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
iii. What is the degree of freedom for this test?
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 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
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
State the conclusion for the test. Give reason for your answer.
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 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?
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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