Standard +0.3 This is a standard two-sample t-test with pooled variance, requiring straightforward application of a learned procedure: calculate pooled variance, compute test statistic, compare to critical value. While it involves multiple steps and careful arithmetic with given summary statistics, it requires no novel insight—just methodical execution of a textbook technique. Slightly easier than average (0.0) because it's a direct application with all necessary information provided clearly.
2 The children at two large schools, \(P\) and \(Q\), are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school \(P\). Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of 2.16 minutes \({ } ^ { 2 }\). A random sample of size 12 is taken from the children at school \(Q\). Their individual times, \(x\) minutes, to complete the puzzle are summarised by
$$\sum x = 99.6 \quad \sum ( x - \bar { x } ) ^ { 2 } = 21.5$$
where \(\bar { x }\) is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance.
Test at the \(5 \%\) significance level whether the population mean time taken to complete the puzzle by children at school \(P\) is greater than the population mean time taken to complete the puzzle by children at school \(Q\).
\(1.339 < 1.725\); Accept \(H_0\) / not significant
M1
Compare with correct tabular value 1.725 and conclusion without context. Condone 'reject \(H_1\)'
Insufficient evidence to suggest that the (mean) time taken at \(P\) is greater than the (mean) time taken at \(Q\)
A1
Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove'), not 'there is no evidence'), no contradictions. A0 if \(H_0\) and \(H_1\) reversed
## Question 2:
| $\bar{x} = \frac{99.6}{12} = 8.3$, $s_x^2 = \frac{21.5}{11} = 1.9545$ or $\frac{43}{22}$ | M1 | Both SOI |
|---|---|---|
| $H_0: \mu_P = \mu_Q \quad H_1: \mu_P > \mu_Q$ | B1 | |
| Pooled estimate $= \frac{9 \times 2.16 + 11 \times 1.9545}{10 + 12 - 2}$ | M1 FT | FT their 1.9545 only |
| $2.047$ $(2.05)$ | A1 | May be implied, allow 2.04 |
| $t = \frac{9.12 - 8.3}{\sqrt{2.047} \times \sqrt{\frac{1}{10} + \frac{1}{12}}}$ | M1 FT | FT their pooled estimate |
| $1.339$ (or $1.338$) | A1 | Accept 1.34 |
| $1.339 < 1.725$; Accept $H_0$ / not significant | M1 | Compare with correct tabular value 1.725 and conclusion without context. Condone 'reject $H_1$' |
| Insufficient evidence to suggest that the (mean) time taken at $P$ is greater than the (mean) time taken at $Q$ | A1 | Correct conclusion in context, following correct work, level of uncertainty in language (not 'prove'), not 'there is no evidence'), no contradictions. A0 if $H_0$ and $H_1$ reversed |
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2 The children at two large schools, $P$ and $Q$, are all given the same puzzle to solve. A random sample of size 10 is taken from the children at school $P$. Their individual times to complete the puzzle give a sample mean of 9.12 minutes and an unbiased variance estimate of 2.16 minutes ${ } ^ { 2 }$. A random sample of size 12 is taken from the children at school $Q$. Their individual times, $x$ minutes, to complete the puzzle are summarised by
$$\sum x = 99.6 \quad \sum ( x - \bar { x } ) ^ { 2 } = 21.5$$
where $\bar { x }$ is the sample mean. Times to complete the puzzle are assumed to be normally distributed with the same population variance.
Test at the $5 \%$ significance level whether the population mean time taken to complete the puzzle by children at school $P$ is greater than the population mean time taken to complete the puzzle by children at school $Q$.\\
\hfill \mbox{\textit{CAIE Further Paper 4 2023 Q2 [8]}}