Question 1062009: A sample of 48 observations is selected from a normal population. The sample mean is 22, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.05 significance level.
H0 : μ ≤ 21
H1 : μ > 21
a. Is this a one- or two-tailed test?
"One-tailed"-the alternate hypothesis is greater than direction.
"Two-tailed"-the alternate hypothesis is different from direction.
b. What is the decision rule? (Round your answer to 3 decimal places.)
H0, when z >
c. What is 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
Reject
There is evidence to conclude that the population mean is greater than 21.
e. What is the p-value? (Round your answer to 4 decimal places.)
p-value
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! One-tailed test. You are interested in one side of the difference, not just whether there is a difference.
Reject when z>1.645, which is the one-sided 95th percentile.
The test stat is z(0.95)=(x bar-mean)/sigma/sqrt(48)
=(22-21)*sqrt (48)/6
=1*6.928/6=1.16
Fail to reject the null hypothesis. Insufficient evidence to state that the mean is greater than 21.
p=0.1241, from table or calculator.
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