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¯¯¯¯?=?$115.41x1¯?=?$115.41 and s₁= 1.12. It is also found that the mean and the standard deviation of the shopping expenses at the 10 Albert’s stores are x2¯¯¯¯?=?$115.27x2¯?=?$115.27 and s₂= 2.39.

(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

Answer:

The given information is,

For Miller's stores:

For Albert's stores:

level of significance,

Let, = population mean of Miller's stores

= population mean of Albert's stores

Hypothesis Testing:

Ans (a):

Calculation for test statistic:

The hypotheses for this test are as below:

There is no any significance difference in the means of the prices charged by two supermarket chains.

i.e.

and There is significance difference in the means of the prices charged by two supermarket chains.

i.e.

Under the assumption of equal variance, the t formula can be stated as

can be estimated by using following formula

From the formula for pooled standard deviation, we get

Putting this value in formula for t, we get

Therefore, the value of test statistic t = 0.1677.

Ans (b):

Here, the level of significance has been specified as 0.05. Then the decision rule is given as:

Value of alpha is 0.05 and the degrees of freedom is 10+10-2=18. The tabular t value is t_0.025,18 = 2.101. The null hypothesis will be rejected if the observed value of t is less than -2.101 or greater than +2.101.

So, critical value is t_0.025,18 = 2.101

Ans (c):

statistical conclusion:

AT the 0.05 significance level, the critical t value from the t distribution table is t_0.025,18 = 2.101. So the observed t value 0.1677 is less than tabular t value +2.101. Hence, the null hypothesis is accepted and alternative hypothesis is rejected. This result shows that there is no any significance difference in the means of the prices charged by two supermarket chains. That is, the prices charged by two supermarket chains are same.