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Chapter 10: Problem 40

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. \(\begin{array}{cl}\text { System A } & \text { System B } \\ n_{1}=120 & n_{2}=100 \\\ \bar{x}_{1}=4.1 \text { minutes } & \bar{x}_{2}=3.4 \text { minutes } \\\ \sigma_{1}=2.2 \text { minutes } & \sigma_{2}=1.5 \text { minutes }\end{array}\) Test at the .05 level of significance to determine whether the population mean checkout times of the two systems differ. Which system is preferred?

Short Answer

Expert verified
System B is preferred as it has a lower mean checkout time.

Step by step solution

01

Define Hypotheses

We start by setting up the null and alternative hypotheses:- Null hypothesis \(H_0\): The population mean checkout times for both systems are equal, \(\mu_1 = \mu_2\).- Alternative hypothesis \(H_a\): The population mean checkout times for both systems are different, \(\mu_1 eq \mu_2\). This is a two-tailed test.
02

Calculate the Test Statistic

For comparing two means (assuming population variances are known), we use the formula for the test statistic:\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - \Delta_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]Substituting the given values, where \(\Delta_0 = 0\):\[ z = \frac{(4.1 - 3.4) - 0}{\sqrt{\frac{2.2^2}{120} + \frac{1.5^2}{100}}} \]Calculate \(\sigma_1^2 = 4.84\) and \(\sigma_2^2 = 2.25\). Then the standard error is:\[ \sqrt{\frac{4.84}{120} + \frac{2.25}{100}} = \sqrt{0.04033 + 0.0225} = \sqrt{0.06283} \approx 0.25066 \]Finally, calculate the test statistic:\[ z = \frac{0.7}{0.25066} \approx 2.79 \]
03

Determine the Critical Value

Since this is a two-tailed test at the 0.05 significance level, we split \(\alpha\) into two tails, giving \(\alpha/2 = 0.025\). We find the critical z-values from the z-table, which are -1.96 and 1.96. So if the test statistic falls outside of \([-1.96, 1.96]\), we reject the null hypothesis.
04

Make a Decision

The calculated z-value is 2.79, which is greater than 1.96. Thus, the test statistic falls into the rejection region.
05

Conclusion

Since we rejected the null hypothesis, there is sufficient evidence to conclude that the population mean checkout times for the two systems are different. System B, with a lower mean checkout time, is preferred given the evidence.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Two-Tailed Test
A two-tailed test is a method used in statistics when we need to determine if there is a significant difference between two sample means in any direction. This test allows for the possibility of observing a difference in both directions, higher or lower.
In our exercise, the two-tailed test examines whether the checkout times between System A and System B differ. We have two hypotheses to consider:
  • The null hypothesis (\( H_0 \)): assumes that there is no difference between the means of the two systems (\( \mu_1 = \mu_2 \)).
  • The alternative hypothesis (\( H_a \)): suggests that there is a difference in the means (\( \mu_1 eq \mu_2 \)).
To perform a two-tailed test, we need to find the test statistic. This will allow us to compare with the critical values on a z-distribution table and determine if the observed difference is statistically significant. When calculating it, we consider the entire rejection region on both ends of the distribution.
Mean Comparison
Mean comparison is the process of determining whether two groups have different average values. In our scenario, we compared the mean checkout times of two systems: System A and System B.
This comparison requires calculating the test statistic using the formula:
\[ z = \frac{(\bar{x}_1 - \bar{x}_2) - \Delta_0}{\sqrt{\frac{\sigma_1^2}{n_1} + \frac{\sigma_2^2}{n_2}}} \]
  • \(\bar{x}_1\) and \(\bar{x}_2\) are the sample means for Systems A and B.
  • \(\sigma_1\) and \(\sigma_2\) are the population standard deviations.
  • \(n_1\) and \(n_2\) represent the sample sizes.
  • \(\Delta_0\) is the assumed difference in population means, which is zero under the null hypothesis.
We calculated the test statistic to be approximately 2.79, which compares the observed mean difference to the variability within the sample data. This statistic helps determine if the observed difference in means is too large to be attributed to chance alone.
Significance Level
The significance level, denoted as \(\alpha\), represents the probability of rejecting the null hypothesis when it is true. It is a boundary that helps us decide if the test result is statistically significant. In simpler terms, it tells us the threshold for doubt.
In the exercise, a significance level of 0.05 implies a 5% risk of concluding that there is a difference when there is none. For our two-tailed test, the 0.05 level is split equally between the two tails of the distribution, hence \(\alpha/2 = 0.025\) for each tail.Calculating the critical values for this level, we look up these values on a z-table which gives us the critical points -1.96 and 1.96. If the test statistic falls outside this range, we reject the null hypothesis.
Since our calculated z-value of 2.79 exceeds 1.96, it falls into the rejection region, leading us to conclude that System A and System B do indeed have different mean checkout times.

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Most popular questions from this chapter

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