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Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $84 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $3.51. Over the first 46 days she was employed at the restaurant, the mean daily amount of her tips was $86.54. At the 0.10 significance level, can Ms. Brigden conclude that her daily tips average more than $84?

a. State the null hypothesis and the alternate hypothesis.

multiple choice 1

H0: μ ≥ 84; H1: μ < 84

H0: μ >84; H1: μ = 84

H0: μ ≤ 84; H1: μ > 84

H0: μ = 84; H1: μ ≠ 84

b. State the decision rule.

multiple choice 2

Reject H0 if z > 1.28

Reject H1 if z < 1.28

Reject H1 if z > 1.28

Reject H0 if z < 1.28

c. Compute the value of the test statistic. (Round your answer to 2 decimal places.)

Answer 1

(a) null hypothesis. We must have equality here so H0 must be mean = 84. H1 (what we are testing ) must be mean > 84. So actually none of the given answers is valid.

At a 0.10 significance level the critical value of Z is 1.28 (we are doing a right tailed test here as the alternative hypothesis has > sign. So Z = 1.28 because the area to the left of Z=1.28 is 1 - 0.10 = 0.90)

If the test statistic is > 1.28 we will reject H0.

Test statistic = (86.54 - 84) / (3.51 / ·√46) = 4.91