Question

At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $80 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $2.81. Over the first 42 days she was employed at the restaurant, the mean daily amount of her tips was $82.74. At the 0.10 significance level, can Ms. Brigden conclude that her daily tips average more than $80?

1. State the null hypothesis and the alternate hypothesis.
2. State the decision rule.
3. Compute the value of the test statistic.
4. What is your decision regarding H₀?
5. What is the p-value?

Answer 1

a)

H0: = 80

Ha: > 80

b)

At 0.10 level, critical value is 1.645

Decision rule = Reject H0 if test statistics z > 1.645.

c)

Test statistics

z = - / / sqrt(n)

= 82.74 - 80 / 2.81 / sqrt(42)

= 6.32

d)

Since test statistics falls in rejection region, we have sufficient evidence to reject H0.

We conclude at 0.05 level that we have enough evidence to support the claim.

e)

p-value = P( Z > z)

= P( Z > 6.32)

= 0