Chapter 10, Problem 38SE

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.

System A	System B
n1 = 120	n2 = 100
${\bar{x}}_1=4.1\text{minutes}$	${\bar{x}}_{\text{2}}=\text{3}.\text{4}\text{minutes}$
σ1 = 2.2 minutes	σ2 = 1.5 minutes

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 hypothesis:

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 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,

$z=\frac{\left({\overline{x}}_1-{\overline{x}}_2\right)-D_0}{\sqrt{\frac{{\sigma }_1^2}{n_1}+\frac{{\sigma }_2^2}{n_2}}}$

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,

$\begin{array}{c}z=\frac{\left(4.1-3.4\right)-0}{\sqrt{\frac{{2.2}^2}{120}+\frac{{1.5}^2}{100}}}\\ =\frac{0.7}{\sqrt{\frac{4.84}{120}+\frac{2.25}{100}}}\\ =\frac{0.7}{\sqrt{0.0403+0.0225}}\\ =\frac{0.7}{0.2506}\end{array}$

  $=2.7933$

Thus, the test statistic is 2.79.

Software procedure:

Step-by-step software procedure to obtain the mean using MINITAB software is as follows,

• Choose Graph > Probability Distribution Plot choose View Probability > OK.
• From Distribution, choose ‘Normal’ distribution.
• Click the Shaded Area tab.
• Choose X Value and Two Tail for the region of the curve to shade.
• Enter the data value as 2.79.
• Click OK.

Output using MINITAB software is as follows:

Thus, the p-value is 0.0052$\left(=0.002635+0.002635\right)$.

Decision rule based on p-value approach:

If p-value ≤ α, then reject the null hypothesis H₀.

If p-value > α, then fail to reject the null hypothesis H₀.

Conclusion:

Here, the p-value 0.0052 is less than or equal to the significance level 0.05.

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

Thus, 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.