In: Statistics and Probability
The null hypothesis and the alternate hypothesis are:
H0: The frequencies are equal.
H1: The frequencies are not equal.
Category | f0 |
A | 10 |
B | 30 |
C | 30 |
D | 10 |
State the decision rule, using the 0.05 significance level. (Round your answer to 3 decimal places.)
Compute the value of chi-square. (Round your answer to 2 decimal place.)
What is your decision regarding H0?
The degrees of freedom = n - 1 = 4 - 1 = 3
The Critical value at = 0.05 is 7.815
(a) Therefore the decision rule is Reject H0, if > 7.815
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(n) Each expected value = (10 + 30 + 30 + 10) / 4 = 80 / 4 = 20
Observed | Expected | O-E | (O-E)2 | (O-E)2/E | |
1 | 10 | 20 | -10 | 100 | 5.000 |
2 | 30 | 20 | 10 | 100 | 5.000 |
3 | 30 | 20 | 10 | 100 | 5.000 |
4 | 10 | 20 | -10 | 100 | 5.000 |
Total | 80 | 80 | 20.00 |
test = 20.00
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(c) Since testis > 7.815, Reject H0.
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The Conclusion: There is sufficient evidence at the 95% level of significance to conclude that the frequencies are not equal.
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