CAIE FP2 2014 June — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2014
SessionJune
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicT-tests (unknown variance)
TypeTwo-sample z-test large samples
DifficultyStandard +0.3 This is a standard two-sample t-test with summary statistics requiring calculation of sample means and variances, then applying the pooled t-test formula. While it involves multiple computational steps and understanding of hypothesis testing framework, it's a routine application of a standard procedure with no conceptual surprises—slightly easier than average due to being a textbook application, but the computational burden and multiple steps keep it near average difficulty.
Spec2.05a Hypothesis testing language: null, alternative, p-value, significance5.05c Hypothesis test: normal distribution for population mean5.05d Confidence intervals: using normal distribution

8 Weekly expenses claimed by employees at two different branches, \(A\) and \(B\), of a large company are being compared. Expenses claimed by an employee at branch \(A\) and by an employee at branch \(B\) are denoted by \(\\) x\( and \)\\( y\) respectively. A random sample of 60 employees from branch \(A\) and a random sample of 50 employees from branch \(B\) give the following summarised data. $$\Sigma x = 6060 \quad \Sigma x ^ { 2 } = 626220 \quad \Sigma y = 4750 \quad \Sigma y ^ { 2 } = 464500$$ Using a \(2 \%\) significance level, test whether, on average, employees from branch \(A\) claim the same as employees from branch \(B\).

Question 8:
AnswerMarks Guidance
\(H_0: \mu_X = \mu_Y\), \(H_1: \mu_X \neq \mu_Y\)B1 State hypotheses
\(S_x^2 = (626220 - 6060^2/60)/59\) Estimate both population variances using two samples
\([= 240\) or \(15.49^2]\) (allow use of biased: \(\sigma_{X,60}^2 = 236\) or \(15.36^2\))
And \(s_Y^2 = (464500 - 4750^2/50)/49\)M1 A1
\([= 270.4\) or \(16.44^2]\) (allow use of biased: \(\sigma_{Y,50}^2 = 265\) or \(16.28^2\))
\(s^2 = s_X^2/60 + s_Y^2/50\)M1 A1 Estimate population variance for combined sample
\(= 9.408\) or \(3.067^2\) (allow \(\sigma_{X,60}^2/60 + \sigma_{Y,50}^2/50\): \(9.233\) or \(3.039^2\))
\(z = (101 - 95)/s\)M1 Calculate value of \(z\) (to 2 d.p., either sign)
\(= 6/3.067 = 1.96\) (or \(1.97\))A1
\(z_{0.99} = 2.326\) or \(2.33\) (allow \(2.36\))B1 State or use correct tabular \(z\)-value (to 2 d.p.)
[Accept \(H_0\)] Claims are the sameB1\(\checkmark\) Correct conclusion (A.E.F., \(\checkmark\) on \(z\)-values)
S.R. Assuming equal population variances:
\(s^2 = (626220 - 6060^2/60 + 464500 - 4750^2/50)/108\)(M1 A1) Find pooled estimate of common variance \(s^2\)
\(= 253.8\) or \(15.93^2\)(A1)
\(z = 6/s\sqrt{(1/60 + 1/50)} = 1.97\)(M1 A1) Calculate value of \(z\)
As above(B1; B1\(\checkmark\)) Tabular value; conclusion
# Question 8:

| $H_0: \mu_X = \mu_Y$, $H_1: \mu_X \neq \mu_Y$ | B1 | State hypotheses |
| $S_x^2 = (626220 - 6060^2/60)/59$ | | Estimate both population variances using two samples |
| $[= 240$ or $15.49^2]$ | | (allow use of biased: $\sigma_{X,60}^2 = 236$ or $15.36^2$) |
| And $s_Y^2 = (464500 - 4750^2/50)/49$ | M1 A1 | |
| $[= 270.4$ or $16.44^2]$ | | (allow use of biased: $\sigma_{Y,50}^2 = 265$ or $16.28^2$) |
| $s^2 = s_X^2/60 + s_Y^2/50$ | M1 A1 | Estimate population variance for combined sample |
| $= 9.408$ or $3.067^2$ | | (allow $\sigma_{X,60}^2/60 + \sigma_{Y,50}^2/50$: $9.233$ or $3.039^2$) |
| $z = (101 - 95)/s$ | M1 | Calculate value of $z$ (to 2 d.p., either sign) |
| $= 6/3.067 = 1.96$ (or $1.97$) | A1 | |
| $z_{0.99} = 2.326$ or $2.33$ (allow $2.36$) | B1 | State or use correct tabular $z$-value (to 2 d.p.) |
| [Accept $H_0$] Claims are the same | B1$\checkmark$ | Correct conclusion (A.E.F., $\checkmark$ on $z$-values) |
| **S.R.** Assuming equal population variances: | | |
| $s^2 = (626220 - 6060^2/60 + 464500 - 4750^2/50)/108$ | (M1 A1) | Find pooled estimate of common variance $s^2$ |
| $= 253.8$ or $15.93^2$ | (A1) | |
| $z = 6/s\sqrt{(1/60 + 1/50)} = 1.97$ | (M1 A1) | Calculate value of $z$ |
| As above | (B1; B1$\checkmark$) | Tabular value; conclusion |
8 Weekly expenses claimed by employees at two different branches, $A$ and $B$, of a large company are being compared. Expenses claimed by an employee at branch $A$ and by an employee at branch $B$ are denoted by $\$ x$ and $\$ y$ respectively. A random sample of 60 employees from branch $A$ and a random sample of 50 employees from branch $B$ give the following summarised data.

$$\Sigma x = 6060 \quad \Sigma x ^ { 2 } = 626220 \quad \Sigma y = 4750 \quad \Sigma y ^ { 2 } = 464500$$

Using a $2 \%$ significance level, test whether, on average, employees from branch $A$ claim the same as employees from branch $B$.

\hfill \mbox{\textit{CAIE FP2 2014 Q8 [9]}}