Question

The Mean Corporation would like to invest in the booming health food industry. It is considering the creation of a health-drink franchise called Goose Juice. The investment department of the Mean Corporation wants to investigate the feasibility of this venture by examining the profits of similar franchises. It believes that the venture will be feasible if an average annual profit of more than $89,000 can be expected from each Goose Juice that is opened. It is known that the annual profits earned by health-drink franchises has a population standard deviation of $8,200.

The Mean Corporation's statisticians would like to construct a hypothesis test for the mean annual profit (μ) earned by health-drink franchises. A random sample of 70 franchises were chosen and their annual profit for the previous financial year was recorded. The mean annual profit for the sample was calculated as $90,250. The hypotheses that will be used by the statisticians are H₀: μ = 89,000 and H_a: μ > 89,000.

A.) Calculate the test statistic (z) that corresponds to the sample and hypotheses. Give your answer to 3 decimal places.

Z =

B.) Using the test statistic for the Mean Corporation's hypothesis test and level α = 0.05, the Mean Corporation should (accept, reject, not reject) the null hypthesis.

C.) If the sample size is increased to 170 (but the sample mean remains unchanged), the Mean Corporation should (accept, reject, not reject) the null hypothesis.

Answer 1

The Hypothesis:

H0: = 30,000: The mean household income is equal to 30,000 dollars.

Ha: > 30,000: The mean household income is equal to 30,000 dollars.

This is a 1 tailed test (Right tailed)

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A) The Test Statistic: The test statistic is given by the equation:

Z observed = 1.275

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B) The p Value: The p value for Z = 1.275, p value = 0.1012

The Critical Value: The critical value (1 Tail) at = 0.05, tcritical= +1.645

The Decision Rule:   If Zobserved is > Zcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision: Since Zobserved (1.275) is < than Z critical, We Do not Reject H0.

Also since P value (0.1012) is > (0.05) , We do not Reject H0.

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C) When n = 170

The Test Statistic:

Z observed = 1.988

The p Value: The p value for Z = 1.988, p value = 0.0234

The Critical Value: The critical value (1 Tail) at = 0.05, tcritical= +1.645

The Decision Rule:   If Zobserved is > Zcritical, Then reject H0.

Also if P value is < , Then Reject H0.

The Decision: Since Zobserved (1.988) is > than Z critical, We Reject H0.

Also since P value (0.0234) is < (0.05) , We Reject H0.

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