Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $85 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.50. Over the first 47 days she was employed at the restaurant, the mean daily amount of her tips was $86.06. At the 0.01 significance level, can Ms. Brigden conclude that her daily tips average more than $85?

  

a.

State the null hypothesis and the alternate hypothesis.

	H0: μ = 85 ; H1: μ ≠ 85
	H0: μ ≥ 85 ; H1: μ < 85
	H0: μ >85 ; H1: μ = 85
	H0: μ ≤ 85 ; H1: μ > 85

  

b.

State the decision rule.

	Reject H0 if z < 2.33
	Reject H1 if z < 2.33
	Reject H1 if z > 2.33
	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?

	Do not reject H0
	Reject H0

   

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

  

p-value

Answer 1

Solution :

= 85

= 86.06

= 4.50

n = 47

a ) This is the right tailed test .

The null and alternative hypothesis is ,

H₀ :   ≤ 85

H_a :   > 85

b

Test statistic = z

= ( - ) / / n

= (86.06- 85) /4.50 / 47

= 1.61

z = 1.61

b ) it is observed that z = 1.615 ≤ zc ​= 2.33, it is then concluded that the null hypothesis is not rejected

c ) Test statistic z is = 161

d ) Do not reject H₀

e ) P(z >1.61) = 1 - P(z < 1.61 ) = 1 - 0.9463 = 0.0537

P-value = 0.0537