Guitars R US has three stores located in three different areas. Random samples o
ID: 3361750 • Letter: G
Question
Guitars R US has three stores located in three different areas. Random samples of the sales of the three stores (in $1000) are shown below. Store 1 80 75 76 89 80 Store 2 85 86 81 Store 79 85 Show the complete ANOVA table. (Do not use Excel) State the null and alternative hypotheses to determine whether there is a significant difference in the means of the three store. a. b. The null hypothesis is to be tested at 95% confidence. Determine the critical value for this test. What do you conclude? Determine the p-value and use it for the test. (Use FDst Function in Excel to find p- value) c. d.Explanation / Answer
Ans:
a)Calculations for Anova tablle:
SST=80914-(984^2/12)=226
SSTR==(400^2/5)+(332^2/4)+(252^2/3)-(984^2/12)=36
SSE=SST-SSTR=226-36=190
df(treatments)=3-1=2
df(error)=12-3=9
MS(between)=36/2=18
MS(within)=190/9=21.11
F=18/21.11=0.853
b)
H0:There is no difference in sales of 3 stores.
Ha:There is difference in sales of 3 stores.
c)Signifcance level=0.05
Fcritical=FINV(0.05,2,9)=4.256
As,F=0.853<4.256,we fail to reject null hypothesis.
There is no significant difference in sales of 3 stores.
d)P-value=FDIST(0.853,2,9)=0.458
As,p-value>0.05,we fail to reject null hypothesis.
There is no significant difference in sales of 3 stores.
Store 1 Store 2 Store 3 80 85 79 75 86 85 76 81 88 89 80 80 Total= 400 332 252 Overall total= 984 square's of all observations 6400 7225 6241 5625 7396 7225 5776 6561 7744 7921 6400 6400 Total=80914 SST= 226 SSTR= 36 SSE= 190 DF(treatments)= 2 DF(error)= 9 SS df MS F p-value Between groups(treatments) 36 2 18 0.853 0.458 within groups(Error) 190 9 21.11 Total 226 11Related Questions
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