Chapter 9.1, Problem 4CYU

a.

To determine

Find the value of t-test statistic.

Answer to Problem 4CYU

The value of t-test statistic is 1.651768.

Explanation of Solution

Calculation:

The given statistics for testing the hypothesis $H_0:{\mu }_1={\mu }_2$ versus $H_1:{\mu }_1\ne {\mu }_2$ are given as follows:

${\overline{x}}_1=73.9$, ${\overline{x}}_2=71.8$, $s_1=4.2$, $s_2=3.8$, $n_1=23$ and $n_2=17$.

Requirements for a small sample t-test:

• The samples must be independent and drawn randomly from the populations.
• Either the sample size must be greater than 30 or the population must be approximately normal.

Here, it is given that independent random samples are drawn from approximately normal populations.

Thus, the conditions are satisfied.

The hypotheses are given below:

Null hypothesis:

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

Alternate hypothesis:

$H_1:{\mu }_1\ne {\mu }_2$

The test statistic for the small sample t is obtained as follows:

$t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-\left({\mu }_1-{\mu }_2\right)}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}$

Under the null hypothesis, $\left({\mu }_1-{\mu }_2\right)=0$.

Therefore the test statistic is,

$\begin{array}{c}t=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-0}{\sqrt{\frac{s_1{}^2}{n_1}+\frac{s_2{}^2}{n_2}}}\\ =\frac{73.9-71.8}{\sqrt{\frac{{4.2}^2}{23}+\frac{{3.8}^2}{17}}}\\ =\frac{2.1}{1.271365}\\ =1.651768\end{array}$

Thus, the test statistic is 1.651768.

b.

To determine

Find the number of degrees of freedom for the test statistic.

Answer to Problem 4CYU

The number of degrees of freedom for the test statistic is 16.

Explanation of Solution

Calculation:

Degrees of freedom:

The degrees of freedom for t using computer package is,

$\Delta =\frac{{\left[\left(\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right)\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_2^2}{n_2}\right)}^2}{n_2-1}}$

The degrees of freedom, when computing by hand is smaller of $n_1-1$ and $n_2-1$.

The degrees of freedom is,

$\begin{array}{c}d.f=\min \left(n_1-1,n_2-1\right)\\ =\min \left(23-1,17-1\right)\\ =\min \left(22,16\right)\\ =16\end{array}$

Thus, the degree of freedom is 16.

c.

To determine

Find the P-value.

Answer to Problem 4CYU

The P-value is 0.11806.

Explanation of Solution

Calculation:

P-value:

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

• Choose Graph > Probability Distribution Plot.
• Choose View Probability > OK.
• From Distribution, choose‘t’ distribution.
• In Degrees of freedom, enter 16.
• Click the Shaded Area tab.
• Choose X value and Both Tail for the region of the curve to shade.
• In X-value enter 1.651768.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the P-value is $0.05903\times 2=0.11806$

Thus, the P-value is 0.11806.

d.

To determine

Interpret the P-value.

Check whether the null hypothesis is rejected at $\alpha =0.05$.

Answer to Problem 4CYU

There is not enough evidence to reject the null hypothesis at $\alpha =0.05$.

Explanation of Solution

Decision rule based on P-value:

If $P-\text{value}\le \alpha$, then reject the null hypothesis $H_0$.

If $P-\text{value}>\alpha$, then fail to reject the null hypothesis $H_0$.

Here, the level of significance is $\alpha =0.05$.

Conclusion based on P-value approach:

The P-value is 0.11806 and $\alpha$ value is 0.05.

Here, P-value is greater than the $\alpha$ value.

That is $0.11806\left(=P-\text{value}\right)>0.05\left(=\alpha \right)$.

By the rejection rule, failed to reject the null hypothesis.

Hence, the null hypothesis cannot be rejected at $\alpha =0.05$.

Thus, it can be concluded that there is sufficient evidence to infer the equality of two population means at $\underset{\_}{\alpha =0.05}$.