CAIE FP2 2018 November — Question 8 9 marks

Exam BoardCAIE
ModuleFP2 (Further Pure Mathematics 2)
Year2018
SessionNovember
Marks9
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicZ-tests (known variance)
TypeTwo-sample t-test with summary statistics
DifficultyStandard +0.3 This is a straightforward two-sample t-test requiring calculation of sample means and standard deviations from summary statistics, then applying a standard hypothesis testing procedure. While it involves multiple computational steps (8 marks), the method is entirely routine for Further Maths students with no conceptual challenges or novel insights required. The second part tests basic understanding of the Central Limit Theorem, making this slightly easier than average overall.
Spec5.05c Hypothesis test: normal distribution for population mean
The weekly salaries of employees at two large electronics companies, \(A\) and \(B\), are being compared. The weekly salary of an employee from company \(A\) and an employee from company \(B\) are denoted by \(\\)x\( and \)\\(y\) respectively. A random sample of 50 employees from company \(A\) and a random sample of 40 employees from company \(B\) give the following summarised data. $$\Sigma x = 5120 \quad \Sigma x^2 = 531000 \quad \Sigma y = 3760 \quad \Sigma y^2 = 375135$$
  1. The population mean salaries of employees from companies \(A\) and \(B\) are denoted by \(\\)\mu_A\( and \)\\(\mu_B\) respectively. Using a 5\% significance level, test the null hypothesis \(\mu_A = \mu_B\) against the alternative hypothesis \(\mu_A \neq \mu_B\). [8]
  2. State, with a reason, whether any assumptions about the distributions of employees' salaries are needed for the test in part (i). [1]
Question 8:
AnswerMarks
8(i)s 2 = (531 000 – 51202/50) / 49 and [= 6712/49]
A
s 2 = (375 135 – 37602/40) / 39 [= 21695/39]
AnswerMarks Guidance
BM1 Estimate both population variances
s 2 = 136.98 and s 2 = 556.28 (to 3 s.f. throughout)
AnswerMarks Guidance
A BA1 (allow biased here: 134.24 and 542.4)
EITHER: s2 = s 2/50 + s 2/40 = 16.65 or 4.082
AnswerMarks Guidance
A BM1A1 EITHER: Estimate combined variance
z = 1.96
AnswerMarks Guidance
0.975*B1 State or use correct tabular z (or t) value
z = (y – x) / s = (102.4 – 94) / s = 2.06
z > tabular value [accept H ]
AnswerMarks Guidance
1M1A1 Calculate value of z (or –z)
(or can compare y – x = 8.4 with 8.0)
so µ ≠ µ or salaries do differ (AEF)
AnswerMarks Guidance
A BB1 Correct conclusion (FT on z, dep *B1)
OR: Assume equal [population] variances
s2 = (49 s 2 + 39 s 2) / 88
A B
AnswerMarks Guidance
or (531 000 – 51202/50 + 375 135 – 37602/40) / 88(B1 OR: State assumption about variances (may be in part (ii))
Find pooled estimate of common variance
(M1 A1 for s 2 and s 2 may be implied here)
A B
AnswerMarks
= 28407/88 or 322.8 or 17.972B1
z = 1.96
AnswerMarks Guidance
0.975*B1 State or use correct tabular z (or t) value
z = 8.4 / s√(1/50 + 1/40) = 2.20M1A1 Calculate value of z (or –z)
(or can compare y – x = 8.4 with 7.47)
z > tabular value [accept H ]
1
so µ ≠ µ or salaries do differ (AEF)
AnswerMarks Guidance
A BB1) Correct conclusion (FT on z, dep *B1)
8
AnswerMarks Guidance
QuestionAnswer Marks
AnswerMarks Guidance
8(ii)No assumption [of normality] needed
since large samples or due to central limit theoremB1 Valid comment on required assumptions (AEF)
(assumption of equal variances may earn B1 above)
1
AnswerMarks Guidance
QuestionAnswer Marks 
Question 8:
--- 8(i) ---
8(i) | s 2 = (531 000 – 51202/50) / 49 and [= 6712/49]
A
s 2 = (375 135 – 37602/40) / 39 [= 21695/39]
B | M1 | Estimate both population variances
s 2 = 136.98 and s 2 = 556.28 (to 3 s.f. throughout)
A B | A1 | (allow biased here: 134.24 and 542.4)
EITHER: s2 = s 2/50 + s 2/40 = 16.65 or 4.082
A B | M1A1 | EITHER: Estimate combined variance
z = 1.96
0.975 | *B1 | State or use correct tabular z (or t) value
z = (y – x) / s = (102.4 – 94) / s = 2.06
z > tabular value [accept H ]
1 | M1A1 | Calculate value of z (or –z)
(or can compare y – x = 8.4 with 8.0)
so µ ≠ µ or salaries do differ (AEF)
A B | B1 | Correct conclusion (FT on z, dep *B1)
OR: Assume equal [population] variances
s2 = (49 s 2 + 39 s 2) / 88
A B
or (531 000 – 51202/50 + 375 135 – 37602/40) / 88 | (B1 | OR: State assumption about variances (may be in part (ii))
Find pooled estimate of common variance
(M1 A1 for s 2 and s 2 may be implied here)
A B
= 28407/88 or 322.8 or 17.972 | B1
z = 1.96
0.975 | *B1 | State or use correct tabular z (or t) value
z = 8.4 / s√(1/50 + 1/40) = 2.20 | M1A1 | Calculate value of z (or –z)
(or can compare y – x = 8.4 with 7.47)
z > tabular value [accept H ]
1
so µ ≠ µ or salaries do differ (AEF)
A B | B1) | Correct conclusion (FT on z, dep *B1)
8
Question | Answer | Marks | Guidance
--- 8(ii) ---
8(ii) | No assumption [of normality] needed
since large samples or due to central limit theorem | B1 | Valid comment on required assumptions (AEF)
(assumption of equal variances may earn B1 above)
1
Question | Answer | Marks | Guidance
The weekly salaries of employees at two large electronics companies, $A$ and $B$, are being compared. The weekly salary of an employee from company $A$ and an employee from company $B$ are denoted by $\$x$ and $\$y$ respectively. A random sample of 50 employees from company $A$ and a random sample of 40 employees from company $B$ give the following summarised data.

$$\Sigma x = 5120 \quad \Sigma x^2 = 531000 \quad \Sigma y = 3760 \quad \Sigma y^2 = 375135$$

\begin{enumerate}[label=(\roman*)]
\item The population mean salaries of employees from companies $A$ and $B$ are denoted by $\$\mu_A$ and $\$\mu_B$ respectively. Using a 5\% significance level, test the null hypothesis $\mu_A = \mu_B$ against the alternative hypothesis $\mu_A \neq \mu_B$. [8]

\item State, with a reason, whether any assumptions about the distributions of employees' salaries are needed for the test in part (i). [1]
\end{enumerate}

\hfill \mbox{\textit{CAIE FP2 2018 Q8 [9]}}
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