The manager of Lone Star Restaurant believes that the average wait time of the Saturday evening meal is 30 minutes. To test this belief, the manager selects 50 customers at random and computes the average wait time to be 34 minutes. Assume a population standard deviation of 12 minutes.

a. Find a 99% confidence interval for the mean wait time of a customer.
b. Do the data indicate that the mean wait time differs from 30 minutes at the 1% significance level?

a. Find table in the back of your statistics text labeled something like "areas under normal distribution" to find the proportion/probability (±.005) and its Z score.

99% = mean ± Z (SD) = 34 ± 2.575(12)

b. Z = (score-mean)/SEm

SEm = SD/√n

Use same table to find probability ≤ .01.

a. To find a 99% confidence interval for the mean wait time of a customer, we can use the formula:

Confidence Interval = Sample Mean ± (Z * (Population Standard Deviation / √n))

Where:
- Sample Mean is the average wait time of the sample (34 minutes)
- Z is the Z-score for the desired confidence level (in this case, 99%)
- Population Standard Deviation is the known standard deviation of the population (12 minutes)
- n is the sample size (50 customers)

First, we need to find the Z-score. The Z-score corresponding to a 99% confidence level is found using a standard normal distribution table or calculator, and it is approximately 2.576.

Plugging in the values, we get:

Confidence Interval = 34 ± (2.576 * (12 / √50))

Calculating this gives us a confidence interval of approximately (31.12, 36.88). Therefore, we can say with 99% confidence that the average wait time of a customer is between 31.12 minutes and 36.88 minutes.

b. To test whether the mean wait time differs from 30 minutes at the 1% significance level, we can use a hypothesis test. The null hypothesis (H0) is that the mean wait time is 30 minutes, and the alternative hypothesis (Ha) is that the mean wait time is different from 30 minutes.

We can use a t-test for this because the population standard deviation is unknown. The t-test statistic is calculated using the formula:

t = (Sample Mean - Population Mean) / (Sample Standard Deviation / √n)

Where:
- Sample Mean is the average wait time of the sample (34 minutes)
- Population Mean is the assumed mean wait time (30 minutes)
- Sample Standard Deviation is the standard deviation of the sample
- n is the sample size (50 customers)

Since the population standard deviation is unknown, we use the sample standard deviation. The formula for sample standard deviation is:

Sample Standard Deviation = Population Standard Deviation / √n

Plugging in the values, we get:

t = (34 - 30) / (12 / √50)

Calculating this gives us a t-value of approximately 1.76.

We then compare this t-value to the critical t-value at the 1% significance level with (n-1) degrees of freedom (49 degrees of freedom in this case). If the calculated t-value is greater than the critical t-value or less than the negative of the critical t-value, we reject the null hypothesis.

Looking up the critical t-value at the 1% significance level with 49 degrees of freedom gives us approximately ±2.68.

Since our calculated t-value (1.76) is not greater than 2.68 or less than -2.68, we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that the mean wait time differs from 30 minutes at the 1% significance level.