Chapter 12, Problem AP4.41CPT

To determine

To find out is there convincing evidence at the $\alpha =0.05$ level that the mean electrical potential differs for normal mice and mice with diabetes.

Answer to Problem AP4.41CPT

There is convincing evidence that at the $\alpha =0.05$ level that the mean electrical potential differs for normal mice and mice with diabetes.

Explanation of Solution

It is given that:

  $\begin{array}{l}{\overline{x}}_1=13090\\ {\overline{x}}_2=10022\\ s_1=4839\\ s_2=2915\\ n_1=24\\ n_2=18\\ \alpha =0.05\end{array}$

The given claim is the difference in mean. Then let us define the null and alternative hypothesis as:

  $\begin{array}{l}{H}_0:{\mu }_1={\mu }_2\\ H_1:{\mu }_1\ne {\mu }_2\end{array}$

Where,

  ${\mu }_1=$ true mean electrical potential for normal mice.

  ${\mu }_2=$ true mean electrical potential for mice with diabetes.

Now, let us assume that all the conditions are met for the test. Thus, the value of test statistics is as:

  $\begin{array}{l}t=\frac{{\overline{x}}_1-{\overline{x}}_2-({\mu }_1-{\mu }_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ =\frac{13090-10022-0}{\sqrt{\frac{{4839}^2}{24}+\frac{{2915}^2}{18}}}\\ =2.550\end{array}$

The degrees of freedom is as:

  $\begin{array}{l}df=\min (n_1-1,n_2-1)\\ =\min (24-1,18-1)\\ =17\end{array}$

Now, we will calculate the P-value by the degrees of freedom as:

  $0.02<P<0.04$

As we know that if the P-value isless than or equal to significance level then the null hypothesis is rejected. Thus, we have,

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

Thus, there is convincing evidence that at the $\alpha =0.05$ level that the mean electrical potential differs for normal mice and mice with diabetes.