Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-BS-2 Probability and Statistics — National Examination, May 2018. Closed book (approved calculator + one hand-written information sheet); any 5 of 8 questions constitute a complete paper (only 5 marked) — all 8 are solved below as a complete study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability and Statistics for Engineers and Scientists (9th ed.), Pearson — used throughout for point-estimation, confidence-interval, and hypothesis-testing formulas; Montgomery & Runger, Applied Statistics and Probability for Engineers, for the regression and correlation-inference formulas in Question 8.
Given. Two independent samples of tire wear (mm), assumed drawn from normal populations.
Given data
Brand A
Brand B
$n$
10
9
Sample mean
8.5 mm
8.7 mm
Sample sd
0.8 mm
1.0 mm
Find. (a) test $H_0:\sigma_A^2=\sigma_B^2$; (b) (conditional on (a)) test $H_0:\mu_A=\mu_B$.
Approach. (a) use the $F$-test for the ratio of two sample variances. Because (a) fails to find a significant variance difference, (b) proceeds with the pooled-variance two-sample $t$-test (equal-variance assumption).
a) $F$-test for equal variances. Placing the larger sample variance in the numerator, $F=s_B^2/s_A^2=(1.0)^2/(0.8)^2=\boxed{1.5625}$, with $df_1=n_B-1=8$ (numerator), $df_2=n_A-1=9$ (denominator). From the $F$ table, $F_{0.025,8,9}=4.10$.
Since $1.5625<4.10$, we fail to reject $H_0:\sigma_A^2=\sigma_B^2$ — the two brands' wear variability is not significantly different. Assumptions: both wear populations are approximately normal, and the two samples are independent random samples.
b) Pooled two-sample $t$-test for the means. Because (a) supports equal population variances, pool the sample variances:
$$s_p^2=\frac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\frac{9(0.64)+8(1.00)}{17}=0.8094\ \text{mm}^2,\qquad s_p=0.8997\ \text{mm}.$$
$$t=\frac{\bar x_A-\bar x_B}{s_p\sqrt{1/n_A+1/n_B}}=\frac{8.5-8.7}{0.8997\sqrt{1/10+1/9}}=\boxed{-0.484},\qquad df=17,\ t_{0.025,17}=2.110.$$
Since $|t|=0.484<2.110$, we fail to reject $H_0:\mu_A=\mu_B$ — the two brands' mean wear is not significantly different.
