| Exam Board | OCR |
|---|---|
| Module | S3 (Statistics 3) |
| Year | 2006 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | T-tests (unknown variance) |
| Type | Find critical alpha or significance level |
| Difficulty | Standard +0.8 This is a multi-part hypothesis testing question requiring calculation of sample statistics, a two-sample test with one known variance, reverse-engineering significance levels, and understanding test assumptions. While the calculations are methodical, part (ii) requires working backwards from critical values, and part (iii) tests deeper conceptual understanding of test validity conditions—going beyond routine application to require statistical reasoning. |
| Spec | 5.05c Hypothesis test: normal distribution for population mean |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(s_B^2 = \frac{1}{49}\left(630.194 - \frac{176.35^2}{50}\right)\) | M1 | Any equivalent formula |
| \(= 0.1675\) | A1 | May be implied by later work |
| \(H_0: \mu_B - \mu_A = 0\), \(H_1: \mu_B - \mu_A > 0\) | B1 | aef |
| \(z = 0.115/\sqrt{(0.049/40 + 0.1675/50)}\) | M1 | Standardising but not from pooled variance estimate |
| \(= 1.700\) | A1 | art \(1.70\) |
| \(z > 1.645\), reject \(H_0\) | M1 | Compare correctly with \(1.645\) |
| and accept that \(\mu_B > \mu_A\) | A1\(\sqrt{}\) | 7 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(z = 0.09/\sqrt{(0.004575)}\) | M1 | Correct form |
| \(= 1.331\) | A1 | |
| \(H_0\) not rejected for \(\alpha < 9.16\) | M1 A1 | 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (a) Not necessary | B1 | Ignore any reason |
| (b) Not necessary since samples large enough for CLT to be applied (normality of sample means giving normality of difference) | M1 | Mention of CLT implied by "sample large" |
| A1 | 3 | Sample mean (approx) normal. Do not award if population or sample said to be normal |
# Question 8(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $s_B^2 = \frac{1}{49}\left(630.194 - \frac{176.35^2}{50}\right)$ | M1 | Any equivalent formula |
| $= 0.1675$ | A1 | May be implied by later work |
| $H_0: \mu_B - \mu_A = 0$, $H_1: \mu_B - \mu_A > 0$ | B1 | aef |
| $z = 0.115/\sqrt{(0.049/40 + 0.1675/50)}$ | M1 | Standardising but not from pooled variance estimate |
| $= 1.700$ | A1 | art $1.70$ |
| $z > 1.645$, reject $H_0$ | M1 | Compare correctly with $1.645$ |
| and accept that $\mu_B > \mu_A$ | A1$\sqrt{}$ | **7** | ft their calculated $z$ |
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# Question 8(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $z = 0.09/\sqrt{(0.004575)}$ | M1 | Correct form |
| $= 1.331$ | A1 | |
| $H_0$ not rejected for $\alpha < 9.16$ | M1 A1 | **4** | Accept $< 9.2$, $\leq 9.2$. M1 for correct method for $9.2$, A1 for inequality |
---
# Question 8(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| (a) Not necessary | B1 | Ignore any reason |
| (b) Not necessary since samples large enough for CLT to be applied (normality of sample means giving normality of difference) | M1 | Mention of CLT implied by "sample large" |
| | A1 | **3** | Sample mean (approx) normal. Do not award if population or sample said to be normal |8 Two machines, $A$ and $B$, produce metal rods. Machine $B$ is new and it is required that its accuracy should be checked against that of machine $A$. The observed variable is the length of a rod. Random samples of rods, 40 from machine $A$ and 50 from machine $B$, are taken and their lengths, $x _ { A } \mathrm {~cm}$ and $x _ { B } \mathrm {~cm}$, are measured. The results are summarised by
$$\Sigma x _ { A } = 136.48 , \quad \Sigma x _ { B } = 176.35 , \quad \Sigma x _ { B } ^ { 2 } = 630.1940 .$$
The variance of the length of the rods produced by machine $A$ is known to be $0.0490 \mathrm {~cm} ^ { 2 }$. The mean lengths of the rods produced by the machines are denoted by $\mu _ { A } \mathrm {~cm}$ and $\mu _ { B } \mathrm {~cm}$ respectively.\\
(i) Test, at the $5 \%$ significance level, the hypothesis $\mu _ { B } > \mu _ { A }$.\\
(ii) Find the set of values of $a$ for which the null hypothesis $\mu _ { B } - \mu _ { A } = 0.025$ would not be rejected in favour of the alternative hypothesis $\mu _ { B } - \mu _ { A } > 0.025$ at the $a \%$ significance level.\\
(iii) For the test in part (i) to be valid,
\begin{enumerate}[label=(\alph*)]
\item state whether it is necessary to assume that the two population variances are equal,
\item state, giving a reason, whether it is necessary to assume that the lengths of rods are normally distributed.
\end{enumerate}
\hfill \mbox{\textit{OCR S3 2006 Q8 [14]}}