Chapter 10, Problem 38SE

Supermarket Checkout Lane Design. Safegate Foods, Inc., is redesigning the checkout lanes in its supermarkets throughout the country and is considering two designs. Tests on customer checkout times conducted at two stores where the two new systems have been installed result in the following summary of the data.

Test at the .05 level of significance to determine whether the population mean checkout times of the two systems differ. Which system is preferred?

To determine

Test whether the population mean checkout times of the two systems differ at .05 level of significance or not.

Suggest the preferable system.

Answer to Problem 38SE

There is sufficient evidence to conclude that, there is a difference between the population mean checkout times of the two systems.

The preferable system is system A.

Explanation of Solution

Calculation:

The results of the tests on customer checkout times at system A and system B are as follows:

System A	System B
$n_1=120$	$n_2=100$
${\overline{x}}_1=4.1\text{minutes}$	${\overline{x}}_2=3.4\text{minutes}$
${\sigma }_1=2.2\text{minutes}$	${\sigma }_2=1.5\text{minutes}$

The level of significance is $\alpha =0.05$.

State the hypotheses.

The test hypotheses are as follows:

Null hypothesis:

$H_0:{\mu }_1-{\mu }_2=0$

That is, there is no difference between the population mean checkout times of the two systems.

Alternative hypothesis:

$H_a:{\mu }_1-{\mu }_2\ne 0$

That is, there is a difference between the population mean checkout times of the two systems.

Test statistic:

The test statistic for hypothesis tests about ${\mu }_1-{\mu }_2$ when ${\sigma }_1$ and ${\sigma }_2$ are known is as follows:

$z=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

Substitute ${\overline{x}}_1$ as 4.1, ${\overline{x}}_2$ as 3.4, ${\sigma }_1$ as 2.2, ${\sigma }_2$ as 1.5, $n_1$ as 120, and $n_2$ as 100 in the above formula.

$\begin{array}{c}z=\frac{\left(4.1-3.4\right)}{\sqrt{\frac{{2.2}^2}{120}+\frac{{1.5}^2}{100}}}\\ =\frac{0.7}{\sqrt{\frac{4.84}{120}+\frac{2.25}{100}}}\\ =\frac{0.7}{\sqrt{0.0403+0.0225}}\\ =\frac{0.7}{0.2506}\end{array}$

  $=2.7933$

Thus, the test statistic is 2.79.

Software procedure:

Step-by-step procedure to obtain the probability value using Excel:

• Open an EXCEL sheet and select the cell A1.
• Enter the formula =NORM.S.DIST(2.79,TRUE) in the cell A1.
• Press Enter.

Output obtained using EXCEL software is given below:

From the output, the p-value for the left tail is 0.997365.

For a two-tailed test, the p-value is two times of upper-tail area.

From the output:

$\begin{array}{c}p\text{-value}=2\left(1-0.997365\right)\\ =2\left(0.0026\right)\\ =0.0052\end{array}$

Thus, the p-value is 0.0052.

Rejection rule:

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

If $p\text{-value}>\alpha$, do not reject the null hypothesis.

Conclusion:

Here, the p-value is less than the level of significance.

That is, $p\text{-value}\left(=0.0052\right)<\alpha \left(=0.05\right)$.

From the rejection rule, the null hypothesis is rejected.

Therefore, there is sufficient evidence to conclude that, there is a difference between the population mean checkout times of the two systems.

Thus, there is a difference between the population mean checkout times of the two systems.

Here, the mean checkout time of system B is less when compared to the mean checkout time of system A.

Therefore, system A is preferable.