,11,9; conclusion	1.69; 3.91; fail to reject (pool)
(b)	pooled s_p; t-stat; t_0.025,20; conclusion	0.587; −2.386; 2.086; reject H₀

Topic: Sequential two-sample inference — F-test for equality of variances, then a pooled or Welch $t$-test for equality of means.

Key relations: $F=S_B^2/S_A^2\sim F_{n_B-1,n_A-1}$ (larger sample variance in the numerator by convention); when variances are judged equal, $S_p^2=\dfrac{(n_A-1)S_A^2+(n_B-1)S_B^2}{n_A+n_B-2}$ and $T=\dfrac{\bar X_A-\bar X_B}{S_p\sqrt{1/n_A+1/n_B}}\sim t_{n_A+n_B-2}$.

Why this works: the F-test is the natural ratio test because a ratio of two independent scaled chi-squares (each sample variance times $(n-1)/\sigma^2$) is exactly F-distributed under equal population variances; once that assumption is not rejected, pooling the two sample variances into one common estimate of $\sigma^2$ is the minimum-variance unbiased combination, which is what lets the pooled $t$ statistic follow the (exact) $t$ distribution.

Common pitfall: putting the smaller sample variance in the numerator of $F$ (the convention is larger-over-smaller, which changes which critical value/table entry applies), or running the pooled $t$-test without first checking the variance assumption in (a).

Source: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed., Ch. 9–10 (Two-Sample Inference).

Question 7 of 8

Question 7 (20 marks)