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:                               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. Estimate the p-value for this test.  ii. State the conclusion for the test. Give the reason for your answer

Glencoe Algebra 1, Student Edition, 9780079039897, 0079039898, 2018
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Author:Carter
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Chapter10: Statistics
Section10.3: Measures Of Spread
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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:

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

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