2022 H2 Mathematics Paper 2 Question 11

Hypothesis Testing
The Normal Distribution

Answers

(a)

0.391.{0.391.}

(b)

x‾=79.21.{\overline{x} = 79.21.}
s2=14.7{s^2 = 14.7}

(c)

p-value=0.130>0.05.{p\textrm{-value} = 0.130 > 0.05.}
There is insufficient evidence at the 5%{5\%} level of significance to conclude whether Zhou's times have reduced

(d)

Tan should use a 2-tail test since he is interested in whether his mean time has changed, so the new mean time could be greater than or less than his original.

(e)

He assumes that the times he takes for the swims are still normally distributed after his diet, and each time is independent from other times.
He assumes that the standard deviation of his times is still the same as before at 3 seconds.

Full solutions

(a)

Let H{H} and T{T} represent the times, in seconds, taken to swim 100 metres by Zhou and Tan respectively.
H∼N(80,4)T∼N(79,9)H−T∼N(1,13)\begin{gather*} H \sim N( 80,4 )\\T \sim N( 79,9 )\\H - T \sim N( 1,13 ) \end{gather*}
P(Zhou wins)=P(H<T)=P(H−T<0)=0.391 (3 sf) ■\begin{align*} &\mathrm{P}(\textrm{Zhou wins}) \\ &=\mathrm{P}(H < T) \\ &=\mathrm{P}(H - T < 0) \\ &= 0.391 \textrm{ (3 sf) } \blacksquare \end{align*}

(b)

Unbiased estimate of population mean=x‾=∑xn=79.21  ■\begin{align*} &\textrm{Unbiased estimate of population mean} \\ &= \overline{x} \\ &= \frac{\sum x}{n} \\ &= 79.21 \; \blacksquare \end{align*}
Unbiased estimate of population variance=s2=1n−1(∑x2−(∑x)2n)=14.7 (3 sf) ■\begin{align*} &\textrm{Unbiased estimate of population variance} \\ &= s^2 \\ &= \frac{1}{n-1} \left( \sum x^2 - \frac{\left( \sum x \right)^2}{n} \right) \\ &= 14.7 \textrm{ (3 sf) } \blacksquare \end{align*}

(c)

Let μ{\mu} denote the population mean of X,{X,} the time Zhou takes to swim 100 metres after a year. Let H0{\mathrm{H}_0} and H1{\mathrm{H}_1} be the null and alternative hypothesis respectively.

H0:μ=80{\textrm{H}_0: \mu = 80}
H1:μ<80{\textrm{H}_1: \mu < 80}

Under H0,{\textrm{H}_0, } test statistic
Z=X‾−μsn∼N(0,1)Z= \frac{\overline{X} - \mu}{\frac{s}{\sqrt{n}}} \sim N(0,1)
approximately by CLT since n=30{n=30} is large
p-value=0.12973{p\textrm{-value} = 0.12973}
Since p-value>0.05,{p\textrm{-value} > 0.05, } we do not reject H0{\textrm{H}_0}
There is insufficient evidence at the 5%{5\%} level of significance to conclude whether Zhou's times have reduced ■{\blacksquare}

(d)

Tan should use a 2-tail test since he is interested in whether his mean time has changed, so the new mean time could be greater than or less than his original.   ■{\; \blacksquare}

(e)

He assumes that the times he takes for the swims are still normally distributed after his diet, and each time is independent from other times. ■{\blacksquare}

He assumes that the standard deviation of his times is still the same as before at 3 seconds. ■{\blacksquare}

Question Commentary

This question combines the normal distribution and hypothesis testing. For the most part, it is relatively straightforward and very similar to questions in the past.