Question
A random variable $X$ has a distribution with mean $\mu$ and variance 4. A random sample of size 100 is to be taken from the distribution of $X$.
Josie takes a different random sample of size 100 to test the null hypothesis that $\mu=60$ against the alternative hypothesis that $\mu>60$ at the $5 \%$ level.
a. State the central limit theorem as applied to a random sample of size $n$, taken from a distribution with mean $\mu$ and variance $\sigma^2$.
b. Jack takes a random sample of size 100 and calculates that $\bar{x}=60.2$. Find an approximate $90 \%$ confidence interval for $\mu$.
c.i. Find the critical region for Josie’s test, giving your answer correct to two decimal places.
c.ii.Write down the probability that Josie makes a Type I error.
c.iiiGiven that the probability that Josie makes a Type II error is 0.25 , find the value of $\mu$, giving your answer correct to three significant figures.
▶️Answer/Explanation
a. for $n$ (sufficiently) large the sample mean $\bar{X}$ approximately
A1
$\sim \mathrm{N}\left(\mu, \frac{\sigma^2}{n}\right)$
A1
Note: Award the first $\boldsymbol{A 1}$ for $n$ large and reference to the sample mean $(\bar{X})$, the second $\boldsymbol{A 1}$ is for normal and the two parameters.
Note: Award the second $\boldsymbol{A 1}$ only if the first $\boldsymbol{A 1}$ is awarded.
Note: Allow ‘ $n$ tends to infinity’ or ‘ $n \geq 30$ ‘ in place of ‘large’.
[2 marks]
b. $[59.9,60.5]$
A1A1
Note: Accept answers which round to the correct 3sf answers.
[2 marks]
c.i. under $H_0, \bar{X} \sim \mathrm{N}\left(60, \frac{4}{100}\right)$
(A1)
required to find $k$ such that $P(\bar{X}>k)=0.05$
(M1)
use of any valid method, eg GDC Inv(Normal) or $k=60+z \frac{\sigma}{\sqrt{n}}$
(M1)
hence critical region is $\bar{x}=60.33$
A1
[4 marks]
c.ii. 0.05
A1
[1 mark]
c.iiiP(Type II error) $=\mathrm{P}\left(H_0\right.$ is accepted $/ H_0$ is false $)$
(R1)
Note: Accept Type II error means $H_0$ is accepted given $H_0$ is false.
$$
\begin{aligned}
& \Rightarrow \mathrm{P}(\bar{X}<60.33)=0.25 \text { when } \bar{X} \sim \mathrm{N}\left(\mu, \frac{4}{100}\right) \\
& \Rightarrow \mathrm{P}\left(\frac{\bar{X}-\mu}{\frac{2}{10}}<\frac{60.33-\mu}{\frac{2}{10}}\right)=0.25 \quad \text { (M1) } \\
& \Rightarrow \mathrm{P}\left(Z<\frac{60.33-\mu}{\frac{2}{10}}\right)=0.25 \text { where } Z \sim \mathrm{N}\left(0,1^2\right) \\
& \frac{60.33-\mu}{\frac{2}{10}}=-0.6744 \ldots \\
& \mu=60.33+\frac{2}{10} \times 0.6744 \ldots \\
& \mu=60.5 \\
& {[5 \text { marks }]}
\end{aligned}
$$
(M1)
A1
[5 marks]
Question
In a large population of hens, the weight of a hen is normally distributed with mean $\mu \mathrm{kg}$ and standard deviation $\sigma \mathrm{kg}$. 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 $x \mathrm{~kg}$ 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 $\mu>1.95$
a. State the distribution of $\bar{X}$ giving its mean and variance.
b. Find an unbiased estimate for $\mu$.
c. Find an unbiased estimate for $\sigma^2$.
d. Find a $90 \%$ confidence interval for $\mu$.
e.i. Find the $p$-value for the test.
e.ii.Write down the conclusion reached.
▶️Answer/Explanation
a.$\bar{X} \sim N\left(\mu, \frac{\sigma^2}{100}\right)$
A1
Note: Accept $n$ in place of 100 .
[1 mark]
b. $\hat{\mu}=\frac{\sum x}{n}=\frac{199.8}{100}=1.998 \quad$ A1
Note: Accept 2.00, 2.0 and 2.
[1 mark]
c. $s_{n-1}^2=\frac{n}{n-1}\left(\frac{\sum x^2}{n}-\bar{x}^2\right)=\frac{100}{99}\left(\frac{407.8}{100}-1.998^2\right)$
(M1)
$$
=0.086864
$$
unbiased estimate for $\sigma^2$ is 0.0869
A1
Note: Accept any answer which rounds to $\mathbf{0 . 0 8 7}$.
[2 marks]
d. $90 \%$ confidence interval is $1.998 \pm 1.660 \sqrt{\frac{0.0869}{100}} \quad$ (M1) $=(1.95,2.05) \quad$ A1A1
Note: $\boldsymbol{F T}$ 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]
e.i. $p$-value is $0.0377 \quad$ A2
Note: Award $\mathbf{A 1}$ for the 2-tail value 0.0754 .
Note: Award A2 for 0.0377 and $\mathbf{A 1}$ for any other value that rounds to 0.038 .
Note: $\boldsymbol{F T}$ their estimated mean from (b), note that 2 gives $p=0.032(0)$.
[2 marks]
e.ii.accept the null hypothesis
A1
Note: $\boldsymbol{F T}$ their $p$-value.
[1 mark]
Question
John rings a church bell 120 times. The time interval, $T_i$, between two successive rings is a random variable with mean of 2 seconds and variance of $\frac{1}{9}$ seconds $^2$.
Each time interval, $T_i$, is independent of the other time intervals. Let $X=\sum_{i=1}^{119} T_i$ be the total time between the first ring and the last ring.
The church vicar subsequently becomes suspicious that John has stopped coming to ring the bell and that he is letting his friend Ray do it. When Ra) rings the bell the time interval, $T_i$ has a mean of 2 seconds and variance of $\frac{1}{25}$ seconds $^2$.
The church vicar makes the following hypotheses:
$H_0$ : Ray is ringing the bell; $H_1$ : John is ringing the bell.
He records four values of $X$. He decides on the following decision rule:
If $236 \leqslant X \leqslant 240$ for all four values of $X$ he accepts $H_0$, otherwise he accepts $H_1$.
a. Find
(i) $\mathrm{E}(X)$;
(ii) $\operatorname{Var}(X)$.
b. Explain why a normal distribution can be used to give an approximate model for $X$.
c. Use this model to find the values of $A$ and $B$ such that $\mathrm{P}(A<X<B)=0.9$, where $A$ and $B$ are symmetrical about the mean of $X$.
d. Calculate the probability that he makes a Type II error.
▶️Answer/Explanation
a.
(i) mean $=119 \times 2=238$
A1
(ii) $\quad$ variance $=119 \times \frac{1}{9}=\frac{119}{9}(=13.2)$
(M1)A1
Note: If 120 is used instead of 119 award $\mathbf{A O}($ M1)AO for part (a) and apply follow through for parts (b)-(d). (b) is unaffected and in (c) the interval becomes $(234,246)$. In (d) the first 2 A1 marks are for $0.3633 \ldots$ and $0.0174 \ldots$ so the final answer will round to 0.017 .
[3 marks]
b. justified by the Central Limit Theorem $\quad \boldsymbol{R} \boldsymbol{1}$
since $n$ is large
A1
Note: Accept $n>30$.
[2 marks]
c. $X \sim N\left(238, \frac{119}{9}\right)$
$Z=\frac{X-238}{\frac{\sqrt{119}}{3}} \sim N(0,1) \quad$ (M1)(A1)
$\mathrm{P}(Z<q)=0.95 \Rightarrow q=1.644 \ldots \quad$ (A1)
so $\mathrm{P}(-1.644 \ldots<Z<1.644 \ldots)=0.9$
$\mathrm{P}\left(-1.644 \ldots<\frac{X-238}{\frac{\sqrt{119}}{3}}<1.644 \ldots\right)=0.9$
interval is $232<X<244$ (3sf) $(A=232, B=244)$
A1A1
Notes: Accept the use of inverse normal applied to the distribution of $X$.
Alternative is to use the GDC to find a pretend $Z$ confidence interval for a mean and then convert by multiplying by 119 .
Either $A$ or $B$ correct implies the five implied marks.
Accept any numbers that round to these 3sf numbers.
d.
$$
\begin{aligned}
& \text { under } \mathrm{H}_1, X \sim N\left(238, \frac{119}{9}\right) \quad \text { (M1) } \\
& \mathrm{P}(236 \leqslant X \leqslant 240)=0.41769 \ldots \quad \text { (A1) }
\end{aligned}
$$
probability that all 4 values of $X$ lie in this interval is
$$
(0.41769 \ldots)^4=0.030439 \ldots \quad \text { (M1)(A1) }
$$
so probability of a Type II error is 0.0304 (3sf)
A1
Note: Accept any answer that rounds to 0.030 .