Academic Integrity: tutoring, explanations, and feedback — we don’t complete graded work or submit on a student’s behalf.

In order to perform the following hypothesis test, report the test statistic, re

ID: 3051230 • Letter: I

Question

In order to perform the following hypothesis test, report the test statistic, reject region and state the test conclusion according to the experiment.
Ho: =12   Ha: <12.

a) =0.05. Suppose the population standard deviation =10, current sample has x =15,s=3,n=25. Use the Z (T/Z) method?

b) =0.05. Suppose the population standard deviation =14, current sample has x =15,s=3,n=25 Use the Z (T/Z) method?
c) =0.001. Suppose the population standard deviation =10, current sample has x =15,s=3,n=100. Use the Z (T/Z) method?
d) =0.001. Suppose the population standard deviation =10, current sample has x =9,s=3,n=100. Use the Z (T/Z) method?
e) =0.05. Current sample has x =15,s=3,n=25. Use the T (T/Z) method?
f) =0.1. Current sample has x =12,s=10,n=25. Use the T (T/Z) method?
g) =0.1. Current sample has x =7,s=20,n=30. Use the T (T/Z) method?

Explanation / Answer

a)

Test statistics, t = (xbar - mu)/(sigma/sqrt(n))
t = (15 - 12)/(10/sqrt(25))
t = 1.5

df = 25 - 1 = 24

p-value = 0.9267
p-value is calculated using standard t-table or excel formula : =T.DIST(1.5,24,TRUE))

As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.

b)


Test statistics, t = (xbar - mu)/(sigma/sqrt(n))
t = (15 - 12)/(14/sqrt(25))
t = 1.0714

df = 25 - 1 = 24

p-value = 0.8527
p-value is calculated using standard t-table or excel formula : =T.DIST(1.0714,24,TRUE))

As p-value is greater than the significance level of 0.05, we fail to reject the null hypothesis.

c)


Test statistics, z = (xbar - mu)/(s/sqrt(n))
z = (15 - 12)/(10/sqrt(100))
z = 3

p-value = 0.9987 (using standard normal z table)

As p-value is greater than the significance level of 0.001, we fail to reject the null hypothesis.

d)


Test statistics, z = (xbar - mu)/(s/sqrt(n))
z = (9 - 12)/(10/sqrt(100))
z = -3

p-value = 0.0013 (using standard normal z table)

As p-value is greater than the significance level of 0.001, we fail to reject the null hypothesis.