Question

Gray squirrel In many parts of the northern United States, two color variants of the Eastern Gray Squirrel— gray and black—are found in the same habitats. A scientist studying squirrels in a large forest wonders if there is a difference in the sizes of the two color variants. He collects random samples of $40$squirrels of each color from a large forest and weighs them. The $40$black squirrels have a mean weight of $20.3$ounces and a standard deviation of $2.1$ounces. The 40 gray squirrels have a mean weight of $19.2$ounces and a standard deviation of 1.9 ounces. Do these data provide convincing evidence at the α=0.01$\frac{305}{1526}=0.200=20.0\%\alpha =0.01$significance level of a difference in the mean weights of all gray and black Eastern Gray Squirrels in this forest?

Short answer

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

Step-by-step solution

Given information

$$\begin{gathered} {\overline{x}}_1=20.3 \\ {\overline{x}}_2=19.2 \\ {n}_1=40 \\ {n}_2=40 \\ {s}_1=2.1 \\ {s}_2=1.9 \\ \alpha =0.01 \end{gathered}$$

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

Explanation

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:

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

Where we have,

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

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

Locate the following test statistics:

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

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: $2\times tcdf(2.457,1\mathrm{E}99,39)$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,

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

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