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¯¯¯¯?=?$114.14x1¯?=?$114.14 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¯¯¯¯?=?$113.14x2¯?=?$113.14 and s₂= 1.67.

(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 0.02 significance level, what it the conclusion?

	Fail to reject
	Reject

Suppose two independent random samples of sizes n₁ = 9 and n₂ = 7 that have been taken from two normally distributed populations having variances σ12σ21 and σ22σ22 give sample variances of s₁² = 117 and s₂² = 19.

(a) Test H₀: σ12σ21 = σ22σ22 versus H_a: σ12σ21 ≠≠ σ22σ22 with σσ = .05. What do you conclude? (Round your answers to 2 decimal places.)

F =     F.025 =

(Click to select)RejectDo not reject H0:σ12σ21 = σ22σ22

(b) Test H₀: σ12σ21< σ22σ22versus H_a: σ12σ21 > σ22σ22 with σσ = .05. What do you conclude? (Round your answers to 2 decimal places.)

F =     F.05 =

(Click to select)Do not rejectReject H0: σ12σ21 < σ22

Answer 1

a) The test statistic

b)

        

       

At 0.02 significance level, the critical values are +/- t_{0.01, 15} = +/- 2.602

c) Since the test statistic value is not greater than the positive critical value(1.57 < 2.602), so we should not reject the null hypothesis.

Fail to reject H₀.

2)a) The test statistic F =

                                    

                                    

F(0.975, 8, 6) = 0.21

F(0.025, 8, 6) = 5.60

Since the test statistic value doesn't lie between the critical values, so we should reject the null hypothesis.

Reject H₀:

b) F(0.05, 8, 6) = 4.15

Since the test statistic value is greater than the critical value(6.16 > 4.15), so we should reject the null hypothesis.

Reject H₀: