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 0.10.

Italian            Seafood     Steakhouse

Sample Mean                25              24                26

Sample Variance          0.7             8.7               3.3

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 ANOVA:

Group	ni	x̅i	S2i	ni*(Xi-Xgrand)2	(ni-1)*S2i
Italian	4	25	0.700	0.000	2.10
Seafood	4	24	8.700	4.000	26.10
Steakhouse	4	26	3.300	4.000	9.90
	grand mean=	25.0000		8.000	38.10
				SSTr	SSE
Source	SS	df	MS	F
between	8.00	2.000	4.0000	0.94
error	38.10	9.000	4.2333
total	46.10	11.0000

a)

degrees of freedom for numerator =2

degrees of freedom for Denominator =9

b)

critical value for this problem =3.01

c) value for the mean of the means =25.00

d) value of SSTR =8

MSTR =4

e) value of SSE =38.10

MSE =SSE/df(SSE)=4.2333

f)

value for the test statistic =0.94

g)

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

h)we can not conclude that average meal price per person is different for three type of restaurants.