Preform a chi square test of independence to determine if Score is depending on
ID: 3357927 • Letter: P
Question
Preform a chi square test of independence to determine if Score is depending on Upper class standing. You will need to recode the scores as letter grades (<60=F, 60-60=D, 70-79=C, 80-89=B, 90-100=A). Use a 10% significance level. Clearly state your hypotheses and all test-statistics or P-values. Interpret your results in a sentence that includes the context of the test.
Student Score Upper Class High Math 1 90 0 1 2 54 0 0 3 56 0 1 4 61 0 1 5 73 1 1 6 81 0 1 7 66 1 1 8 74 0 1 9 71 0 1 10 60 1 0 11 85 1 1 12 44 1 0 13 53 0 1 14 98 1 1 15 61 1 1 16 73 0 0 17 69 0 0 18 81 0 1 19 72 1 1 20 68 0 0 21 78 1 1 22 80 0 0 23 36 1 0 24 59 1 1 25 87 1 1 26 70 0 1 27 79 1 1 28 76 1 1 29 57 0 0 30 81 0 1 31 75 0 0 32 68 1 1 33 75 1 0 34 86 1 1 35 62 0 1 36 61 0 0 37 73 1 0 38 39 1 1 39 82 1 1 40 75 1 1 41 53 1 1 42 74 1 1 43 65 1 0 44 91 0 1 45 76 0 1 46 89 0 1 47 79 1 1 48 74 1 0 49 60 1 0 50 80 0 1 51 94 1 1 52 75 0 0 53 65 0 1 54 64 1 1 55 69 1 0 56 39 0 0 57 72 0 1 58 70 1 1 59 68 1 1 60 74 1 1 61 62 1 0 62 70 1 0 63 45 1 0 64 83 1 1 65 92 1 1 66 71 0 1 67 68 0 1 68 79 1 0 69 89 1 1 70 68 1 0 71 82 1 0 72 54 0 1 73 56 0 0 74 95 0 1 75 69 1 0 76 80 0 1 77 81 0 1 78 82 1 1 79 77 1 0 80 69 1 1 81 65 1 0 82 68 1 0 83 56 1 0 84 62 0 0 85 72 1 1 86 70 0 0 87 82 1 1 88 74 0 0 89 79 1 1 90 56 1 0 91 69 1 0 92 74 1 1 93 64 1 1 94 83 1 1 95 83 0 1 96 66 0 1 97 83 1 1 98 86 1 1 99 56 1 0 100 58 0 1 101 79 0 1 102 100 1 1 103 75 1 1 104 75 1 1 105 43 0 0 106 64 0 1 107 71 0 1 108 85 1 0 109 71 1 1 110 88 0 1 111 65 1 0 112 72 0 0 113 60 1 0 114 74 1 0 115 83 0 0 116 77 1 0 117 67 1 1 118 77 0 0 119 70 1 1 120 48 1 1 121 79 0 1 122 77 1 0 123 72 0 0 124 96 0 1 125 62 1 0 126 70 1 0 127 74 1 0 128 73 0 0 129 92 1 1 130 65 1 0 131 62 1 0 132 77 0 0 133 89 1 1 134 76 1 1 135 79 1 1 136 56 1 1 137 66 1 1 138 94 0 1 139 77 1 1 140 75 1 1 141 67 1 0 142 74 1 1 143 67 0 0 144 59 0 0 145 71 0 1 146 76 1 1 147 78 1 0 148 70 0 1 149 62 0 0 150 72 1 0 151 65 1 0 152 64 0 0 153 60 0 0 154 75 1 0 155 69 0 1 156 88 0 1 157 80 1 0 158 68 1 0 159 71 1 1 160 79 0 1 161 74 0 1 162 71 1 1 163 68 0 0 164 76 1 0 165 75 1 1 166 65 0 0 167 66 0 1 168 68 1 1 169 74 1 0 170 80 0 0 171 58 1 0 172 79 0 1 173 62 1 1 174 72 1 0 175 63 0 1 176 75 1 1 177 73 1 0 178 75 1 1 179 72 0 1 180 69 0 1 181 83 1 0 182 77 0 0 183 60 0 0 184 51 1 1 185 65 0 1 186 69 0 1 187 60 1 0 188 74 1 1 189 84 1 1 190 84 1 1Explanation / Answer
Solution:
Here, we have to use chi square test for independence of two categorical variables. The null and alternative hypothesis for this test is given as below:
Null hypothesis: H0: The two categorical variables whether student from an Upper class and score of the student are independent from each other.
Alternative hypothesis: Ha: The two categorical variables whether student from an Upper class and score of the student are not independent from each other.
We are given a level of significance or alpha value for this test as 10% or 0.10.
The test statistic formula for this test is given as below:
Chi square = [(O - E)^2/E]
Where O is observed frequencies and E is expected frequencies.
Expected frequencies are calculated as below:
E = Row total * Column total / Grand total
Required categorical data and calculations table for this test are given as below:
Chi-Square Test
Observed Frequencies
Upper Class
Calculations
Score
No
Yes
Total
(O - E)
A
5
5
10
0.947368
-0.94737
B
13
18
31
0.436842
-0.43684
C
26
47
73
-3.58421
3.584211
D
23
30
53
1.521053
-1.52105
F
10
13
23
0.678947
-0.67895
Total
77
113
190
Expected Frequencies
Upper Class
Score
No
Yes
Total
(O - E)^2/E
A
4.052632
5.947368
10
0.221463
0.150908
B
12.56316
18.43684
31
0.01519
0.010351
C
29.58421
43.41579
73
0.434237
0.295896
D
21.47895
31.52105
53
0.107715
0.073399
F
9.321053
13.67895
23
0.049455
0.033699
Total
77
113
190
From above calculations, we have
Chi square = [(O - E)^2/E] = 1.392312
Number of columns = c = 2
Number of rows = r = 5
Degrees of freedom = (c – 1)*(r – 1) = 1*4 = 4
P-value = 0.84553
P-value > = 0.10
So, we do not reject the null hypothesis that the two categorical variables whether student from an Upper class and score of the student are independent from each other.
There is sufficient evidence to conclude that the two categorical variables whether student from an Upper class and score of the student are independent from each other.
Chi-Square Test
Observed Frequencies
Upper Class
Calculations
Score
No
Yes
Total
(O - E)
A
5
5
10
0.947368
-0.94737
B
13
18
31
0.436842
-0.43684
C
26
47
73
-3.58421
3.584211
D
23
30
53
1.521053
-1.52105
F
10
13
23
0.678947
-0.67895
Total
77
113
190
Expected Frequencies
Upper Class
Score
No
Yes
Total
(O - E)^2/E
A
4.052632
5.947368
10
0.221463
0.150908
B
12.56316
18.43684
31
0.01519
0.010351
C
29.58421
43.41579
73
0.434237
0.295896
D
21.47895
31.52105
53
0.107715
0.073399
F
9.321053
13.67895
23
0.049455
0.033699
Total
77
113
190