Question
In: Statistics and Probability
Use the following contingency table to complete (a) and (b) below. A B C Total 1...
Use the following contingency table to complete (a) and (b) below.
A B C Total
1 15 35 40 90
2 30 35 45 110
Total 45 70 85 200
(a). Compute the expected frequencies for each cell.
(b) Compute χ2STAT. Is it significant at α=0.01?
Solutions
Expert Solution
We have to test the hypothesis that whether fitting of distibution is good or not.
a) Observed Frequencies table.
| A | B | C | Total | |
| 1 | 15 | 35 | 40 | 90 |
| 2 | 30 | 35 | 45 | 110 |
| Total | 45 | 70 | 85 | 200 |
Expected Frequency Table.
| A | B | C | Total | |
| 1 | 20.25 ( 90* 45 /200) | 31.50(90*70/200) | 38.25(90*85/200) | 90 |
| 2 | 24.75(110*45/200) | 38.50(110*70/200) | 46.75(110*85/2000 | 110 |
| Total | 45 | 70 | 85 | 200 |
b) Value of Chi-square Statistic
Where Oi = Observed Frequency and Ei = expected frequency
r= number of rows = 2
and c= number of columns. = 3
d.f. = 1 * 2 = 2.
| Oi | Ei | (Oi-Ei)^2/Ei |
| 15 | 20.25 | 1.3611 |
| 35 | 31.5 | 0.3889 |
| 40 | 38.25 | 0.0801 |
| 30 | 24.75 | 1.1136 |
| 35 | 38.5 | 0.3182 |
| 45 | 46.75 | 0.0655 |
| Total | 3.3274 | |
Value of Chi-square -Statistic is
Critical value.
Alpha = level of significance = 0.01
d.f. = 2
Since Critical value s greater than calculated value, the test is not significant. We failed to reject Ho at null hypothesis.
Conclusion : Fitting of distribution is good.
orchestra answered 2 years agoRelated Solutions
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Use the following contingency table to complete (a) and (b)
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A
B
C
Total
1
15
30
45
90
2
45
50
55
150
Total
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80
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A
B
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150
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