Question

A four-year university has decided to implement a new approach to teaching Statistics. Full-time and adjunct...

A four-year university has decided to implement a new approach to teaching Statistics. Full-time and adjunct professors were surveyed to determine whether they preferred the traditional lecture approach or a computer approach to teaching Statistics.

status prefers lectures prefers computers no preference
full time 15 13 2
adjunct 11 22 10

1. For a hypothesis test to test the independence of opinion between the two groups, identify the correct alternate hypothesis

a. H0: The opinion is independent of status

b. HA: The opinion is independent of status

c. HA: The opinion is dependent upon status

d. HA: The status is dependent upon opinion

2. For a hypothesis test to test the independence of opinion between the two groups, find the expected value (from the expected matrix) of a full-time professor who prefers lecture. i.e. E1,1. Round to 3 decimal places.


a. 20.616

b. 10.685

c. 14.384

d. 15.315

3. For a hypothesis test to test the independence of opinion between the two groups, calculate the p-value and round to 3 significant decimal places

4. For the status versus opinion data,, for α=5%, the correct decision about H0 is..

a. Accept H0

b. Fail to reject H0

c. Reject H0

d.Fail to Accept H0

e. Accept HA

Homework Answers

Answer #1

Ans:

1)Correct option is:

HA: The opinion is dependent upon status.

2)row sum=15+11=26

column sum=15+13+2=30

Overall sum=73

Expected count(full time,prefer lecture)=26*30/73=10.685

Option b is correct(10.685)

3)

Obsered(fo)
prefer lecture prefer computer no preference Total
Full time 15 13 2 30
adjunct 11 22 10 43
Total 26 35 12 73
Expected(fe)
prefer lecture prefer computer no preference Total
Full time 10.685 14.384 4.932 30
adjunct 15.315 20.616 7.068 43
Total 26 35 61
Chi square=(fo-fe)^2/fe
prefer lecture prefer computer no preference Total
Full time 1.743 0.133 1.743 3.618
adjunct 1.216 0.093 1.216 2.524
Total 2.958 0.226 2.958 6.143

Test statistic:

Chi square=6.143

df=(2-1)*(3-1)=2

p-value=CHIDIST(6.143,2)=0.046

4)

As,p-value<0.05,Reject H0

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