A pizza lover wants to compare the average delivery times for four local pizza r
ID: 3364638 • Letter: A
Question
A pizza lover wants to compare the average delivery times for four local pizza restaurants. Over the course of a few weeks, he orders a number of pizzas from each restaurant, and he records the time it takes for each pizza to be delivered.
a) When performing an ANOVA with this data, what is the alternative hypothesis?
All of the restaurants have different mean delivery times
One of the restaurants has a different mean delivery time than the others
At least two of the restaurants have different mean delivery times
Two of the restaurants have different mean delivery times
b) A partial ANOVA table for his data is shown below. What is the value of B?
c) What is the value of C in the ANOVA table?
d) What is the value of D in the ANOVA table? Give your answer to three decimal places.
e) What is the value of E in the ANOVA table? Give your answer to three decimal places.
f) What is the value of F in the ANOVA table? Give your answer to two decimal places.
g) What is the value of G in the ANOVA table? Give your answer to four decimal places.
h) Using a 0.05 level of significance, what should his conclusion be in this case?
He should conclude that at least two of the restaurants have different mean delivery times because the P-value is greater than 0.05
He should conclude that at least two of the restaurants have different mean delivery times because the P-value is less than 0.05.
He should fail to reject the claim that at all of the restaurants have the same mean delivery times because the P-value is greater than 0.05.
He should conclude that at all of the restaurants have the same mean delivery times because the P-value is less than 0.05.
Source DF SS MS F P-value Treatment B 28.891 D F G Error C 16.56 E Total 20 45.451Explanation / Answer
a. C: At least two of the restaurants have different mean delivery times.
Why? - As the null hypothesis remains that mean of all the different treatment are same, the alternative has to be that at least one pair has different mean as complement space.
b. B=3, as there are 4 treatments, the degree of freedom is 3. As we loose one of the degree for the contrast.
c. C = 17. As total degree of freedom is 20 and treatment has 3 degree of freedom, that's why the error has to have 17 degree of freedom.
d. Value of D is 9.630333. As Mean sum of square is (Sum of square)/degree of freedom. So 28.891/3 = 9.630333
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