Using Excel, the solution is as follows:
Store A Store B Store C
29 4 2
20 45 22
8 34 17
30 12 6
40 9 39
25 43
Anova: Single Factor
SUMMARY
Groups Count Sum Average Variance
Store A 6 152 25.33 115.87
Store B 6 147 24.50 333.90
Store C 5 86 17.20 213.70
ANOVA
Source of Variation SS df MS F P-value
Between Groups 212.25 2 106.12 0.479 0.629
Within Groups 3103.63 14 221.69
Total 3315.88 16
•Perform a one-way analysis of variance on these data, assuming a = 0.05:
See above.
•State the null and alternate hypotheses:
H0: The amount purchased at each store equal.
H1: The amount purchased at each store is not equal (i.e., greater than or less than) among the stores.
•Calculate the sums of squares SS(total), SS(factor), and SS(error)
See 'SS' column above. SS(error) is SS labeled 'Within Groups.'
•Calculate the degrees of freedom df(total), df(factor), and df(error)
df(total = n-1 = 17-1 = 16
df(factor) = k-1 = 3-1 = 2
df(error) = (n-1) - (k-1) = n - k = 17-3 = 14
•Calculate the mean square for factor, and the mean square for error
See column labeled 'MS' above.
•Calculate the F-statistic
See column labled 'F' above.
•Determine the critical value(s)
You can look these up in an F table.
•State your decision: Should the null hypothesis be rejected?
No. You should accept the null hypothesis because P-value is NOT less than 0.05. These data indicate the shopper spent about the same at each store.
Done