## Question

Eleven students who had under-performed in a philosophy practice examination were given extra tuition before their final examination. The differences between their final examination marks and their practice examination marks were

\[10,{\text{ }} – 1,{\text{ }}6,{\text{ }}7,{\text{ }} – 5,{\text{ }} – 5,{\text{ }}2,{\text{ }} – 3,{\text{ }}8,{\text{ }}9,{\text{ }} – 2.\]

Assume that these differences form a random sample from a normal distribution with mean \(\mu \) and variance \({\sigma ^2}\).

Determine unbiased estimates of \(\mu \) and \({\sigma ^2}\).

(i) State suitable hypotheses to test the claim that extra tuition improves examination marks.

(ii) Calculate the \(p\)-value of the sample.

(iii) Determine whether or not the above claim is supported at the \(5\% \) significance level.

**Answer/Explanation**

## Markscheme

unbiased estimate of \(\mu \) is \(2.36(36 \ldots )\;\;\;(26/11)\) *(M1)A1*

unbiased estimate of \({\sigma ^2}\) is \(33.65(45 \ldots ) = ({5.801^2})\;\;\;(1851/55)\) *(M1)A1*

**Note:** Accept any answer that rounds correctly to \(3\) significant figures.

**Note:** Award ** M1A0** for any unbiased estimate of \({\sigma ^2}\) that rounds to \(5.80\).

**[4 marks]**

(i) \({{\text{H}}_0}:\mu = 0;{\text{ }}{{\text{H}}_1}:\mu > 0\) *A1A1*

**Note:** Award ** A1A0** if an inappropriate symbol is used for the mean,

*eg*, \(r\), \({\rm{\bar d}}\).

(ii) attempt to use *t*-test *(M1)*

\(t = 1.35\) *(A1)*

\({\text{DF}} = 10\) *(A1)*

\(p\)-value \( = 0.103\) *A1*

**Note:** Accept any answer that rounds correctly to \(3\) significant figures.

(iii) \(0.103 > 0.05\) *A1*

there is insufficient evidence at the \(5\% \) level to support the claim (that extra tuition improves examination marks)

**OR**

the claim (that extra tuition improves examination marks) is not supported at the \(5\% \) level (or equivalent statement) *R1*

**Note:** Follow through the candidate’s \(p\)-value.

**Note:** Do not award ** R1** for Accept \({{\text{H}}_0}\) or Reject \({{\text{H}}_1}\).

**[8 marks]**

**Total [12 marks]**

## Examiners report

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the *p*-value. Instead, many candidates found the *p*-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

Almost every candidate gave the correct estimate of the mean but some chose the wrong variance from their calculators to estimate \({\sigma ^2}\). In (b)(i), the hypotheses were sometimes incorrectly written, usually with an incorrect symbol instead of \(\mu \), for example \(d\), \(\bar x\) and ‘mean’ were seen. Many candidates failed to make efficient use of their calculators in (b)(ii). The intention of the question was that candidates should simply input the data into their calculators and use the software to give the \(p\)-value. Instead, many candidates found the \(p\)-value by first evaluating \(t\) using the appropriate formula. This was a time consuming process and it gave opportunity for error. In (b)(iii), candidates were expected to refer to the claim so that the answers ‘Accept \({H_0}\)’ or ‘Reject \({H_1}\)’ were not accepted.

## Question

The random variable *X* has a binomial distribution with parameters \(n\) and \(p\).

Let \(U = nP\left( {1 – P} \right)\).

Show that \(P = \frac{X}{n}\) is an unbiased estimator of \(p\).

Show that \({\text{E}}\left( U \right) = \left( {n – 1} \right)p\left( {1 – p} \right)\).

Hence write down an unbiased estimator of Var(*X*).

**Answer/Explanation**

## Markscheme

\({\text{E}}\left( P \right) = {\text{E}}\left( {\frac{X}{n}} \right) = \frac{1}{n}{\text{E}}\left( X \right)\) **M1**

\( = \frac{1}{n}\left( {np} \right) = p\) **A1**

so *P* is an unbiased estimator of \(p\) **AG**

**[2 marks]**

\({\text{E}}\left( {nP\left( {1 – P} \right)} \right) = {\text{E}}\left( {n\left( {\frac{X}{n}} \right)\left( {1 – \frac{X}{n}} \right)} \right)\)

\( = {\text{E}}\left( X \right) = \frac{1}{n}{\text{E}}\left( {{X^2}} \right)\) **M1A1**

use of \({\text{E}}\left( {{X^2}} \right) = {\text{Var}}\left( X \right) + {\left( {{\text{E}}\left( X \right)} \right)^2}\) **M1**

**Note:** Allow candidates to work with *P* rather than *X* for the above 3 marks.

\( = np – \frac{1}{n}\left( {np\left( {1 – p} \right) + {{\left( {np} \right)}^2}} \right)\) **A1**

\( = np – p\left( {1 – p} \right) – n{p^2}\)

\( = np\left( {1 – p} \right) – p\left( {1 – p} \right)\) **A1**

**Note:** Award * A1* for the factor of \(\left( {1 – p} \right)\).

\( = \left( {n – 1} \right)p\left( {1 – p} \right)\) **AG**

**[5 marks]**

an unbiased estimator is \(\frac{{{n^2}P\left( {1 – P} \right)}}{{n – 1}}\left( { = \frac{{nU}}{{n – 1}}} \right)\) **A1**

**[1 mark]**

## Examiners report

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