Question 251241
Using Excel, the solution is as follows:
<pre>
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			
</pre>

•Perform a one-way analysis of variance on these data, assuming a = 0.05:
See above.
<br />
•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.
<br />
•Calculate the sums of squares SS(total), SS(factor), and SS(error)
See 'SS' column above.  SS(error) is SS labeled 'Within Groups.'
<br />
•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
<br />
•Calculate the mean square for factor, and the mean square for error
See column labeled 'MS' above.
<br />
•Calculate the F-statistic
See column labled 'F' above.
<br />
•Determine the critical value(s)
You can look these up in an F table.
<br />
•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.
<br />
Done