Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $71 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $4.48. Over the first 33 days she was employed at the restaurant, the mean daily amount of her tips was $73.33. At the 0.05 significance level, can Ms. Brigden conclude that her daily tips average more than $71?

State the null hypothesis and the alternate hypothesis.

H0: μ ≥ 71; H1: μ < 71
H0: μ >71; H1: μ = 71
H0: μ ≤ 71; H1: μ > 71
H0: μ = 71; H1: μ ≠ 71

State the decision rule.

Reject H0 if z > 1.65
Reject H1 if z < 1.65
Reject H1 if z > 1.65
Reject H0 if z < 1.65

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

What is your decision regarding H0?

Reject H0
Do not reject H0

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

Answer 1

Soution :

The null and alternative hypothesis is ,

=71

= 73.33

= 4.48

n = 33

This will be a right tailed test because the alternative hypothesis is showing a specific direction

This is the right tailed test .

The null and alternative hypothesis is ,

H0 :   = 71

Ha : > 71

= 0.05

Z =Z 0.05 = 1.65

Reject H0 if z > 1.65

Test statistic = z

= ( - ) / / n

= (73.33 - 71 / 4.48 / 33

= 2.99

2.99 > 1.65

Test statistic > Critical value

Reject H0

p(Z > 2.99 )

= 1-P (Z < 2.99 )

= 1- 0.9986

= 0.0014

P-value = 0.0014