Question

Cottonmouth Litter Size. Refer to Problem 8. Find a $98\%$confidence interval for the difference between the mean litter sizes of cottonmouths in Florida and Virginia. Interpret your result.

Short answer

Yes

Step-by-step solution

Given Information

To calculate at$98\%$significance level, the difference between the mean litter sizes of cottonmouths in Florida and 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:

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

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

After this, calculate the degrees of freedom

$$\begin{gathered} \Delta =\frac{{\left(\frac{s_1^2}{n_1}+\frac{{\frac{2}{2}}_2^{n_2}}{}\right)}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{z_2^2}{n_2}\right)}^2}{n_2-1}} \\ =\frac{{\left(\frac{(1.59{)}^2}{24}+\frac{(2.68{)}^2}{44}\right)}^2}{\frac{{\left(\frac{(1.59{)}^2}{24}\right)}^2}{24-1}+\frac{{\left(\frac{(2.68{)}^2}{44}\right)}^2}{44-1}} \\ \approx 65 \end{gathered}$$

Now, determine $t_{\frac{a}{2}}$ with$df=65$

$t_{0.01}=2.385$

The confidence interval's endpoints for ${\mu }_1-{\mu }_2$ can be calculated as:

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)-t_{\frac{a}{2}}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\ =(5.46-7.59)-2.385\times \sqrt{\frac{(1.59{)}^2}{24}+\frac{(2.68{)}^2}{44}} \\ =-3.366 \end{gathered}$$

$$\begin{gathered} \left({\overline{x}}_1-{\overline{x}}_2\right)+t_{\frac{a}{2}}\times \sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}} \\ =(5.46-7.59)+2.385\times \sqrt{\frac{(1.59{)}^2}{24}+\frac{(2.68{)}^2}{44}} \\ =-0.894 \end{gathered}$$

It can be said with$98\%$confidence that the no: of a young per litter of cottonmouths in Florida is between $0.894$and $3.366$less than in Virginia.