Question 7 of 8: Two-Sample Tests — Comparing Light Bulb Makes
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, May 2013 — 04-BS-2 Probability and Statistics (2 hours, closed book except one hand-written information sheet and an approved calculator; any 5 of the 8 questions constitute a complete paper — all 8 are answered below as a full study resource).
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists, 9th ed. (sampling distributions, estimation, hypothesis testing, correlation/regression); standard Normal, t, χ² and F tables as supplied with the exam.
Check: the printed sample-size line for Question 7 reads "thirteen bulbs were randomly selected...one result had to be discarded" immediately above a table showing nA=13, nB=12. Read as: Make A's sample stayed at 13 (no discard), Make B's original 13 dropped to 12 after the discard — the table's printed nA=13, nB=12 are used as given.
Find. (a) F-test for equal variances at α=0.05; (b) t-test for equal means at α=0.05 (choosing pooled or Welch form based on part (a)'s conclusion).
Approach. Test variances first with an F-test since the correct form of the mean-comparison test depends on whether the two population variances can be assumed equal; both samples are independent random samples assumed drawn from Normal populations (necessary for both the F and t procedures at these small sample sizes).
(a) F-test for equal variances, H₀: σA²=σB² vs H₁: σA²≠σB². Assumption: both samples are independent random samples from Normal populations. With the larger variance in the numerator,
$$F=\dfrac{s_A^2}{s_B^2}=\dfrac{110.0^2}{85.0^2}=\dfrac{12{,}100}{7{,}225}=\boxed{1.675}.$$
Degrees of freedom $df_A=12$, $df_B=11$; critical value $F_{0.025,12,11}=3.430$. Since $F=1.675<3.430$, fail to reject H₀: the two variances are not significantly different at α=0.05.
(b) Pooled-variance t-test for equal means, H₀: μA=μB vs H₁: μA≠μB. Since (a) supports equal variances, pool the two sample variances:
$$s_p^2=\dfrac{(n_A-1)s_A^2+(n_B-1)s_B^2}{n_A+n_B-2}=\dfrac{12(12{,}100)+11(7{,}225)}{23}=\dfrac{224{,}675}{23}=9{,}768.5\ \Rightarrow\ s_p=98.84.$$
$$t=\dfrac{m_A-m_B}{s_p\sqrt{1/n_A+1/n_B}}=\dfrac{10{,}000-9{,}900}{98.84\sqrt{1/13+1/12}}=\boxed{2.527}.$$
Degrees of freedom $n_A+n_B-2=23$; critical value $t_{0.025,23}=2.069$. Since $|t|=2.527>2.069$, reject H₀: the mean life of Make A is significantly greater than that of Make B at α=0.05.
