The performance of students at a local community was recorded and categorized based on student’s faculty. The gathered data was then analyzed by a statistician and the results obtained using MINITAB are shown below: Expected counts are printed below observed counts                               Low             Average         High                  Total Science                    12                  30                52                      94                             (15.67)             (29.24)          (49.09) Humanities             14                   20                    30                   **                             (10.67)             (19.91)             (33.42) Law                               4                   6                    12                22                                    (*)                 (6.84)           (11.49)      Total                          30                   56                 94                 180 Chi-sq =      0.858   +      0.020   +    ***     +                     1.042           0.000    +   0.350     +                               0.030  +     0.104    +   0.023         = 2.600 i. Define the null and alternative hypotheses of the χ2 test for the above output. ii. find the missing values '*', '***', showing necessary working. iii. what is the degree of freedom for this test?       iv. Estimate the p-value for this test.  v. State the conclusion for the test. Give reason for your answer.

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Sub-Part (iv &  v)

The performance of students at a local community was recorded and categorized based on student’s faculty. The gathered data was then analyzed by a statistician and the results obtained using
MINITAB are shown below:

Expected counts are printed below observed counts

                              Low             Average         High                  Total

Science                    12                  30                52                      94

                            (15.67)             (29.24)          (49.09)

Humanities             14                   20                    30                   **

                            (10.67)             (19.91)             (33.42)

Law                               4                   6                    12                22

                                   (*)                 (6.84)           (11.49)     

Total                          30                   56                 94                 180

Chi-sq =      0.858   +      0.020   +    ***     +

                    1.042           0.000    +   0.350     +         

                     0.030  +     0.104    +   0.023         = 2.600

i. Define the null and alternative hypotheses of the χ2 test for the above output.

ii. find the missing values '*', '***', showing necessary working.

iii. what is the degree of freedom for this test?

 

    iv. Estimate the p-value for this test. 
v. State the conclusion for the test. Give reason for your answer.    

 

              

 
 
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