04-BS-2, December 2016, 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.

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Note $n_A=12$ (not 11) despite the prose stating "one result had to be discarded" from an original twelve tests — the table's own sample-size row prints $n_A=12,\ n_B=11$, so the discarded result belongs to Make B's batch (which is why $n_B=11$); Make A's own batch of 12 is used intact. Both sample sizes are taken directly from the printed table.

Given. Make A: $n_A=12$, $\bar x_A=416.0$ kPa, $s_A=2.5$ kPa. Make B: $n_B=11$, $\bar x_B=420.0$ kPa, $s_B=2.0$ kPa.

Find. (a) test $H_0:\sigma_A=\sigma_B$; (b) test $H_0:\mu_A=\mu_B$ (assuming, per part (a)'s outcome, equal population variances).

Approach. Compare the two variances with an $F$-test first (this determines whether a pooled or Welch $t$-test is appropriate); since (a) fails to reject equal variances, pool the variances and run a two-sample pooled-variance $t$-test for the means.

(a) F-test for equality of variances (α = 0.05, two-tail). $F=\dfrac{s_A^2}{s_B^2}=\dfrac{2.5^2}{2.0^2}=\dfrac{6.25}{4.00}=\boxed{1.5625}$, with $df_1=n_A-1=11$, $df_2=n_B-1=10$. Critical values $F_{0.025,11,10}=3.665$ and $F_{0.975,11,10}=0.284$. Since $0.284<1.5625<3.665$, $\boxed{\text{fail to reject }H_0}$ — the two variances are NOT significantly different, so pooling them in part (b) is justified. Assumption: both Make A and Make B measurements are independent samples from normal populations.
(b) Pooled two-sample t-test for the means (α = 0.05, two-tail). Pooled variance: $s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{11(6.25)+10(4.00)}{21}=\dfrac{68.75+40.00}{21}=\boxed{5.179}$. $t=\dfrac{\bar x_B-\bar x_A}{\sqrt{s_p^2\left(\frac1{n_A}+\frac1{n_B}\right)}}=\dfrac{420.0-416.0}{\sqrt{5.179\left(\frac1{12}+\frac1{11}\right)}}=\dfrac{4.0}{0.9499}=\boxed{4.211}$, with $df=n_A+n_B-2=21$. Critical $t_{0.025,21}=2.080$. Since $|t|=4.211>2.080$, $\boxed{\text{reject }H_0}$ — the mean shear strengths ARE significantly different (Make B's glue tests significantly stronger than Make A's).

Question 7 — final results
Part	Quantity	Value
(a)	F, critical range (11,10 df)	1.5625, (0.284, 3.665) → fail to reject H₀
(b)	Pooled s²_p; t; t-crit (21 df)	5.179; 4.211; 2.080 → reject H₀

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

Key relations: $F=\dfrac{s_1^2}{s_2^2}\sim F_{n_1-1,n_2-1}$; pooled variance $s_p^2=\dfrac{(n_1-1)s_1^2+(n_2-1)s_2^2}{n_1+n_2-2}$; $t=\dfrac{\bar x_1-\bar x_2}{\sqrt{s_p^2(1/n_1+1/n_2)}}\sim t_{n_1+n_2-2}$.

Why this works: the F-test exploits the ratio of two independent scaled chi-square variables (each sample variance, scaled by its own population variance, is chi-square); when population variances are equal this ratio has the standard F-distribution. Once equal variances are supported, pooling both samples' variance estimates into one gives a more precise, higher-df estimate of the common variance for the t-test.

Common pitfall: running the pooled t-test without first checking (or explicitly assuming) equal variances — if the F-test had rejected $H_0$, a Welch (unequal-variance) t-test with the Welch-Satterthwaite df would be required instead.

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

Question 7 of 8

Question 7 (20 marks)