SOLUTION: Please help me solve this problem.. A sample of 32 observations is selected from a normal population. The sample mean is 32, and the population standard deviation is 6. Conduct

Question 1042849: Please help me solve this problem..
A sample of 32 observations is selected from a normal population. The sample mean is 32, and the population standard deviation is 6. Conduct the following test of hypothesis using the 0.05 significance level.

H0 : μ ≤ 29
H1 : μ > 29

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.)

(Click to select)RejectDo not reject 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?

Reject
Do not reject

There is (Click to select)insufficientsufficient evidence to conclude that the population mean is greater than 29.

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

p-value

Answer by rothauserc(4718) (Show Source): You can put this solution on YOUR website!
a) this is a one-tailed test since H1 does not contain = sign
:
b) At a 5% significance level, the critical value for a one tailed test is found from the table of z-scores to be 1.645
Reject when z > 1.645
:
c) standard error of the mean (SE) = population standard deviation / square root(sample size)
SE = 6 / square root(32) approx 1.061
z value = (32 - 29) / 1.061 approx 2.83
:
d) reject H0 since 2.83 > 1.645
:
e) a z-score of 2.83 has a p-value of 0.0026
:

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