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The body’s natural electrical field helps wounds heal. If diabetes changes this field, it might explain why people with diabetes heal more slowly. A study of this idea compared randomly selected normal mice and randomly selected mice bred to spontaneously develop diabetes. The investigators attached sensors to the right hip and front feet of the mice and measured the difference in electrical potential (in millivolts) between these locations. Graphs of the data for each group reveal no outliers or strong skewness. The computer output below provides numerical summaries of the $dat{a}^{26}$.

The researchers want to know if there is evidence of a significant difference in mean electrical potentials between normal mice and mice with diabetes. Carry out a test using a $5\%$level of significance and report your conclusion.

Step by step solution

01

Given Information

We need to find if there is evidence of a significant difference in mean electrical potentials between normal mice and mice with diabetes or not.

02

Simplify

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

Determine the test statistic:

$t=\frac{{\stackrel{-}{x}}_1−{\stackrel{-}{x}}_2}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}=\frac{13.090−10.022}{\sqrt{\frac{{4.839}^2}{24}+\frac{{2.915}^2}{18}}}≈2.550$

Determine the degrees of freedom:

$\text{Δ}=\frac{{\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)}^2}{\frac{{\left(s_1^2/n_1\right)}^2}{n_1−1}+\frac{{\left(s_2^2/n_2\right)}^2}{n_2−1}}=\frac{{\left(\frac{{4.839}^2}{24}+\frac{{2.915}^2}{18}\right)}^2}{\frac{{\left({4.839}^2/24\right)}^2}{24−1}+\frac{{\left({2.915}^2/18\right)}^2}{18−1}}≈38.45983$

Determine the corresponding P-value from table B:

$0.01=2×0.005<P<2×0.01=0.02$

If the P-value is less than or equal to the significance level, then the null hypothesis is rejected:

$P<0.05=5\%⇒Reject{H}_0$

Here is significant to support the claim of a difference.

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