Question 7 of 8: Two-Sample F-Test and t-Test for Drill Lifetimes

Question 7: Two-Sample F-Test and t-Test for Drill Lifetimes (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.

Find. (a) test $H_0:\sigma_A=\sigma_B$; (b) test $H_0:\mu_A=\mu_B$.

Approach. Test the variances first with an $F$-test (needed to choose pooled-vs-Satterthwaite for the mean test); assuming both populations are Normal and independent, then run the two-sample $t$-test using whichever variance assumption (a) supports.

1. (a) $F$-test for equal variances. Put the larger sample variance on top: $F=\dfrac{s_B^2}{s_A^2}=\dfrac{35^2}{26^2}=\dfrac{1{,}225}{676}=\boxed{1.812}$, with $df_1=n_B-1=8$ (numerator) and $df_2=n_A-1=7$ (denominator). From the $F$-table, $F_{0.025,8,7}\approx4.90$. Since $1.812\lt4.90$: fail to reject $H_0:\sigma_A=\sigma_B$ — no significant difference in variability; it is reasonable to assume equal population variances for part (b).
2. (b) Pooled two-sample $t$-test. With equal variances assumed (justified by (a)), pool the sample variances: $$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{7(676)+8(1{,}225)}{15}=\dfrac{4{,}732+9{,}800}{15}=968.8,\quad s_p=31.13$$ $$t=\dfrac{m_B-m_A}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{739-725}{31.13\sqrt{1/8+1/9}}=\dfrac{14}{15.13}=\boxed{0.926}$$ with $df=n_A+n_B-2=15$, giving $t_{0.025,15}=2.131$. Since $0.926\lt2.131$: fail to reject $H_0:\mu_A=\mu_B$ — the mean lifetimes of Make A and Make B are not significantly different.