The MacBurger restaurant chain claims that the waiting time of customers for service is

normally distributed, with a mean of 3 minutes and a 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?

You can use a one-sample z-test on this data.

Null hypothesis:
Ho: µ = 3 -->meaning: population mean is equal to 3 minutes
Alternate hypothesis:
Ha: µ < 3 -->meaning: population mean is less than 3 minutes

Using the z-test formula to find the test statistic:
z = (sample mean - population mean)/(standard deviation divided by the square root of the sample size)
z = (2.75 - 3)/(1/√50)

I'll let you finish the calculation.

Check a z-table for your critical or cutoff value at .05 level of significance for a one-tailed test (the alternate hypothesis is showing a specific direction; therefore, the test is one-tailed). Compare the test statistic you calculated to the critical value from the table. If the test statistic exceeds the critical value, reject the null and conclude that the mean waiting time is less than 3 minutes. If the test statistic does not exceed the critical value from the table, you cannot reject the null and conclude a difference.

I hope this will help.

The MacBurger restaurant chain claims that the waiting time of customers for service is

normally distributed, with a mean of 3 minutes and a 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?

To answer this question, we can conduct a hypothesis test. Here's how you can do it:

Step 1: State the hypotheses:
- Null hypothesis (H0): The mean waiting time is equal to 3 minutes.
- Alternative hypothesis (H1): The mean waiting time is less than 3 minutes.

Step 2: Determine the significance level:
The significance level, denoted as alpha (α), is given in the problem as 0.05. This represents a 5% level of significance.

Step 3: Calculate the test statistic:
To calculate the test statistic, we need to use the formula for a z-test:
z = (sample mean - population mean) / (population standard deviation / square root of sample size)

Given:
- Sample mean (x̄) = 2.75 minutes
- Population mean (μ) = 3 minutes
- Population standard deviation (σ) = 1 minute
- Sample size (n) = 50

Substituting these values into the formula, we get:
z = (2.75 - 3) / (1 / √50)

Step 4: Find the critical value:
Since the alternative hypothesis is that the mean waiting time is less than 3 minutes, we are conducting a one-tailed test. This means that we need to find the critical value corresponding to a 5% significance level in the left tail of the standard normal distribution.

Using a standard normal distribution table or a calculator, we find that the critical value for a 5% significance level is approximately -1.645.

Step 5: Make a decision:
If the test statistic (z) is less than the critical value (-1.645), we reject the null hypothesis. Otherwise, we fail to reject the null hypothesis.

Step 6: Calculate the test statistic:
Substitute the values into the formula to find the test statistic (z):
z = (2.75 - 3) / (1 / √50)
z = -1.7678 (approximately)

Step 7: Make a decision:
Since the test statistic (-1.7678) is less than the critical value (-1.645), we reject the null hypothesis.

Step 8: Interpret the result:
Based on the sample data, at a 5% significance level, we have enough evidence to conclude that the mean waiting time is less than 3 minutes at the Warren Road MacBurger.

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