Exam-Style Problem

9231 P41 - Jun 2021 - Q4 - 8 marks

6794

A scientist is investigating the lengths of the leaves of birch trees in different regions. He takes a random sample of 50 leaves from birch trees in region \(A\) and a random sample of 60 leaves from birch trees in region \(B\). He records their lengths in \(\mathrm{cm}, x\) and \(y\), respectively. His results are summarised as follows.
\(\sum x=282 \quad \sum x^{2}=1596 \quad \sum y=328 \quad \sum y^{2}=1808\)

The population mean lengths of leaves from birch trees in regions \(A\) and \(B\) are \(\mu_{A} \mathrm{~cm}\) and \(\mu_{B} \mathrm{~cm}\) respectively.

Carry out a test at the \(5 \%\) significance level to test the null hypothesis \(\mu_{A}=\mu_{B}\) against the alternative hypothesis \(\mu_{A} \neq \mu_{B}\).

Solution

Answer: Let

\(H_0: \mu_A=\mu_B\)

\(H_1: \mu_A\neq\mu_B\)

Sample means:

\(\bar x=\frac{282}{50}=5.64\), \(\bar y=\frac{328}{60}=5.4667\)

Sample variances:

\(s_x^2=\frac{1}{49}\left(1596-\frac{282^2}{50}\right)=0.11265\)

\(s_y^2=\frac{1}{59}\left(1808-\frac{328^2}{60}\right)=0.25311\)

Standard error of \(\bar x-\bar y\):

\(\sqrt{\frac{s_x^2}{50}+\frac{s_y^2}{60}}=\sqrt{0.006471\ldots}=0.08044\)

Test statistic:

\(z=\frac{5.64-5.4667}{0.08044}=2.155\)

For a two-tailed test at the 5% level, the critical value is \(1.96\).

Since \(|z|=2.155\gt1.96\), reject \(H_0\).

There is sufficient evidence at the 5% significance level to conclude that the population mean leaf lengths in regions \(A\) and \(B\) are different.

We test whether the population mean leaf lengths are the same in the two regions.

Take

\(H_0: \mu_A=\mu_B\)

\(H_1: \mu_A\neq\mu_B\)

This is a two-tailed test.

First find the sample means.

For region \(A\), with \(n_A=50\),

\(\bar x=\frac{\sum x}{50}=\frac{282}{50}=5.64\)

For region \(B\), with \(n_B=60\),

\(\bar y=\frac{\sum y}{60}=\frac{328}{60}=5.4667\)

Now find the unbiased sample variances.

For region \(A\),

\(s_x^2=\frac{1}{50-1}\left(\sum x^2-\frac{(\sum x)^2}{50}\right)\)

\(=\frac{1}{49}\left(1596-\frac{282^2}{50}\right)\)

\(=\frac{1}{49}(1596-1590.48)=0.11265\)

For region \(B\),

\(s_y^2=\frac{1}{60-1}\left(\sum y^2-\frac{(\sum y)^2}{60}\right)\)

\(=\frac{1}{59}\left(1808-\frac{328^2}{60}\right)\)

\(=\frac{1}{59}(1808-1793.0667)=0.25311\)

Since both sample sizes are large, we use a large-sample test for the difference of means.

Under \(H_0\), the standard error of \(\bar x-\bar y\) is

\(\sqrt{\frac{s_x^2}{50}+\frac{s_y^2}{60}}\)

\(=\sqrt{\frac{0.11265}{50}+\frac{0.25311}{60}}\)

\(=\sqrt{0.002253+0.0042185}\)

\(=\sqrt{0.0064715}=0.08044\)

So the test statistic is

\(z=\frac{\bar x-\bar y}{\sqrt{\frac{s_x^2}{50}+\frac{s_y^2}{60}}}\)

\(=\frac{5.64-5.4667}{0.08044}\)

\(=2.155\)

At the 5% significance level for a two-tailed test, the critical values are \(\pm 1.96\).

Since \(2.155\gt1.96\), the test statistic lies in the critical region.

Therefore we reject \(H_0\).

There is sufficient evidence to suggest that \(\mu_A\neq\mu_B\). In context, the mean leaf lengths in regions \(A\) and \(B\) are significantly different.