Question

A marketing research firm wishes to compare the prices charged by two supermarket chains—Miller’s and Albert’s. The research firm, using a standardized one-week shopping plan (grocery list), makes identical purchases at 10 of each chain’s stores. The stores for each chain are randomly selected, and all purchases are made during a single week. It is found that the mean and the standard deviation of the shopping expenses at the 10 Miller’s stores are x1⎯⎯⎯⎯?=?$129.94x1¯?=?$129.94 and s₁= 1.55. It is also found that the mean and the standard deviation of the shopping expenses at the 10 Albert’s stores are x2⎯⎯⎯⎯?=?$107.92x2¯?=?$107.92 and s₂= 1.75.

(a) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.)

Test statistic          

(b) Calculate the critical value. (Round your answer to 2 decimal places.)

Critical value          

(c) At the 0.05 significance level, what it the conclusion?

Answer 1

here we use t-test with

null hypothesis H0:µ1=µ2 and alternate hypothesis H1: µ1≠µ2

(a) statistic t=|(mean1-mean2)|/((s_p*(1/n1 +1/n2)^1/2) =29.79 with df is n=n1+n2-2-18 and

s_p²=((n1-1)s1²+(n2-1)s2²)/n=2.7325

(b) critical value=2.10 ( two tailed t(0.05,18)=2.10)

(c)since calcuated/observed t=29.79 is more than critical t=2.10,so we reject the null hypothesis and conclude that there is difference in price between the two stores.

or , 2-tailed p-value is less than typical level of significance , we reject the null hypothesis and conclude that there is difference in price between the two stores

t-test
	sample	mean	s	sample variance=s2	sample size=n	(n-1)s2
	Miller	129.9400	1.5500	2.4025	10	21.6225
	albert	107.9200	1.7500	3.0625	10	27.5625
	difference=	22.0200	sum=	5.4650	20	49.1850
	sp2=	2.7325
	sp=	1.6530
	SE=(sp*(1/n1 +1/n2)1/2)=	0.7393
	t=	29.7867
two tailed	p-value=	0.0000
two tailed critical	t(0.05)	2.1009