Question

Cottonmouth Litter Size. In the article "The Eastern Cottonmouth (Agkistrodon piscivorous) at the Northern Edge of its Range" (Journal of Herpetology. Vol. 29, No. 3, pp. 391-398), C. Blem and 1. Blem examined the reproductive characteristics of the eastern cottonmouth. The data in the following table is based on the results of the researchers" study. give the number of young per litter for 24 female cottonmouths in Florida and 44 female cottonmouths in Virginia.

Preliminary data analyses indicate that you can reasonably presume that litter sizes of cottonmouths in both states are approximately normally distributed. At the $1\%$ significance level, do the data provide sufficient evidence to conclude that, on average, the number of a young per litter of cottonmouths in Florida is less than that in Virginia? Do not assume that the population standard deviations are equal. (Note:$x_1=5.46,s_1=1.59,x_2=7.59$, and$\left.s_2=2.68.\right)$

Short answer

Yes

Step-by-step solution

Given Information

To calculate at $1\%$a significance level, if the data provide sufficient evidence to conclude that the number of young cottonmouths in Florida is less than in Virginia.

Explanation

As the given data is normally distributed and the given population standard deviations are not equal, the unpooled t-test can be used as follows:

$H_0:{\mu }_1={\mu }_2$

$H_{\alpha }:{\mu }_1<{\mu }_2$

Now, determine the test statistic:

$$\begin{gathered} \mathrm{t}=\frac{\frac{-1}{1}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}} \\ =\frac{5.46-7.59}{\sqrt{\frac{(1.59{)}^2}{24}}+\frac{(2.65{)}^2}{44}} \\ \approx -4.11 \end{gathered}$$

After this, calculate the degrees of freedom

$$\begin{gathered} \Delta =\frac{{\left(\frac{s_1^2}{n_1}+\frac{2^2}{n_2}\right)}^2}{\frac{{\left(\frac{J_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{z_2^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{{\left(\frac{(1.59{)}^2}{24}+\frac{(2.0.8{)}^2}{44}\right)}^2}{\frac{{\left(\frac{(1.59{)}^2}{24}\right)}^2}{24-1}+\frac{{\left(\frac{(2.65{)}^2}{44}\right)}^2}{44-1}} \\ \approx 65 \end{gathered}$$

The corresponding P-value can be found in the table

$P<0.005$

As in the given case, the P-value is smaller than the significance level, therefore the null hypothesis can be rejected:

$$\begin{gathered} P<0.01=1\% \\ \Rightarrow \text{Reject}{H}_0 \end{gathered}$$