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 $H_0, \bar{X} \sim \mathrm{N}\left(60, \frac{4}{100}\right)$
required to find $k$ such that $P(\bar{X}>k)=0.05$
(M1)
use of any valid method, eg GDC $\operatorname{lnv}\left(\right.$ 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.
$$
\begin{aligned}
& \Rightarrow \mathrm{P}(\bar{X}<60.33)=0.25 \text { when } \bar{X} \sim \mathrm{N}\left(\mu, \frac{4}{100}\right) \\
& \Rightarrow \mathrm{P}\left(\frac{\bar{X}-\mu}{\frac{2}{10}}<\frac{60.33-\mu}{\frac{2}{10}}\right)=0.25 \quad \text { (M1) } \\
& \Rightarrow \mathrm{P}\left(Z<\frac{60.33-\mu}{\frac{2}{10}}\right)=0.25 \text { where } Z \sim \mathrm{N}\left(0,1^2\right) \\
& \frac{60.33-\mu}{\frac{2}{10}}=-0.6744 \ldots \\
& \mu=60.33+\frac{2}{10} \times 0.6744 \ldots \\
& \mu=60.5
\end{aligned}
$$
[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.$\bar{X} \sim N\left(\mu, \frac{\sigma^2}{100}\right) \quad \boldsymbol{A 1}$
Note: Accept $n$ in place of 100 .
[1 mark]
b. $\hat{\mu}=\frac{\sum x}{n}=\frac{199.8}{100}=1.998$
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 0.087 .
[2 marks] $=(1.95,2.05) \quad$ 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
A2
Note: Award A1 for the 2-tail value 0.0754.
Note: Award $\mathbf{A 2}$ for 0.0377 and $\boldsymbol{A 1}$ for any other value that rounds to 0.038 .
Note: $F T$ their estimated mean from (b), note that 2 gives $p=0.032(0)$.
[2 marks]
e.ii.accept the null hypothesis
A1
Note: $F T$ their $p$-value.
[1 mark]
Question
The times $t$, in minutes, taken by a random sample of 75 workers of a company to travel to work can be summarized as follows
$$
\sum t=2165, \sum t^2=76475 .
$$
Let $T$ be the random variable that represents the time taken to travel to work by a worker of this company.
a.i. Find unbiased estimates of the mean of $T$.
a.ii.Find unbiased estimates of the variance of $T$.
b. Assuming that $T$ is normally distributed, find
(i) the $90 \%$ confidence interval for the mean time taken to travel to work by the workers of this company,
(ii) the $95 \%$ confidence interval for the mean time taken to travel to work by the workers of this company.
c. Before seeing these results the managing director believed that the mean time was 26 minutes.
Explain whether your answers to part (b) support her belief.
▶️Answer/Explanation
a.i.
$$
\bar{t}=\frac{\sum_{i=1}^{75} t_i}{75}=28.886 \ldots=28.9
$$
A1
[1 mark]
a.ii. $s_{n-1}{ }^2=\frac{75}{74}\left(\frac{\sum_{i=1}^{75} t_i^2}{75}-\bar{t}^2\right)=188.9009 \ldots=189 \quad$ (M1)A1
Note: Accept all answers that round to 28.9 and 189.
Note: Award MO if division by 75.
[2 marks]
b. attempting to find a confidence interval. (M1)
(i) 90\% interval: $(26.2,31.5) \quad \mathbf{A 1}$
(ii) 95\% interval: $(25.7,32.0) \quad \boldsymbol{A 1}$
Note: Accept any values which round to within 0.1 of the correct value.
Note: Award M1A1AO if only confidence limits are given in the form $28.9 \pm 2.6$.
[3 marks]
c. 26 lies within the $95 \%$ interval but not within the $90 \%$ interval
R1
Note: Award $\mathbf{R 1}$ for considering whether or not one or two of the intervals contain 26.
the belief is supported at the $5 \%$ level (accept 95\%)
A1
the belief is not supported at the $10 \%$ level (accept $90 \%$ )
A1
Note: $F T$ their intervals but award R1A1AO if both intervals give the same conclusion.
[3 marks]
Question
A smartphone’s battery life is defined as the number of hours a fully charged battery can be used before the smartphone stops working. A company claims that the battery life of a model of smartphone is, on average, 9.5 hours. To test this claim, an experiment is conducted on a random sample of 20 smartphones of this model. For each smartphone, the battery life, $b$ hours, is measured and the sample mean, $\bar{b}$, calculated. It can be assumed the battery lives are normally distributed with standard deviation 0.4 hours.
It is then found that this model of smartphone has an average battery life of 9.8 hours.
a. State suitable hypotheses for a two-tailed test.
b. Find the critical region for testing $\bar{b}$ at the $5 \%$ significance level.
c. Find the probability of making a Type II error.
d. Another model of smartphone whose battery life may be assumed to be normally distributed with mean $\mu$ hours and standard deviation 1.2 hours is tested. A researcher measures the battery life of six of these smartphones and calculates a confidence interval of $[10.2,11.4]$ for $\mu$.
Calculate the confidence level of this interval.
▶️Answer/Explanation
a.Note: In question 3 , accept answers that round correctly to 2 significant figures.
$\mathrm{H}_0: \mu=9.5 ; \mathrm{H}_1: \mu \neq 9.5$
A1
[1 mark]
b. Note: In question 3, accept answers that round correctly to 2 significant figures. $\begin{array}{lll}\text { the critical values are } 9.5 \pm 1.95996 \ldots \times \frac{0.4}{\sqrt{20}} & \text { (M1)(A1) }\end{array}$ i.e. $9.3247 \ldots, 9.6753 \ldots$
the critical region is $\bar{b}<9.32, \bar{b}>9.68$
A1A1
Note: Award $\boldsymbol{A 1}$ for correct inequalities, $\boldsymbol{A 1}$ for correct values.
Note: Award MO if $t$-distribution used, note that $t(19)_{97.5}=2.093 \ldots$
[4 marks]
c. Note: In question 3, accept answers that round correctly to 2 significant figures.
$$
\begin{aligned}
& \bar{B} \sim \mathrm{N}\left(9.8,\left(\frac{0.4}{\sqrt{20}}\right)^2\right) \quad \text { (A1) } \\
& \mathrm{P}(9.3247 \ldots<\bar{B}<9.6753 \ldots) \quad \text { (M1) } \\
& =0.0816 \quad \text { A1 }
\end{aligned}
$$
(M1)
A1
Note: FT the critical values from (b). Note that critical values of 9.32 and 9.68 give 0.0899 .
[3 marks]
d. Note: In question 3, accept answers that round correctly to 2 significant figures.
METHOD 1
$$
\begin{aligned}
& X \sim \mathrm{N}\left(10.8, \frac{1.2^2}{6}\right) \quad \text { (M1)(A1) } \\
& \mathrm{P}(10.2<X<11.4)=0.7793 \ldots
\end{aligned}
$$
confidence level is $77.9 \%$
A1
Note: Accept 78\%.
METHOD 2
$11.4-10.2=2 z \times \frac{1.2}{\sqrt{6}}$
(M1)
$z=1.224 \ldots \quad$ (A1)
$\mathrm{P}(-1.224 \ldots<Z<1.224 \ldots)=0.7793 \ldots$
confidence level is $77.9 \%$
A1
Note: Accept $78 \%$.