Problem 57 - Basic Statistics for Business and Economics with Student

Question

A national grocer's magazine reports the typical shopper spends eight minutes in line waiting to check out. A sample of 24 shoppers at the local Farmer Jack's showed a mean of 7.5 minutes with a standard deviation of 3.2 minutes. Is the waiting time at the local Farmer Jack's less than that reported in the national magazine? Use the .05 significance level.

Step-by-Step Solution

Answer

The waiting time at Farmer Jack's is not significantly less than 8 minutes.

Step 1: State the Hypotheses

We are testing if the mean waiting time at Farmer Jack's is less than the national average of 8 minutes. The null hypothesis (\( H_0 \)) is that the mean waiting time is \( \mu \geq 8 \), and the alternative hypothesis (\( H_1 \)) is that \( \mu < 8 \).

Step 2: Identify the Significance Level

The significance level, \( \alpha \), is given as 0.05. This is the probability of rejecting the null hypothesis when it is true.

Step 3: Calculate the Test Statistic

To calculate the test statistic, we use the formula for the t-statistic: \[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \] where \( \bar{x} = 7.5 \) (sample mean), \( \mu = 8 \) (population mean), \( s = 3.2 \) (standard deviation), and \( n = 24 \) (sample size). Substitute the values into the formula:\[ t = \frac{7.5 - 8}{3.2/\sqrt{24}} \approx \frac{-0.5}{0.653} \approx -0.766 \]

Step 4: Determine the Critical Value

The critical value for a one-tailed t-test at a significance level of 0.05 with \( n-1 = 23 \) degrees of freedom can be found using a t-distribution table or calculator: \( t_{critical} = -1.714 \).

Step 5: Compare the Test Statistic and Critical Value

Since the calculated t-statistic (\( t \approx -0.766 \)) is greater than the critical value (\( -1.714 \)), we fail to reject the null hypothesis.

Step 6: Conclusion

There is not sufficient evidence to conclude that the mean waiting time at the local Farmer Jack's is significantly less than 8 minutes at the 0.05 significance level.

Key Concepts

Significance Level

The significance level, often represented by the Greek letter \( \alpha \), is an important concept in hypothesis testing. It helps us determine how strong the evidence must be before we can reject the null hypothesis. A common choice for the significance level is 0.05, as in this exercise.
This means there is a 5% risk of concluding that a difference exists, when in fact there is no actual difference. In simpler terms, it is the probability of making a type I error, which is rejecting a true null hypothesis.

• Example: In our exercise, the significance level of 0.05 suggests that there is a 5% chance we might falsely say that the average waiting time at Farmer Jack's is less.

When conducting an experiment or a test, selecting the significance level is a key decision. It influences how you interpret the results. Lower alpha values mean more demanding standards for evidence against the null hypothesis.
Thus, choosing \( \alpha = 0.05 \) shows a moderate level of tolerance for error, balancing the need for evidence with the possibility of errors.

t-statistic

The t-statistic is a value derived from your sample data during hypothesis testing, allowing you to compare your sample result with what we would expect under the null hypothesis. It is calculated using the formula:
\[ t = \frac{\bar{x} - \mu}{s/\sqrt{n}} \]
where:

• \( \bar{x} \) is the sample mean
• \( \mu \) is the population mean according to the null hypothesis
• \( s \) is the sample standard deviation
• \( n \) is the sample size

In our example, the calculated t-statistic was approximately \( -0.766 \).
This result shows how many standard deviations the sample mean is from the population mean stated in the null hypothesis. Essentially, if the t-statistic falls outside of the critical region (determined by significance level and degrees of freedom), you may reject the null hypothesis.
However, if it lies within the critical bounds, you fail to reject it, as happened here.

Critical Value

The critical value acts as a threshold in hypothesis testing. It helps decide whether to reject the null hypothesis by comparing it to the calculated t-statistic. The critical value is determined by the chosen significance level and degrees of freedom (\( n-1 \), where \( n \) is sample size).
In our case, for a one-tailed test at \( \alpha = 0.05 \) with 23 degrees of freedom, the critical value was \( -1.714 \).
Here's how it works:

• If the t-statistic is less than this critical value, it prompts rejecting of the null hypothesis, suggesting significant results.
• If the t-statistic is greater, as in our exercise, the null hypothesis is not rejected.

Thus, the critical value helps provide a clear rule to guide decision-making in hypothesis tests, ensuring consistency across studies.

Null Hypothesis

The null hypothesis, denoted as \( H_0 \), is a fundamental part of hypothesis testing. It sets a baseline we test against, often implying 'no effect' or 'no difference.'
For our exercise, the null hypothesis states that the mean waiting time at Farmer Jack's is 8 minutes or more, \( \mu \geq 8 \).
The idea is to gather enough evidence to reject this hypothesis by showing that the alternative hypothesis (\( H_1 \)) could be true. In this case, the alternative hypothesis is that the mean waiting time is less than 8 minutes (\( \mu < 8 \)).
To test the null hypothesis, compare the calculated t-statistic with the critical value derived from the chosen significance level and distribution. Failing to reject \( H_0 \) suggests insufficient evidence to support the alternative claim—in our scenario, indicating no significant reduction in wait times at Farmer Jack's.
This process helps ensure decisions are based on statistical evidence rather than assumptions.