Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $77 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.48. Over the first 39 days she was employed at the restaurant, the mean daily amount of her tips was $78.46. At the 0.01 significance level, can Ms. Brigden conclude that her daily tips average more than $77?

  

a.	State the null hypothesis and the alternate hypothesis.
	a.) H0: μ ≤ 77 ; H1: μ > 77 b.) H0: μ = 77 ; H1: μ ≠ 77 c.) H0: μ >77 ; H1: μ = 77 d.) H0: μ ≥ 77 ; H1: μ < 77

  

b.	State the decision rule.
	a.) Reject H0 if z < 2.33 b.) Reject H1 if z < 2.33 c.) Reject H1 if z > 2.33 d.) Reject H0 if z > 2.33

  

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

  

Value of the test statistic:

  

d.	What is your decision regarding H0?
	a.) Reject H0 b.) Do not reject H0

   

e.	What is the p-value? (Round your answer to 4 decimal places.)

  

p-value:

Answer 1

Solution :

This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   = 77

H_a : > 77

= 78.46

= 77

= 4.48

n = 39

= 0.01

Z = Z_0.01 = 2.33

c.) Reject H1 if z > 2.33

Test statistic = z

= ( - ) / / n

= (78.46 - 77) / 4.48 / 39

= 2.04

Test statistic = 2.04

This is the right tailed test .

P(z > 2.04) = 1 - P(z < 2.04) = 1 - 0.9793 = 0.0207

P-value = 0.0207

= 0.01

P-value >

Do not reject H0