a. A consumer organization wants to know whether there is a difference in the pr
ID: 3268432 • Letter: A
Question
a. A consumer organization wants to know whether there is a difference in the price of a particular toy at three different types of stores. The price of the toy was checked in a sample of five discount stores, five variety stores, and five department stores. Is there is any significant difference among these stores on the price. Test this claim at 0.05 significance level and use the 5-step procedure for testing a hypothesis. Price in RM are shown below. Discount stores: 12 13 14 12 15 Variety stores: 15 17 14 18 17 Departmental stores: 19 17 16 20 19 b. A financial analyst wants to compare the turnover rates, in percent, for shares of oil and gas related stocks versus other stocks. Analyst selected 32 oil related stocks and 49 other stocks. The mean turnover rate of oil related stocks is 31.4 percent and the population standard deviation is 5.1 percent. For the other stocks, the mean rate is 34.9 percent and the population standard deviation is 6.7 percent. Is there any significant difference in the turnover rates of the two types of stock at 1% level of significance? i. Construct null and alternative hypothesis. ii. Compute the value of test statistic. iii. What is your decision regarding the null hypothesis?Explanation / Answer
a.
H0: 1 = 2 = 3
H1: j not all equal
Discount stores= 12 13 14 12 15, X1=13.2
Variety stores= 15 17 14 18 17, X2=16.2
Departmental stores= 19 17 16 20 19, X3=18.2
X= 15.87, n= 15, c= 3
SSA= 5(13.2-15.87)2+5(16.2-15.87)2+5(18.2-15.87)2=63.33
SSW= (12-13.2)2……..+ (19-18.2)2=28.4
MSA=63.33/ (3-1) =31.67
MSW=28.4/ (15-3) =2.37
FSTAT=31.67/2.37=13.38
b)
1)
H0: 1 - 2 = 0 i.e. (1 = 2)
H1: 1 - 2 0 i.e. (1 2)
2)
Assuming population variances are equal, we would have to calculate pooled-variance t-Test
Sp^2= (n1-1)S1^2+(n2-1)S2^2/(n1-1)+(n2-1)
= (32-1)*5.1^2+(49-1)*6.7^2/31+48
=37.48
tSTAT=(X1-X2)-(µ1-µ2)/Sp^2(1/n1+1/n2)
=(31.4-34.9)-0/37.48(1/32+1/49)
=-3.5/1.39
=-2.515
3)
tCRIT is =+/-2.64 and hence cannot reject the null hypothesis because the value of test statistic doesnot fall within the rejection region.
Anova: Single Factor SUMMARY Groups Count Sum Average Variance Discount stores 5 66 13.2 1.7 Variety stores 5 81 16.2 2.7 Departmental stores 5 91 18.2 2.7 ANOVA Source of Variation SS df MS F P-value F crit Between Groups 63.33333333 2 31.66666667 13.38028169 0.000880536 3.885293835 Within Groups 28.4 12 2.366666667 Total 91.73333333 14Related Questions
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