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 $121.92 and $1.40, respectively. It is also found that the mean and the standard deviation of the shopping expenses at the 10 Albert's stores are $114.81 and $1.84, respectively. Assuming normality, test to see if the corresponding population standard deviations differ by setting a equal to .05. Is it reasonable to use the equal variances procedure to compare population means? (a) Calculate the value of the test statistic. (Do not round intermediate calculations. Round your answer to 2 decimal places.) Test statistic 9.73 (b) Calculate the critical value. (Round your answer to 2 decimal places.) Critical value 2.10 (c) At the 0.05 significance level, what it the conclusion? O Reject Ho O Fail to reject Ho

Answer 1

Solution:

a. 9.73

b.2.11

c. Reject Ho

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Explanation:-

Null Hypothesis (Ho): $\mu_{1}=\mu_{2}$

Alternative Hypothesis (Ha): $\mu_{1}\neq\mu_{2}$

Test Statistics

$t = \frac{(\bar{X}_{1}-\bar{X}_{2})-(\mu_{1}-\mu_{2})}{\sqrt{s_{1}^2/n_{1}+s_{2}^2/n_{2}}}$

$t = \frac{(121.92-114.81)-(0)}{\sqrt{1.40^2/10+1.84^2/10}}$

t = 9.725

Degrees of freedom, $df$ is calculated using the formula

2 2 2 п, п, df 2 2 1 1 n1-1n n2-1n п,- п,

$df = \frac{(1.4^2/10+1.84^2/10)^2}{1/9(1.4^2/10)^2+1/9(1.84^2/10)^2}$

$df =$16.8 ~ 17

Using t-tables, the critical value at a/2 = 0.05/2 = 0.025 with 17 df is 2.11.

Since test statistics is greater than the critical value, we reject the null hypothesis.