Question

For the following scenarios create an appropriate decision rule.

a. A manufacturer wants to know if there is a difference in the average number of defects produced on the day shift and the night shift. He conducts a hypothesis test at the .05 significance level with the following hypotheses:

H0: µd = 0

H1: µd ≠ 0

A paired sample of 10 days of the two shifts yields a mean difference of 4.32 with standard deviation 3.4.

b. A real estate agent wants to compare the variation in the selling price of homes on the oceanfront with those one to three blocks from the ocean. A sample of 25 oceanfront homes sold within the last year revealed the standard deviation of the selling prices was $32,100. A sample of 18 homes, also sold within the last year, that were one to three blocks from the ocean revealed that the standard deviation was $23,400. Conduct a hypothesis test at the .05 significance level to test the claim that there is greater variation in the selling price of oceanfront homes than in those one to three blocks from the ocean.

(Please make sure you are focusing on the numbers mentioned above in the question)

Answer 1

(a) From the research hypothesis, this is a two tailed test

The degrees of freedom = n - 1 = 10 - 1 = 9

At = 0.05, the critical values are -2.262 and +2.262

Therefore the rejection region / rule is Reject H0, if t test is > 2.262 or if t test is < -2.262

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(b) We use the F test for two variances here

For this we conduct an F test for 2 variances (Since we want to test variation)

Given

s₁ = 32100 s₁² = 1030410000, n₁ = 25, df₁ = 25-1 = 24

s₂ = 23400, s₂² = 547560000, n₂ = 18, df₂ = 18 -1 = 17

The Hypothesis:

H0:

Ha:

This is a Right Tailed Test.

The Test Statistic:

F = s₁² / s₂² = 1030410000 / 547560000 = 1.88

The Critical Value at = 0.05, df1 = 24, df2 = 17 is 2.19

The p value for F= 1.88, df1 = 24, df2 = 17 is 0.0917

The Decision Rule: If F observed is > F critical, Then reject H0.

If P value is < , Then Reject H0

The Decision: Since F observed (1.88) is < F critical (2.19), We Fail to reject H0.

Since p value (0.0917) is > (0.05), We Fail to Reject H0.

The Conclusion: There is isn’t sufficient evidence at the 95% significance level to conclude that there is a difference in variance between selling price on the oceanfront between 1 and 3 blocks from the ocean.