SOLUTION: A class survey in a large class for first-year college students asked, "About how many hours do you study during a typical week?" The mean response of the 492 students was x = 15

Question 1061709: A class survey in a large class for first-year college students asked, "About how many hours do you study during a typical week?" The mean response of the 492 students was
x = 15.3 hours. Suppose that we know that the study time follows a Normal distribution with standard deviation
σ = 8.5 hours in the population of all first-year students at this university.
Regard these students as an SRS from the population of all first-year students at this university. Does the study give good evidence that students claim to study more than 15 hours per week on the average?
(1) State null and alternative hypotheses in terms of the mean study time in hours for the population.
a) H0: μ = 15 hours
b) Ha: μ > 15 hours
c) H0: μ = 15 hours
d) Ha: μ ≠ 15 hours
e) H0: μ = 15 hours
f) Ha: μ < 15 hours
g) H0: μ = 15 hours
h) Ha: μ = 15 hours
(2) What is the value of the test statistic z? (Give your answer to two decimal places.)
z = _____

(3) What is the P-value of the test?
a) less than 0.001
b) between 0.001 and 0.01
c) between 0.01 and 0.025
d) between 0.025 and 0.05
e) larger than 0.05
Can you conclude that students do claim to study more than 15 hours per week on average?
a) Yes, there is sufficient evidence that the average amount of time students claim to study is more than 15 hours per week.
b) No, there is little evidence that the average amount of time students claim to study is more than 15 hours per week.

Answer by Boreal(15235) (Show Source): You can put this solution on YOUR website!
I don't like any of the first choices. This is a one-sided hypothesis test, whether the students study more than 15 hours.
it should be
Ho: mu <=15 hours
Ha: mu >15 hours.
A/B are not bad, but both hypotheses must together cover all situations.
C/D is best, but that tests for a change, not a unidirectional one.
------------------
The test statistic is z= (x bar-mean)/s/sqrt (n)
This is 0.3/8.5/sqrt (492)=0.3* sqrt(492)/8.5=0.78=z
------------------
The p-value is >0.05
NO, There is insufficient evidence to say that the average amount of time is >15 hours/wk.
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