Question
In: Statistics and Probability
The calculator at this link will allow you to perform a one-way chi-square or “goodness of...
The calculator at this link will allow you to perform a one-way chi-square or “goodness of fit test”: http://vassarstats.net/csfit.html. Fifty students can choose between four different professors to take Introductory Statistics. The number choosing each professor is shown below. Use the calculator above to test the null hypothesis that there is no preference for professors -- that there is an equal chance of choosing each of them. Report your results including chi-square, degrees of freedom, p-value and your interpretation. Use an alpha level of .05. Be careful not to over interpret – state only what the test result tells you.
Professor N
Dr. Able 20
Dr. Baker 8
Dr. Chavez 14
Dr. Davis 8
Solutions
Expert Solution
as per given instruction i have performed the chi square test.
the findings are as followed:-
interpretation:
p value=0.0477<0.05, so,we have enough evidence to reject our null hypothesis.
so,here is not sufficient evidence to claim the null hypothesis,i.e, there is no preference for professors -- that there is an equal chance of choosing each of them.
***if you have any doubt regarding the problem please write it in the comment box.if you are satisfied please give me a LIKE if possible
orchestra answered 1 year agoRelated Solutions
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This is the information provided.
Column1
Column2
Column3
Approving
Disapproving
Unsure/NoData
85
10
3
85
10
3
87
8
4
84
11
3
86
9
3
87
8
3
88
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2
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Approving
49
51
51
53
52
51
51
50
51
49
50
50
52
51
49
48
48
48
50
50
50
47
49
52
52
51
49
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51
51
49
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49
48
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16
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