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.

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