Chapter 10.2, Problem 60E

To determine

To explain do these data provide convincing evidence at the $\alpha =0.01$ significance level of a difference in the mean weights of all grey and black Eastern Gray squirrels in the forest.

Answer to Problem 60E

There is convincing evidence of a difference in the mean weights of all grey and black Eastern Gray squirrels in the forest.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=20.3\\ {\overline{x}}_2=19.2\\ n_1=40\\ n_2=40\\ s_1=2.1\\ s_2=1.9\end{array}$

  $\alpha =0.01$

The given claim that: difference in the means.

Now, we have to find out the appropriate hypotheses for performing a significance test.

Thus, the claim is either the null hypothesis or the alternative hypothesis. The null hypothesis states that the population proportions are equal. If the null hypothesis is the claim then the alternative hypothesis states the opposite of the null hypothesis.

Therefore, the appropriate hypotheses for this is:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_a:{\mu }_1\ne {\mu }_2\end{array}$

Where we have,

  ${\mu }_1=$ the true mean weights of all grey Eastern Gray squirrels in the forest.

  ${\mu }_2=$ the true mean weights of all black Eastern Gray squirrels in the forest.

Now, find the test statistics:

  $\begin{array}{l}t=\frac{({\overline{x}}_1-{\overline{x}}_2)-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{20.3-19.2-0}{\sqrt{\frac{{2.1}^2}{40}+\frac{{1.9}^2}{40}}}\\ =2.457\end{array}$

Now, the degree of freedom will be:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (56-1,56-1)\\ =55\end{array}$

Since student’s T distribution table in the appendix does not contains the value of $df=39$ so we will take the nearest value $df=30$ . So the P -value will be:

  $P=2(0.01)=0.02$

On the other hand by using the calculator command: $2\times \text{tcdf(2}\text{.457,1E99,39)}$ which results in the P -values as $0.01856$

  $0.80506$ .

And we know that if the P -value is less than or equal to the significance level then the null hypothesis is rejected, then,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude that there is convincing evidence of a difference in the mean weights of all grey and black Eastern Gray squirrels in the forest.