Question 1090847: Emma's On-the-Go, a large convenience store that makes a good deal of money from magazine sales, has three possible locations in the store for its magazine rack: in the front of the store (to attract "impulse buying" by all customers), on the left-hand side of the store (to attract teenagers who are on that side of the store looking at the candy and soda), and in the back of the store (to attract the adults searching through the alcohol cases). The manager at Emma's experiments over the course of several months by rotating the magazine rack among the three locations, choosing a sample of 47 days at each location. Each day, the manager records the amount of money brought in from the sale of magazines.
Below are the sample mean daily sales (in dollars) for each of the locations, as well as the sample variances:
Group: Front
Sample size: 47
Sample mean: 215.2
Sample variance: 270.3
Group: Left-hand side
Sample size: 47
Sample mean: 212.0
Sample variance: 456.3
Group:Right-hand side
Sample size: 47
Sample mean: 219.3
Sample variance: 486.6
Suppose that we were to perform a one-way, independent-samples ANOVA test to decide if there is a significant difference in the mean daily sales among the three locations.
Answer the following, carrying your intermediate computations to at least three decimal places and rounding your responses to at least one decimal place.
What is the value of the "between groups"and "within group" mean square that would be reported in the ANOVA test?
Answer by jim_thompson5910(35256)   (Show Source):
You can put this solution on YOUR website!
There's a typo in the problem. It first mentions the back of the store, but then never mentions it again when it comes to the table of values. Instead it brings up the right side of the store. I'm going to assume that "right side of the store" should be "back of the store"
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I'm going to redefine the labels
Group Front will be changed to Group 1
Group Left will be changed to Group 2
Group Back (or Group Right) will be changed to Group 3
We have this given information
Group 1:
Sample size: n1 = 47
Sample mean: xbar1 = 215.2
Sample variance: s1 = 270.3
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Group 2:
Sample size: n2 = 47
Sample mean: xbar2 = 212.0
Sample variance: s2 = 456.3
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Group 3:
Sample size: n3 = 47
Sample mean: xbar3 = 219.3
Sample variance: s3 = 486.6
Let L1,L2, and L3 be three lists with 47 items each
L1 = {a1,a2,...a47}
L2 = {b1,b2,...b47}
L3 = {c1,c2,...c47}
Summing the lists gives
Sum[L1] = a1+a2+...+a47
Sum[L2] = b1+b2+...+b47
Sum[L3] = c1+c2+...+c47
Which allows us to say
xbar1 = SampleMean[L1] = Sum[L1]/n1
xbar2 = SampleMean[L2] = Sum[L2]/n2
xbar3 = SampleMean[L3] = Sum[L3]/n3
For each equation, let's isolate the "Sum[...]" portion like so
xbar1 = Sum[L1]/n1
n1*xbar1 = Sum[L1]
Sum[L1] = n1*xbar1
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xbar2 = Sum[L2]/n2
n2*xbar2 = Sum[L2]
Sum[L2] = n2*xbar2
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xbar3 = Sum[L3]/n3
n3*xbar3 = Sum[L3]
Sum[L3] = n3*xbar3
So far we know that
Sum[L1] = n1*xbar1
Sum[L2] = n2*xbar2
Sum[L3] = n3*xbar3
Which allows us to compute the grand mean G
G = (Sum[L1]+Sum[L2]+Sum[L3])/(n1+n2+n3)
G = (n1*xbar1+n2*xbar2+n3*xbar3)/(n1+n2+n3)
G = (47*215.2+47*212.0+47*219.3)/(47+47+47)
G = 215.5
As expected, the grand mean is fairly close to the sample means
Other useful information is that we have
k = 3 groups (front, left, back)
N = n1+n2+n3 = 47+47+47 = 141 total items being sampled
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Now that we know the grand mean, we can use it to find the sum of squares between groups
SS_B = Sum of Squares Between groups
SS_B = n1*(xbar1-G)^2 + n2*(xbar2-G)^2 + n3*(xbar3-G)^2
SS_B = 47*(215.2-215.5)^2 + 47*(212.0-215.5)^2 + 47*(219.3-215.5)^2
SS_B = 1258.66
Similarly, we calculate the sum of squares within groups like so
SS_W = Sum of Squares Within groups
SS_W = (n1-1)*(s1)^2 + (n2-1)*(s2)^2 + (n3-1)*(s3)^2
SS_W = (47-1)*(270.3)^2 + (47-1)*(456.3)^2 + (47-1)*(486.6)^2
SS_W = 23,830,361.64
Use the SS_W and SS_B values to compute the corresponding MS_B and MS_W values
MS_B = Mean Square Between groups
MS_B = (SS_B)/(k-1)
MS_B =(1258.66)/(3-1)
MS_B = 629.33 which rounds to 629.3
MS_W = Mean Square Within groups
MS_W = (SS_W)/(N-k)
MS_W = (23,830,361.64)/(141-3)
MS_W = 172683.78 which rounds to 172683.8
So that wraps up the answer
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Side Note:
Taking the ratio of MS_B and MS_W gives
F = (MS_B)/(MS_W) = (629.33)/(172683.78) = 0.0036444071354
where F is the F statistic
Use a calculator or table to find the F critical value to be 3.0617
The fact that this F statistic (0.0036444071354) is smaller than the F critical value (3.0617) indicates that we fail to reject the null hypothesis. The F statistic is not in the rejection region. We don't have enough statistically significant evidence to overturn the null. We have no choice but to conclude that mu1 = mu2 = mu3. So there is no statistical difference between the three groups.
Alternatively, you can use a calculator or table to find the p value to be 0.9964 which is very large. This is more evidence that we fail to reject the null. You can only reject the null if the p value is less than alpha.
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