Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Examinations, 04-BS-2 Probability and Statistics — May 2015. Closed book (approved calculator + one hand-written information sheet); statistical tables of the Normal, t, chi-square and F distributions supplied. The paper states that only 5 of the 8 questions constitute a complete paper — every question is solved in full as a study resource.
Reference texts: Walpole, Myers, Myers & Ye, Probability & Statistics for Engineers and Scientists (9th ed.) — sampling distributions, binomial/Poisson/hypergeometric models, confidence intervals, hypothesis testing, correlation and simple linear regression.
Given. (A) $n=450$, 315 plan to join the labour force ($\hat p=315/450$). (B) $n=13$ paired before/after scores; $\sum d_i=6$, $\sum d_i^2=14$.
Find. (A) Test $H_0:p=0.75$ at $\alpha=0.05$. (B)(a) $m_d$, $s_d$; (b) test whether training raised performance ($H_1:\mu_d>0$) at $\alpha=0.05$.
Approach. (A) is a large-sample proportion test using the Normal approximation. (B) is a paired-difference $t$-test, one-tailed since the alternative asserts a specific direction of improvement.
(A) Large-sample test of a proportion. $\hat p=315/450=0.700$.
$$z=\frac{\hat p-p_0}{\sqrt{p_0(1-p_0)/n}}=\frac{0.700-0.75}{\sqrt{0.75(0.25)/450}}=\boxed{-2.449}$$
Critical value $z_{0.025}=1.960$. Since $|{-2.449}|>1.960$, reject $H_0$: the proportion planning to join the labour force IS significantly different from 0.75 (it is lower).
(B)(a) Mean and standard deviation of the paired differences.
$$m_d=\frac{\sum d_i}{n}=\frac{6}{13}=\boxed{0.4615}$$
$$s_d^2=\frac{\sum d_i^2-(\sum d_i)^2/n}{n-1}=\frac{14-6^2/13}{12}=\boxed{0.9359}\qquad s_d=\boxed{0.9674}$$
(B)(b) One-tailed paired $t$-test. $H_0:\mu_d=0$ (no change) vs $H_1:\mu_d>0$ (Y > X, i.e. performance improved after training).
$$t=\frac{m_d}{s_d/\sqrt n}=\frac{0.4615}{0.9674/\sqrt{13}}=\boxed{1.720}$$
Critical value (one-tailed) $t_{0.05,12}=1.782$. Since $1.720<1.782$, fail to reject $H_0$: at $\alpha=0.05$ there is insufficient evidence that the training programme significantly increased performance — the sample mean improvement (0.46) is in the expected direction, but the sample-to-sample variability is too large to rule out chance at the 5% level. This is a genuinely close call ($t$ is within 0.06 of the critical value); a larger sample or a lower significance requirement (e.g. $\alpha=0.10$, critical value 1.356) would have reversed the conclusion.
