Exam-Style Problem

9231 P43 - Jun 2024 - Q6 - 10 marks

6672

Seva is investigating the lengths of the tails of adult wallabies in two regions of Australia, \(X\) and \(Y\). He chooses a random sample of 50 adult wallabies from region \(X\) and records the lengths, \(x \mathrm{~cm}\), of their tails. He also chooses a random sample of 40 adult wallabies from region \(Y\) and records the lengths, \(y \mathrm{~cm}\), of their tails. His results are summarised as follows.
\(\sum x=1080 \quad \sum x^{2}=23480 \quad \sum y=940 \quad \sum y^{2}=22220\)

It cannot be assumed that the population variances of the two distributions are the same.
(a) Find a \(90 \%\) confidence interval for the difference between the population mean lengths of the tails of adult wallabies in regions \(X\) and \(Y\).

The population mean lengths of the tails of adult wallabies in regions \(X\) and \(Y\) are \(\mu_{X} \mathrm{~cm}\) and \(\mu_{Y} \mathrm{~cm}\) respectively.
(b) Test, at the \(10 \%\) significance level, the null hypothesis \(\mu_{Y}-\mu_{X}=1.1\) against the alternative hypothesis \(\mu_{Y}-\mu_{X}\gt 1.1\). State your conclusion in the context of the question.

Solution

Answer: (a) The sample means are \(\bar x=\frac{1080}{50}=21.6\) and \(\bar y=\frac{940}{40}=23.5\), so the estimated difference is \(\bar y-\bar x=1.9\text{ cm}\).

A \(90\%\) confidence interval for \(\mu_Y-\mu_X\) is

\(1.9 \pm 1.645\sqrt{\frac{3.102}{50}+\frac{3.333}{40}}\)

which gives \((1.27,\,2.53)\).

So the \(90\%\) confidence interval is \(1.27\text{ cm} \lt \mu_Y-\mu_X \lt 2.53\text{ cm}\).

(b) Test statistic:

\(z=\frac{1.9-1.1}{\sqrt{\frac{3.102}{50}+\frac{3.333}{40}}}=2.10\) (approximately).

Since \(2.10\gt 1.282\), reject \(H_0\) at the \(10\%\) level.

There is sufficient evidence to suggest that the mean tail length in region \(Y\) is more than \(1.1\) cm greater than the mean tail length in region \(X\).

(a) First find the sample means.

For region \(X\): \(n_X=50\), so

\(\bar x=\frac{\sum x}{50}=\frac{1080}{50}=21.6\)

For region \(Y\): \(n_Y=40\), so

\(\bar y=\frac{\sum y}{40}=\frac{940}{40}=23.5\)

Hence

\(\bar y-\bar x=23.5-21.6=1.9\)

Now find the sample variances.

For region \(X\):

\(s_X^2=\frac{1}{49}\left(\sum x^2-\frac{(\sum x)^2}{50}\right)=\frac{1}{49}\left(23480-\frac{1080^2}{50}\right)\)

\(=\frac{1}{49}(23480-23328)=\frac{152}{49}\approx 3.102\)

For region \(Y\):

\(s_Y^2=\frac{1}{39}\left(\sum y^2-\frac{(\sum y)^2}{40}\right)=\frac{1}{39}\left(22220-\frac{940^2}{40}\right)\)

\(=\frac{1}{39}(22220-22090)=\frac{130}{39}\approx 3.333\)

Because the population variances cannot be assumed equal, use

\(\mathrm{Var}(\bar y-\bar x)\approx \frac{s_Y^2}{40}+\frac{s_X^2}{50}\)

\(\frac{3.333}{40}+\frac{3.102}{50}\approx 0.1454\)

and the standard error is

\(\sqrt{0.1454}\approx 0.3813\)

For a \(90\%\) confidence interval, the critical value is \(1.645\).

Therefore the confidence interval for \(\mu_Y-\mu_X\) is

\((\bar y-\bar x)\pm 1.645\times \text{SE}=1.9\pm 1.645(0.3813)\)

\(=1.9\pm 0.627\)

So the interval is

\((1.273,\,2.527)\)

which to 3 significant figures is \((1.27,\,2.53)\).

(b) We test

\(H_0: \mu_Y-\mu_X=1.1\)

against

\(H_1: \mu_Y-\mu_X\gt 1.1\)

The observed value of \(\bar y-\bar x\) is \(1.9\), and from part (a) the standard error is

\(\sqrt{\frac{3.102}{50}+\frac{3.333}{40}}\approx 0.3813\)

So the test statistic is

\(z=\frac{1.9-1.1}{0.3813}\approx 2.10\)

For a one-tailed test at the \(10\%\) significance level, the critical value is \(1.282\).

Since

\(2.10\gt 1.282\)

the test statistic lies in the critical region, so we reject \(H_0\).

Conclusion: there is sufficient evidence, at the \(10\%\) level, to suggest that the mean tail length in region \(Y\) is more than \(1.1\) cm longer than the mean tail length in region \(X\).