91Ó°ÊÓ

$$\begin{gathered} {\stackrel{¯}{x}}_1=20.3 \\ {\stackrel{¯}{x}}_2=19.2 \\ {n}_1=40 \\ {n}_2=40 \\ {s}_1=2.1 \\ {s}_2=1.9 \\ \mathrm{Î\pm }=0.01 \end{gathered}$$

The given claim is that there is a disparity in the means.

Now we must determine the most appropriate hypotheses for a significance test.

As a result, either the null hypothesis or the alternative hypothesis is the claim. According to the null hypothesis, the population proportions are equal. If the claim is the null hypothesis, the alternative hypothesis is the polar opposite of the null hypothesis.

The appropriate hypotheses for this are:

$$\begin{gathered} H_0:{\mathrm{μ}}_1={\mathrm{μ}}_2 \\ {H}_a:{\mathrm{μ}}_{1}notequalto{\mathrm{μ}}_2 \end{gathered}$$

Where we have,

${\mathrm{μ}}_1=$the true mean weights of the forest's grey Eastern Gray squirrels.

${\mathrm{μ}}_2=$the forest's true mean weights of all black Eastern Gray squirrels

Locate the following test statistics:

$$\begin{gathered} t=\frac{({\stackrel{¯}{x}}_1-{\stackrel{¯}{x}}_2)-({\mathrm{μ}}_1-{\mathrm{μ}}_2)}{\sqrt{{\displaystyle \frac{{s_1}^2}{n_1}}+{\displaystyle \frac{{s_2}^2}{n_2}}}} \\ =\frac{20.3-19.2-0}{\sqrt{{\displaystyle \frac{2.1^2}{40}}+{\displaystyle \frac{1.9^2}{40}}}} \\ =2.457 \end{gathered}$$

The degree of liberty will now be:

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

Since the student's T distribution table in the appendix does not contain 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: $2×tcdf(2.457,1\mathrm{E}99,39)$which results in the $P$-values as $0.80506$
And we know that if the $P$-value is less than or equal to the significance level then the null hypothesis is rejected, then,

$P<0.05⇒\text{Reject}{H}_0$

Therefore, We conclude that there is compelling evidence of a weight difference between all grey and black Eastern Gray squirrels in the forest.