# The performance of students at

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

1. Define the null and alternative hypotheses of the χ 2 test for the above output.
2. Find the missing values ‘*’, ‘**’, ‘***’, Showing necessary working

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

1. Estimate the p-value for this test.
2. State the conclusion for the test. Give a reason for your answer.

# The performance of students at

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.

# The performance of students at

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 (15.67) 30 (29.24) 52 (49.09) 94 Humanities 14 (10.67) 20 (19.91) 30 (33.42) ** 64 Law 4 (*) 3.67 6 (6.84) 12 (11.49) 22 Total 30 56 94 180

chi – sq =     0.858  +  0.020  +  ***(0.173)  +

1.042  +  0.000  +  0.350  +

0.030  +  0.104  +  0.023  =  2.600

1. Estimate the p-value for this test.

# The performance of students at

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

Science        Low            Average              High           Total

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

# The performance of students at

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

Science        Low            Average              High           Total

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

Find the missing values ‘*’, ‘**’, ‘***’, Showing necessary working

# The performance of students at

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

Science        Low            Average              High           Total

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

What is the degree of freedom for this test?

# The performance of students at

question 4

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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