Suppose Sally was interested in determining whether there is a significant diffe

Question

Suppose Sally was interested in determining whether there is a significant difference in the price charged for different foods at three grocery stores (Provigo, Metro, and Adonis). Sally selected five foods and recorded the price for each item in each store. After calculating the appropriate values, the two-way analysis of variance table was created:

Store SS: 0.9053, MS: 0.4527
Food item SS: 1823.348, MS: 455.8371
Total SS: 1828.058

a.) Based on the test statistic (alpha of 0.05) for the price of foods, what does this tell you about the appropriateness of the experimental design? Explain your reasoning. Be sure to provide the critical value and calculated value.

b) Is there a significant difference in the prices charged by the three stores at the 0.05 level of significance?

---------Please explain the steps, it is my second time posting this question.Thank you!-----
This is the answer posted before.

From above information we got

SSerror = 3.8047

d.f. for store = 2

d.f. for Food = 4

d.f. for error = 8

So MSerror = 3.8047 / 8 = 0.4756

a) The hypothesis are ,

H0 : Prices of food are appropriate.

H1 : Prices of food are not appropriate.

                                 = 958.45

Critical value = F0.05, 4,8 = 3.84

Here calculated value of F > critical value of F

Hence we reject null hypothesis.

Conclusion : Prices of food are not appropriate.

b) The hypothesis are :

H0 : There is no significant difference in the prices charged by the three stores.

H1 : There is a significant difference in the prices charged by the three stores.

                                 = 0.9519

Critical value = F0.05,2,8 = 4.46

Here calculated value of F < critical value of F

There is no significant difference in the prices charged by the three stores.

Explanation / Answer

All formulas and calculations are included in the updated solution.

We are given

SS store = 0.9053

MS store = 0.4527

SS food = 1823.348

MS store = 455.8371

Total SS = 1828.058

The formula for SSerror is given as below:

SSerror = Total SS – SS food – SS store

Now, plug all values in the above formula and calculate the SSerror given as below:

SSerror = 1828.058 – 1823.348 – 0.9053

SSerror = 3.8047

There are total three stores, so n = 3

d.f. for store = n – 1 = 3 – 1 = 2

d.f. for store = 2

There are five types of foods, so n = 5

d.f. for food = n – 1 = 5 – 1 = 4

d.f. for Food = 4

d.f. for error = total d.f. – d.f. for food – d.f. for store

total d.f. = 5*3 – 1 = 15 – 1 = 14

d.f. for error = 14 – 4 – 2 = 14 – 6 = 8

d.f. for error = 8

MSerror = SS error / d.f. error = 3.8047 / 8 = 0.4756

MS food = SS food / d.f. food = 1823.348 / 4 = 455.8371

MS store = SS store / d.f. store = 0.9053 / 2 = 0.4527

H0 : Prices of food are appropriate.

H1 : Prices of food are not appropriate.

Calculated test statistic = F = MS food / MS error = 455.8371 / 0.4756 = 958.45

Critical value = F0.05, 4,8 = 3.84 by using F-distribution or F-table

Decision Rule: We reject the null hypothesis if F test statistic > critical value and we do not reject the null hypothesis if F-test statistic < critical value

Here calculated value of F test statistic> critical value of F

Hence we reject null hypothesis.

Conclusion: Prices of food are not appropriate.

b) The hypothesis are :

H0 : There is no significant difference in the prices charged by the three stores.

H1 : There is a significant difference in the prices charged by the three stores.

Calculated value = F = MS store / MS error = 0.4527 / 0.4756 = 0.9519

Critical value = F0.05,2,8 = 4.46

Here calculated value of F < critical value of F

There is no significant difference in the prices charged by the three stores.

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