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 for collected data in dollars. The MSE value is 7. Allow for an error of 10% with 12 degrees of freedom.

Italian            Seafood     Steakhouse

Sample Mean                14              17                20

a What is the critical value for the given level of confidence?

b. What is the test statistic for means A & B ? (allow for a difference in values of + or - 0.01)

c. What is the test statistic for means A & C? (allow for a difference in values of + or - 0.01)

d. What is the test statistic for means B & C?  (allow for a difference in values of + or - 0.01)

e. What is your decision (with respect to the null hypotheses) after performing the test for means A & B, A & C and B & C respectively?

f. What is your conclusion after performing these tests with regards to which is different and which is the same?

Answer 1

Answer:

a)

To determine the critical value for given level of significance

Given,

n1 = n2 = 5

Significance level = 0.01

degree of freedom = n₁ + n₂ - 2

= 5 + 5 - 2

degree of freedom = 8

tcritical = 2.8965

b)

To determine the test statistic for means A & B

consider,

t = (x1bar - x2bar) / SE -------->(1)

where as

SE = sqrt(MSE)

= sqrt(7)

= 2.6458

the t statistic comparing A&B

substitute values in (1)

t = 14 - 17 / 2.6458

t = - 1.1339

c)

To determine the test statistic for means A & C

t = (x1bar - x2bar) / SE

substitute values

t = 14 - 20 / 2.6458

t = - 2.2678

d)

Now t statistic comparing B&C

t = (x1bar - x2bar) / SE

substitute values

t = 17 - 20 / 2.6458

t = - 1.1339

e)

Comparision can be given below

Comparison |t statistic| t critical |Decision

A&B 1.1339 < 2.8965 Do not reject Nullhypothesis

A&C 2.2678 < 2.8965 Do not reject Null hypothesis

B&C 1.1339 < 2.8965 Do not reject Null hypothesis

f)

Here we can conclude that all of the means are same at the 1% level of significance.