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 5 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 0.01.

Italian            Seafood     Steakhouse

Sample Mean                14              17                20

Sample Variance          2.5             2.5               16

a. How many degrees of freedom are there in the problem?

b. What is the critical value for this problem?

c. What is the value for the mean of the means?

d. What is the value of SSTR and MSTR respectively?

e. What is the value of SSE and MSE respectively?

f. What is the value for the test statistic?

g. What is your decision after performing the test?

h. What is your conclusion after the test? Be specific and relate the conclusion to the problem.

Answer 1

applying one way ANOVA:

Group	ni	x̅i	S2i	ni*(Xi-Xgrand)2	(ni-1)*S2i
Italian	5	14	2.500	45.000	10.00
Seafood	5	17	2.500	0.000	10.00
Steakhouse	5	20	16.000	45.000	64.00
	grand mean=	17.0000		90.000	84.00
				SSTr	SSE
Source	SS	df	MS	F
between	90.00	2.000	45.0000	6.43
error	84.00	12.000	7.0000
total	174.00	14.0000

a)

degrees of freedom for numerator =2

degrees of freedom for denomintor =12

b)

critical value at 0.01 level and above degrees of freedom=6.93

c)

value for the mean of the means =17

d)

value of SSTR =90

MSTR=90/2 =45

e)

value of SSE =84

MSE=84/12 =7

f)

value for the test statistic =MSTR/MSE =6.429

g)

as test statistic is not higher than critical value we can not reject Ho

h)

we can not conclude that average meal price per person differs for three restaurants,