Problem 6 - Basic Statistics for Business and Economics with Student

Question

The MacBurger restaurant chain claims that the mean waiting time of customers is 3 minutes with a population standard deviation of 1 minute. The quality- assurance department found in a sample of 50 customers at the Warren Road MacBurger that the mean waiting time was 2.75 minutes. At the .05 significance level, can we conclude that the mean waiting time is less than 3 minutes?

Step-by-Step Solution

Answer

Yes, we can conclude that the mean waiting time is less than 3 minutes.

Step 1: Formulate the Hypotheses

To conduct a hypothesis test, first, we need to set up the null and alternative hypotheses. The null hypothesis \( H_0 \) is that the mean waiting time is 3 minutes (\( \mu = 3 \)). The alternative hypothesis \( H_a \) is that the mean waiting time is less than 3 minutes (\( \mu < 3 \)).

Step 2: Determine the Significance Level and Critical Value

The significance level is given as \( \alpha = 0.05 \). This is a one-tailed test since we are checking if the mean is less than a certain value. Using a standard normal distribution table, we find that the critical z-value for \( \alpha = 0.05 \) is \( z_{0.05} = -1.645 \).

Step 3: Calculate the Test Statistic

The test statistic for a sample mean is calculated using the formula \[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \] where \( \bar{x} = 2.75 \), \( \mu = 3 \), \( \sigma = 1 \), and \( n = 50 \). Plugging in these values gives us \[ z = \frac{2.75 - 3}{\frac{1}{\sqrt{50}}} \approx -1.77. \]

Step 4: Make the Decision

Compare the calculated test statistic \(-1.77\) with the critical value \(-1.645\). Since \(-1.77 < -1.645\), we reject the null hypothesis.

Step 5: State the Conclusion

Since we rejected the null hypothesis, we have statistical evidence at the \(0.05\) significance level to support the claim that the mean waiting time at the Warren Road MacBurger is less than 3 minutes.

Key Concepts

Null Hypothesis

In hypothesis testing, the null hypothesis is a statement we aim to test. It represents a theory that there is no effect or no difference in the situation being analyzed. Essentially, it's the default position that there is nothing of interest occurring. In our exercise concerning the MacBurger restaurant chain, the null hypothesis (denoted as \( H_0 \)) claims that the mean waiting time for customers is 3 minutes. This suggests that any deviation observed in the sample mean, such as the 2.75 minutes recorded, is due to random sampling error and not due to an actual difference in the population mean.

When conducting hypothesis testing, the null hypothesis serves as a starting point for analysis. We either seek evidence to reject it or fail to reject it. Rejecting the null hypothesis provides evidence that the effect or difference we suspect is real. It's crucial, however, not to interpret failure to reject the null hypothesis as evidence that the null is true. It simply means there isn't enough evidence against it at the chosen significance level.

Alternative Hypothesis

The alternative hypothesis represents the statement you're testing against the null hypothesis. It indicates the presence of an effect or a difference. In simpler terms, it reflects the possibility that what you suspect is actually true. In our example of customer waiting time at MacBurger, the alternative hypothesis (denoted as \( H_a \)) is that the mean waiting time is less than 3 minutes.

The alternative hypothesis is crucial because it tells us what kind of evidence we are looking for in the data. When our test statistic indicates that the probability of observing such a sample mean under the null hypothesis is very low, we say there is enough evidence to reject the null hypothesis in favor of the alternative hypothesis.

It's essential to define the type of test as part of formulating the alternative hypothesis. In this case, it's a one-tailed test since we are only interested in whether the mean is significantly less than 3 minutes and not whether it is simply different from 3 minutes.

Significance Level

The significance level, represented by \( \alpha \), is a threshold used to determine whether to reject the null hypothesis. It defines the probability of rejecting a true null hypothesis, known as Type I error. In most scientific studies, a significance level of \( 0.05 \) is commonly used, indicating a 5% risk of concluding a difference when there is none.

The significance level sets the critical value which the calculated test statistic must exceed in order to reject the null hypothesis. In our MacBurger exercise, a significance level of \( \alpha = 0.05 \) means we are willing to accept a 5% risk of error when claiming that the mean waiting time is less than 3 minutes.

This level influences the critical value and thereby helps in comparing the computed test statistic to make a decision. A lower \( \alpha \) reduces the chance of making a Type I error, but it also makes it more challenging to reject the null hypothesis.

Test Statistic

The test statistic is a standardized value used in hypothesis testing to determine the distance of the sample mean from the population mean under the null hypothesis, in terms of standard deviations. It helps us decide whether to reject the null hypothesis or not. In our situation, we calculate the test statistic using a formula based on a z-score, given the known population standard deviation.

For our MacBurger exercise, the test statistic \( z \) is computed using the formula:\[ z = \frac{\bar{x} - \mu}{\frac{\sigma}{\sqrt{n}}} \]
where \( \bar{x} \) is the sample mean (2.75 minutes), \( \mu \) is the population mean under the null hypothesis (3 minutes), \( \sigma \) is the population standard deviation (1 minute), and \( n \) is the sample size (50).

This calculation gives us a z-score of approximately -1.77. By comparing this calculated z-score to the critical value determined by the significance level, we can decide whether there is enough evidence to reject the null hypothesis. If the test statistic falls beyond the critical value (like our calculated \( -1.77 \) compared to the \( -1.645 \) critical value), it indicates that the sample mean is sufficiently unlikely under the null hypothesis, warranting its rejection.