Question

In: Statistics and Probability

Dominick Aldo owns and operates Carolina’s which is an Italian restaurant in New York. The file Single Population.xlsx is shown below and contains the amount of time that table times varied from table to table.

Dominick Aldo owns and operates Carolina’s which is an Italian restaurant in New York. The file Single Population.xlsx is shown below and contains the amount of time that table times varied from table to table. A) Test the hypothesis that the average table time exceeds 98 minutes using 0.05 ? = . B) What is the p-value? Interpret the results. Be sure that your project shows the following steps 1. Null and alternative hypothesis 2. Determine which distribution to use for the test statistic. 3. Using data provided, calculate necessary sample statistics. 4. Draw a conclusion and interpret the decision.

Time

70, 80, 100, 90, 75, 60, 70, 75, 110, 90, 100, 75 ,60, 75, 80, 75 ,165, 60, 75, 90,

9-2 Chapter 9

110, 95, 110, 150, 50, 70, 90, 110, 165, 60, 145, 80, 115, 75, 50, 90, 110, 90, 100, 110, 110, 85, 70, 145, 120, 130, 80, 90, 105, 105, 100, 95, 80, 100, 120, 130, 75, 90, 70, 125, 80 90 95

Solutions

Expert Solution

Let be the population average table time. We don't know the value of . We wish to test hypothesis whether exceeds 98 minutes or not.

So, we want to test

1. Null Hypothesis:   

VERSUS

Alternative Hypothesis :

2. As we have sample of size n = 63, which is sufficiently large enough to assume central limit theorem, we can always safely use Normal distribution for the test-statistic.

But, there is a strong evidence that the data follows a normal distribution from the Q-Q plot of the data. Here is the Q-Q plot. (Write the code: ### qqnorm(unique(time)) ### in R to get the Q-Q plot)

Because, the plot looks almost in a straight line, there is a strong evidence that the data follows a normal distribution with mean ad standard deviation unknown.

So, it is better to use the distribution of test-statistic as a student's t-distribution with n-1 = 63-1 = 62 degrees of freedom.

3) Sample mean = , sample standard deviation =

So, the test-statistic is

p-value =

As p-value is much greater than 0.05, we fail to reject null hypothesis.

Thus we conclude there is no evidence that the average table time exceeds 98 minutes.

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70 80 100 90 75 60 70 75 110 90 100 75 60 75 80 75 165 60 75 90 110 95
110 150 50 70 90 110 165 60 145 80 115 75 50 90 110 90 100 110 110 85 70 145
120 130 80 90 105 105 100 95 80 100 120 130 75 90 70 125 80 90 95

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