Chapter 24, Problem 24.63SE

To determine

To find: The value of x and s for the insulin concentrations in the two types of mice.

To test: Whether there is a significant difference in the mean insulin concentrations in the two types of mice or not.

To check: The P-value confirm the claim in the report that $P<0.005$.

Answer to Problem 24.63SE

The mean for Wild type is 59 and the mean for $aP{2}^{-/-}$ is 0.75.

The value of s for Wild type is 2.85.

The value of s for $aP{2}^{-/-}$ is 0.632.

The conclusion is that there is a significant difference in the mean insulin concentrations in the two types of mice.

Yes, the P-value confirm the claim in the report that $P<0.005$.

Explanation of Solution

Given info:

The data shows that the insulin concentrations in the two types of mice.

In the given information, $n_{\text{Wild}}=10$, $n_{aP{2}^{-/-}}=10$, ${\overline{x}}_{\text{Wild}}=5.9$, ${\overline{x}}_{aP{2}^{-/-}}=0.75$, $SE{M}_{\text{Wild}}=0.9$ and $SE{M}_{aP{2}^{-/-}}=0.2$.

Calculation:

Mean:

From the information in Exercise 62, the mean for Wild type is 59 and the mean for $aP{2}^{-/-}$ is 0.75.

Standard deviation for Wild type:

$\begin{array}{c}SE{M}_{\text{Wild}}=\frac{s_{\text{Wild}}}{\sqrt{n_{\text{Wild}}}}\\ s_{\text{Wild}}=SE{M}_{\text{Wild}}\sqrt{n_{\text{Wild}}}\\ =0.9\left(\sqrt{10}\right)\\ =2.85\end{array}$

Thus, the value of s for Wild type is 2.85.

Standard deviation for $aP{2}^{-/-}$:

$\begin{array}{c}SE{M}_{aP{2}^{-/-}}=\frac{s_{aP{2}^{-/-}}}{\sqrt{n_{aP{2}^{-/-}}}}\\ s_{aP{2}^{-/-}}=SE{M}_{aP{2}^{-/-}}\sqrt{n_{aP{2}^{-/-}}}\\ =0.2\left(\sqrt{10}\right)\\ =0.632\end{array}$

Thus, the value of s for $aP{2}^{-/-}$ is 0.632.

PLAN:

Check whether or not there is a significant difference in the mean insulin concentrations in the two types of mice.

State the test hypotheses.

Null hypothesis:

$H_0:{\mu }_{\text{1}}={\mu }_{\text{2}}$

Alternative hypothesis:

$H_a:{\mu }_{\text{1}}\ne {\mu }_{\text{2}}$

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

• Choose Stat > Basic Statistics > 2-Sample t.
• Choose Summarized data.
• In first, enter Sample size as 10, Mean as 5.9, Standard deviation as 2.85.
• In second, enter Sample size as 10, Mean as 0.75, Standard deviation as 0.632.
• Choose Options.
• In Confidence level, enter 95.
• In Alternative, select not equal.
• Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

From the MINITAB output, the value of the t-statistic is 5.58 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, $0.000\left(=P\text{-value}\right)<0.05\left(=\alpha \right)$.

Therefore, by the rejection rule, it can be concluded that there is evidence to reject $H_0$ at $\alpha =0.05$.

Thus, there is a significant difference in the mean insulin concentrations in the two types of mice.

Justification:

From the MINITAB output, the P-value is 0.000. Hence, the P-value confirm the claim in the report that $P<0.005$.