Use homogeneity data and relative and conditional distribution data to create an
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Question
Use homogeneity data and relative and conditional distribution data to create analogy for medical field (placebo-based blind studies)????????? Help
Table I
M&M’s color
Observed count (Oi)
Frequency
Homogeneity %
Expected count (Ei)
(Oi - Ei)
(Oi - Ei)2
red
88
0.1306
0.1667
112.3333
-24.3333
592.1111
5.271
orange
154
0.2285
0.1667
112.3333
41.6667
1736.111
15.455
yellow
83
0.1231
0.1667
112.3333
-29.3333
860.4444
7.6597
green
115
0.1706
0.1667
112.3333
2.6667
7.1111
0.0633
blue
138
0.2047
0.1667
112.3333
25.6667
658.7778
5.8645
brown
96
0.1424
0.1667
112.3333
-16.3333
266.7778
2.3749
Total
674
36.6884
Now we determine the critical value using Table VIII. All goodness-of-fit tests are right-tailed tests, so the critical value is with k – 1 degrees of freedom.
To minimize Type I error, we use 0.10, 0.05, and 0.01 level of significance with df = 6 – 1 = 5 degrees of freedom.
1) 0.10 level of significance
9.236
2) 0.05 level of significance
3) 0.01 level of significance
Because the test statistic, 36.6884 does lie in the critical region, we reject the null hypothesis. That means, the sample evidence suggests there is sufficient evidence at the = 0.10, = 0.05, and = 0.001 level of significance to conclude the distribution of M&M’s colors does not follow Homogeneity distribution.
Table IV
M&M’s
Type
M&M’s color count
Total
Relative frequency
red
orange
yellow
green
blue
brown
Plain
88
154
83
115
138
96
674
0.606115
Peanut
41
109
78
81
81
48
438
0.393885
Total
129
263
161
196
219
144
1112
Relative frequency
0.116
0.236
0.145
0.176
0.197
0.129
1
A conditional distribution lists the relative frequency of each category of the response variable, given a specific value of the explanatory variable in the contingency table.
First, we compute the relative frequency for each type of M&M’s, given the color of M&M’s in each cell.
Table V
M&M’s
Type
M&M’s color count
Total
red
orange
yellow
green
blue
brown
Plain
=0.682
=0.586
=0.516
0.587
=0.630
=0.666
674
Peanut
=0.318
=0.414
=0.484
=0.413
=0.369
= 0.333
438
Total
129
263
161
196
219
144
1112
Use homogeneity data and relative and conditional distribution data to create analogy for medical field (placebo-based blind studies)????????? Help
M&M’s color
Observed count (Oi)
Frequency
Homogeneity %
Expected count (Ei)
(Oi - Ei)
(Oi - Ei)2
red
88
0.1306
0.1667
112.3333
-24.3333
592.1111
5.271
orange
154
0.2285
0.1667
112.3333
41.6667
1736.111
15.455
yellow
83
0.1231
0.1667
112.3333
-29.3333
860.4444
7.6597
green
115
0.1706
0.1667
112.3333
2.6667
7.1111
0.0633
blue
138
0.2047
0.1667
112.3333
25.6667
658.7778
5.8645
brown
96
0.1424
0.1667
112.3333
-16.3333
266.7778
2.3749
Total
674
36.6884
Explanation / Answer
10% - 9.23635
5% - 11.07049
1% - 15.08627
since TS = 36.688 > critical values
we reject the nulll at all alpha
, the sample evidence suggests there is sufficient evidence at the = 0.10, = 0.05, and = 0.001 level of significance to conclude the distribution of M&M’s colors does not follow Homogeneity distribution.
p Oi Ei (Oi-Ei)^2/Ei 0.166667 88 112.333333 5.271018793 0.166667 154 112.333333 15.45499505 0.166667 83 112.333333 7.659742829 0.166667 115 112.333333 0.06330366 0.166667 138 112.333333 5.864490603 0.166667 96 112.333333 2.37487636 1 674 674 36.6884273 10% 1% 5% critical value 11.07049769 9.2363569 15.08627 p-value 6.91557E-07Related Questions
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