Question

A reporter with the Saint Pete Times is working on a story about the main factors making restaurants in the St. Pete area different from each other. The variable he is considering is the average meal price per person. The reporter selects a sample of 4 restaurants serving Italian food, Seafood and Steaks. The reporter believes that average price meal per person in the St Pete area is about the same independently of the type of restaurant. The table below shows sample mean and variances for collected data in dollars. Allow for an alpha of 0.1

	ITALIAN	SEAFOOD	STEAKHOUSE
SAMPLE MEAN	24	24	27
SAMPLE VARIANCE	0.7	8.7	0.7

5.B) What is the critical value for this problem?5.A) How many degrees of freedom are there in the problem?

5.C) What is the value for the mean of the means?

5.D) What is the value of SSTR and MSTR respectively?

5.E) What is the value of SSE and MSE respectively?

5.F) What is the value for the test statistic?

5.G) What is your decision after performing the test?

5.H) What is your conclusion after the test? Be specific and relate the conclusion to the problem.

Answer 1

	Italian	Seafood	Steakhouse
Count, n	4	4	4
Average, x̅	24	24	27
Sum	96	96	108
Variance, s2	0.7	8.7	0.7

5. A) Number of treatment, k = 3
Total sample Size, N = 4+4+4 = 12
      
df(between) = k-1 = 2
df(within) = N-k = 9
df(total) = N-1 = 11

B) Critical value, Fc = F.INV.RT(0.1, 2, 9) = 3.006

C) Mean of the means, Grand mean = (x̅1+x̅2+x̅3)/3 = 25

D) SSTR = (x̅1)²*n1 + (x̅2)²*n2 + (x̅3)²*n3 - (Grand Mean)²*N = 24

MSTR = SSTR/ df(between) = 24/2 = 12

E) SSE = (n1-1)*s1² + (n2-1)*s2² + (n3-1)*s3² = 30.3

MSE = SSE/df(within) = 3.3667

F) Test statistic:

F = MSTR/MSE = 3.5644

G) As F = 3.5644 > 3.006, we reject the null hypothesis.

H) There is enough evidence to conclude that the mean of the three restaurant are different at 0.1 significance level.

The reporter claim that average price meal per person in the St Pete area is about the same independently of the type of restaurant is rejected.