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
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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