Question 3 - A-Level Maths

OCR MEI S3 2006 June — Question 3 18 marks

Exam Board	OCR MEI
Module	S3 (Statistics 3)
Year	2006
Session	June
Marks	18
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Type I/II errors and power of test
Type	Carry out hypothesis test
Difficulty	Moderate -0.3 This is a standard S3 question testing routine sampling methods and a one-sample t-test. Part (v) requires straightforward application of the t-test formula and confidence interval with given summary statistics—no conceptual challenges or novel problem-solving. The sampling discussion parts are bookwork recall. Slightly easier than average due to being mostly procedural with clear signposting.
Spec	2.01c Sampling techniques: simple random, opportunity, etc 2.01d Select/critique sampling: in context 5.05c Hypothesis test: normal distribution for population mean 5.05d Confidence intervals: using normal distribution

3 An employer has commissioned an opinion polling organisation to undertake a survey of the attitudes of staff to proposed changes in the pension scheme. The staff are categorised as management, professional and administrative, and it is thought that there might be considerable differences of opinion between the categories. There are 60,140 and 300 staff respectively in the categories. The budget for the survey allows for a sample of 40 members of staff to be selected for in-depth interviews.

Explain why it would be unwise to select a simple random sample from all the staff.
Discuss whether it would be sensible to consider systematic sampling.
What are the advantages of stratified sampling in this situation?
State the sample sizes in each category if stratified sampling with as nearly as possible proportional allocation is used. The opinion polling organisation needs to estimate the average wealth of staff in the categories, in terms of property, savings, investments and so on. In a random sample of 11 professional staff, the sample mean is \(\pounds 345818\) and the sample standard deviation is \(\pounds 69241\).
Assuming the underlying population is Normally distributed, test at the \(5 \%\) level of significance the null hypothesis that the population mean is \(\pounds 300000\) against the alternative hypothesis that it is greater than \(\pounds 300000\). Provide also a two-sided \(95 \%\) confidence interval for the population mean.
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Question 3:

Part (i)

Answer	Marks	Guidance
Answer	Mark	Guidance
Simple random sample might not be representative	E1
e.g. it might contain only managers.	E1	Or other sensible comment.

Part (ii)

Answer	Marks	Guidance
Answer	Mark	Guidance
Presumably there is a list of staff, so systematic sampling would be possible.	E1
List is likely to be alphabetical, in which case systematic sampling might not be representative.	E1
But if the list is in categories, systematic sampling could work well.	E1	Or other sensible comments.

Part (iii)

Answer	Marks	Guidance
Answer	Mark	Guidance
Would cover the entire population.	E1
Can get information for each category.	E1

Part (iv)

Answer	Marks	Guidance
Answer	Mark	Guidance
5, 11, 24	B1	(4.8, 11.2, 24)

Part (v)

Answer	Marks	Guidance
Answer	Mark	Guidance
\(\bar{x} = 345818,\ s_{n-1} = 69241\); Underlying Normality; \(H_0: \mu = 300000,\ H_1: \mu > 300000\)		All given in the question.
Test statistic is \(\dfrac{345818 - 300000}{\dfrac{69241}{\sqrt{11}}} = 2.19(47)\)	M1	Allow alternatives: \(300000 + (\text{c's}\ 1.812) \times \frac{69241}{\sqrt{11}}\) (= 337829) for subsequent comparison with 345818. Or \(345818 - (\text{c's}\ 1.812) \times \frac{69241}{\sqrt{11}}\) (= 307988) for comparison with 300000. Use of \(\mu - \bar{d}\) scores M1A0, but ft.
A1	c.a.o. but ft from here in any case if wrong.
Refer to \(t_{10}\). Upper 5% point is 1.812.	M1, A1	No ft from here if wrong.
Significant.	A1	ft only c's test statistic. Special case: (\(t_{11}\) and 1.796) can score 1 of these last 2 marks if either form of conclusion is given.
Evidence that mean wealth is greater than 300000.	A1	ft only c's test statistic.
CI is given by \(345818 \pm 2.228 \times \dfrac{69241}{\sqrt{11}}\)	M1, B1, M1
\(= 345818 \pm 46513.84 = (299304(.2),\ 392331(.8))\)	A1	c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to \(t_{10}\) is OK.

# Question 3:

## Part (i)
| Answer | Mark | Guidance |
|--------|------|----------|
| Simple random sample might not be representative | E1 | |
| e.g. it might contain only managers. | E1 | Or other sensible comment. |

## Part (ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Presumably there is a list of staff, so systematic sampling would be possible. | E1 | |
| List is likely to be alphabetical, in which case systematic sampling might not be representative. | E1 | |
| But if the list is in categories, systematic sampling could work well. | E1 | Or other sensible comments. |

## Part (iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| Would cover the entire population. | E1 | |
| Can get information for each category. | E1 | |

## Part (iv)
| Answer | Mark | Guidance |
|--------|------|----------|
| 5, 11, 24 | B1 | (4.8, 11.2, 24) |

## Part (v)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\bar{x} = 345818,\ s_{n-1} = 69241$; Underlying Normality; $H_0: \mu = 300000,\ H_1: \mu > 300000$ | | All given in the question. |
| Test statistic is $\dfrac{345818 - 300000}{\dfrac{69241}{\sqrt{11}}} = 2.19(47)$ | M1 | Allow alternatives: $300000 + (\text{c's}\ 1.812) \times \frac{69241}{\sqrt{11}}$ (= 337829) for subsequent comparison with 345818. Or $345818 - (\text{c's}\ 1.812) \times \frac{69241}{\sqrt{11}}$ (= 307988) for comparison with 300000. Use of $\mu - \bar{d}$ scores M1A0, but ft. |
| | A1 | c.a.o. but ft from here in any case if wrong. |
| Refer to $t_{10}$. Upper 5% point is 1.812. | M1, A1 | No ft from here if wrong. |
| Significant. | A1 | ft only c's test statistic. Special case: ($t_{11}$ and 1.796) can score 1 of these last 2 marks if either form of conclusion is given. |
| Evidence that mean wealth is greater than 300000. | A1 | ft only c's test statistic. |
| CI is given by $345818 \pm 2.228 \times \dfrac{69241}{\sqrt{11}}$ | M1, B1, M1 | |
| $= 345818 \pm 46513.84 = (299304(.2),\ 392331(.8))$ | A1 | c.a.o. Must be expressed as an interval. ZERO/4 if not same distribution as test. Same wrong distribution scores maximum M1B0M1A0. Recovery to $t_{10}$ is OK. |

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3 An employer has commissioned an opinion polling organisation to undertake a survey of the attitudes of staff to proposed changes in the pension scheme. The staff are categorised as management, professional and administrative, and it is thought that there might be considerable differences of opinion between the categories. There are 60,140 and 300 staff respectively in the categories. The budget for the survey allows for a sample of 40 members of staff to be selected for in-depth interviews.\\
(i) Explain why it would be unwise to select a simple random sample from all the staff.\\
(ii) Discuss whether it would be sensible to consider systematic sampling.\\
(iii) What are the advantages of stratified sampling in this situation?\\
(iv) State the sample sizes in each category if stratified sampling with as nearly as possible proportional allocation is used.

The opinion polling organisation needs to estimate the average wealth of staff in the categories, in terms of property, savings, investments and so on. In a random sample of 11 professional staff, the sample mean is $\pounds 345818$ and the sample standard deviation is $\pounds 69241$.\\
(v) Assuming the underlying population is Normally distributed, test at the $5 \%$ level of significance the null hypothesis that the population mean is $\pounds 300000$ against the alternative hypothesis that it is greater than $\pounds 300000$. Provide also a two-sided $95 \%$ confidence interval for the population mean.\\[0pt]
[10]

\hfill \mbox{\textit{OCR MEI S3 2006 Q3 [18]}}

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