19M.3.AHL.TZ0.HSP_3

pestleMathematicsAIHLPaper 319M· ahl-4-14-linear-transformation-of-a-single-rv-e(x)-and-var(x)-unbiased-estimatorssource ↗

In a large population of hens, the weight of a hen is normally distributed with mean $\mu$ kg and standard deviation $\sigma$ kg. A random sample of 100 hens is taken from the population.

The mean weight for the sample is denoted by $\bar X$ .

The sample values are summarized by $\sum {x = 199.8}$ and $\sum {{x^2} = 407.8}$ where $x$ kg is the weight of a hen.

It is found that $\sigma$ = 0.27 . It is decided to test, at the 1 % level of significance, the null hypothesis $\mu$ = 1.95 against the alternative hypothesis $\mu$ > 1.95.

State the distribution of $\bar X$ giving its mean and variance.

[1]

Find an unbiased estimate for $\mu$ .

[1]

Find an unbiased estimate for ${\sigma ^2}$ .

[2]

Find a 90 % confidence interval for $\mu$ .

[3]

Find the $p$ -value for the test.

[2]

e.i.

Write down the conclusion reached.

[1]

e.ii.

Markscheme / solution

* This question is from an exam for a previous syllabus, and may contain minor differences in marking or structure.

$\bar X \sim N\left( {\mu {\text{,}}\,\,\frac{{{\sigma ^2}}}{{100}}} \right)$ A1

Note: Accept $n$ in place of 100.

[1 mark]

$\hat \mu = \frac{{\sum x }}{n} = \frac{{199.8}}{{100}} = 1.998$ A1

Note: Accept 2.00, 2.0 and 2.

[1 mark]

${s_{n - 1}}^2 = \frac{n}{{n - 1}}\left( {\frac{{\sum {{x^2}} }}{n} - {{\bar x}^2}} \right) = \frac{{100}}{{99}}\left( {\frac{{407.8}}{{100}} - {{1.998}^2}} \right)$ (M1)

= 0.086864

unbiased estimate for ${\sigma ^2}$ is 0.0869 A1

Note: Accept any answer which rounds to 0.087.

[2 marks]

90 % confidence interval is $1.998 \pm 1.660\sqrt {\frac{{0.0869}}{{100}}}$ (M1)

= (1.95, 2.05) A1A1

Note: FT their $\sigma$ from (c).

Note: Condone the use of the $z$ -value 1.645 since $n$ is large.

Note: Accept any values that round to 1.95 and 2.05.

[3 marks]

$p$ -value is 0.0377 A2

Note: Award A1 for the 2-tail value 0.0754.

Note: Award A2 for 0.0377 and A1 for any other value that rounds to 0.038.

Note: FT their estimated mean from (b), note that 2 gives $p$ = 0.032(0).

[2 marks]

e.i.

accept the null hypothesis A1

Note: FT their $p$ -value.

[1 mark]

e.ii.

Examiners’ report

[N/A]

e.i.

[N/A]

e.ii.