CAIE Further Paper 4 2020 June — Question 2 5 marks

Exam BoardCAIE
ModuleFurther Paper 4 (Further Paper 4)
Year2020
SessionJune
Marks5
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicLinear combinations of normal random variables
TypeStandard CI with summary statistics
DifficultyStandard +0.8 This is a Further Maths statistics question requiring construction of a confidence interval for the difference of two means using summary statistics. While the procedure is standard (pooled variance, t-distribution), it involves multiple steps including calculating standard error, finding critical values, and correct interpretation. The Further Maths context and need for careful handling of two-sample inference places it moderately above average difficulty.
Spec5.05c Hypothesis test: normal distribution for population mean

2 A random sample of 40 observations of a random variable \(X\) and a random sample of 50 observations of a random variable \(Y\) are taken. The resulting values for the sample means, \(\bar { x }\) and \(\bar { y }\), and the unbiased estimates, \(\mathrm { s } _ { \mathrm { x } } ^ { 2 }\) and \(\mathrm { s } _ { \mathrm { y } } ^ { 2 }\), for the population variances are as follows. $$\bar { x } = 24.4 \quad \bar { y } = 17.2 \quad s _ { x } ^ { 2 } = 10.2 \quad s _ { y } ^ { 2 } = 11.1$$ Find a \(90 \%\) confidence interval for the difference between the population means of \(X\) and \(Y\).

Question 2:
AnswerMarks Guidance
AnswerMarks Guidance
\(s^2 = \frac{10.2}{40}+\frac{11.1}{50} = 0.477\)M1A1
\(\text{CI} = (24.4-17.2) \pm zs\)M1
\(= (24.4-17.2) \pm 1.645\sqrt{0.477}\)A1
\(= [6.06,\ 8.34]\)A1
Total5
## Question 2:

| Answer | Marks | Guidance |
|--------|-------|----------|
| $s^2 = \frac{10.2}{40}+\frac{11.1}{50} = 0.477$ | M1A1 | |
| $\text{CI} = (24.4-17.2) \pm zs$ | M1 | |
| $= (24.4-17.2) \pm 1.645\sqrt{0.477}$ | A1 | |
| $= [6.06,\ 8.34]$ | A1 | |
| **Total** | **5** | |
2 A random sample of 40 observations of a random variable $X$ and a random sample of 50 observations of a random variable $Y$ are taken. The resulting values for the sample means, $\bar { x }$ and $\bar { y }$, and the unbiased estimates, $\mathrm { s } _ { \mathrm { x } } ^ { 2 }$ and $\mathrm { s } _ { \mathrm { y } } ^ { 2 }$, for the population variances are as follows.

$$\bar { x } = 24.4 \quad \bar { y } = 17.2 \quad s _ { x } ^ { 2 } = 10.2 \quad s _ { y } ^ { 2 } = 11.1$$

Find a $90 \%$ confidence interval for the difference between the population means of $X$ and $Y$.\\

\hfill \mbox{\textit{CAIE Further Paper 4 2020 Q2 [5]}}