Chapter 11, Problem 1SR

Tom Sevits is the owner of the Appliance Patch. Recently Tom observed a difference in the dollar value of sales between the men and women he employs as sales associates. A sample of 40 days revealed the men sold a mean of $1,400 worth of appliances per day. For a sample of 50 days, the women sold a mean of $1,500 worth of appliances per day. Assume the population standard deviation for men is $200 and for women $250. At the .05 significance level, can Mr. Sevits conclude that the mean amount sold per day is larger for the women?

1. (a) State the null hypothesis and the alternate hypothesis.
2. (b) What is the decision rule?
3. (c) What is the value of the test statistic?
4. (d) What is your decision regarding the null hypothesis?
5. (e) What is the p-value?
6. (f) Interpret the result.

To determine

State the null and the alternative hypothesis.

Answer to Problem 1SR

The null hypothesis is $H_0:{\mu }_1\le {\mu }_2$

The alternative hypothesis is $H_1:{\mu }_1>{\mu }_2$

Explanation of Solution

In this context, ${\mu }_1$ denotes the mean amount of sales per day by women and ${\mu }_2$ denotes the mean amount of sales per day by men.

The hypotheses are given below:

Null hypothesis:

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

Alternative hypothesis:

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

Thus, the null hypothesis is $H_0:{\mu }_1\le {\mu }_2$ and alternative hypothesis is $H_1:{\mu }_1>{\mu }_2$.

To determine

Determine the decision rule.

Explanation of Solution

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

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 0 and Standard deviation value as 1
• Click the Shaded Area tab.
• Choose Probability and Right Tails for the region of the curve to shade.
• Enter the probability value as 0.05.
• Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the critical value is 1.645.

The decision rule is as follows:

If $\text{z}\text{>}\text{1}\text{.645}$, then reject the null hypothesis $H_0$.

To determine

Find the value of the test statistics.

Answer to Problem 1SR

The value of the test statistics is 2.108.

Explanation of Solution

The test statistics is obtained as follows:

$\begin{array}{c}z=\frac{{\overline{X}}_1-{\overline{X}}_2}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}\\ =\frac{1,500-1,400}{\sqrt{\frac{{250}^2}{50}+\frac{{200}^2}{40}}}\\ =\frac{100}{47.4342}\\ =2.108\end{array}$

Thus, the value of test statistics is 2.108.

To determine

Determine the decision regarding $H_0$.

Answer to Problem 1SR

The decision is that reject the null hypothesis $\left(H_0\right)$.

Explanation of Solution

Decision:

The critical value is 1.645 and the value of test statistic is 2.108.

The value of test statistic is greater than the critical value.

That is, $\text{Test statistic}\left(2.108\right)>\text{Critical}\text{value}\left(1.645\right)$.

From the decision rule, reject the null hypothesis.

To determine

Find the p-value.

Answer to Problem 1SR

The p-value is 0.0175.

Explanation of Solution

p-value:

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

• Choose Graph > Probability Distribution Plot choose View Probability> OK.
• From Distribution, choose ‘Normal’ distribution.
• Enter the Mean as 0 and Standard deviation value as 1.
• Click the Shaded Area tab.
• Choose X Value and right tails for the region of the curve to shade.
• Enter the X value as 2.108.
• Click OK.

Output obtained using MINITAB software is given below:

From the MINITAB output, the p-value is 0.0175.

Thus, the p-value is 0.0175.

To determine

Interpret the result.

Explanation of Solution

Interpretation:

Here, the null hypothesis is rejected.

Therefore, there is evidence that the mean amount of sales per day by woman is greater than that of women.