Question 1039543: I'm still having trouble with figuring out these formula's to do these problems. I must be missing something! Any help would be greatly appreciated!
At the time she was hired as a server at the Grumney Family Restaurant, Beth Brigden was told, “You can average $72 a day in tips.” Assume the population of daily tips is normally distributed with a standard deviation of $2.45. Over the first 34 days she was employed at the restaurant, the mean daily amount of her tips was $73.07. At the 0.10 significance level, can Ms. Brigden conclude that her daily tips average more than $72?
H0: μ ≤ 72 ; H1: μ > 72
Reject H0 if z > 1.28
Compute the value of the test statistic. (Round your answer to 2 decimal places.)
What is the p-value? (Round your answer to 4 decimal places.)
Answer by rothauserc(4718)   (Show Source):
You can put this solution on YOUR website! Typically we only use = in stating H0, so I would say that H0 should be
:
H0: x = 72 and H1: x > 72
:
Since H1 contains a strict inequality, we use a one-tailed test
:
Since we are using a 10% significance level, our alpha = 0.10
:
Since the population is normally distributed, we use the standard normal distribution and its z-tables
:
The standard error(SE) is standard deviation / square root of sample size
SE = 2.45 / 5.83 = 0.42
:
******************************************
test z-value = (73.07 - 72) / 0.42 = 2.55
******************************************
:
:
an alpha of 0.10 corresponds to a z-value of 1.28
:
********************************************
since our test z-value > 1.28, we reject H0
********************************************
:
a z-value of 2.55 has a p-value of 0.0054
note that since the p-value < 0.10, we reject H0
:
|
|
|
