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12.3 TESTS OF GOODNESS-OF-FIT 609
uniform distribution of flu cases during the flu season in southern
Nevada.
9. Conclusion. We conclude that the occurrence of flu cases does not follow
a uniform distribution.
10. p value. From the MINITAB output we see that p .000 (i.e., .001).
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EXAMPLE 12.3.5
A certain human trait is thought to be inherited according to the ratio 1:2:1 for
homozygous dominant, heterozygous, and homozygous recessive. An examination of
a simple random sample of 200 individuals yielded the following distribution of the
trait: dominant, 43; heterozygous, 125; and recessive, 32. We wish to know if these
data provide sufficient evidence to cast doubt on the belief about the distribution of
the trait.
Solution:
1. Data. See statement of the example.
2. Assumptions. We assume that the data meet the requirements for the application
of the chi-square goodness-of-fit test.
3. Hypotheses.
H 0 : The trait is distributed according to the ratio 1:2:1 for homozygous
dominant, heterozygous, and homozygous recessive.
: The trait is not distributed according to the ratio 1:2:1.
H A
4. Test statistic. The test statistic is
1O -
X 2 E22
= a c d
E
5. Distribution of test statistic. If H is true, X 2
0 is distributed as chi-square
with 2 degrees of freedom.
6. Decision rule. Suppose we let the probability of committing a type I error
be .05. Reject H if the computed value of X 2
0 is equal to or greater than
5.991.
7. Calculation of test statistic. If H 0 is true, the expected frequencies for
the three manifestations of the trait are 50, 100, and 50 for dominant,
heterozygous, and recessive, respectively. Consequently,
X 2 = 143 - 502 2 >50 + 1125 - 10022>100 + 132 - 502 2 >50 = 13.71
8. Statistical decision. Since 13.71 7 5.991, we reject H 0 .
9. Conclusion. We conclude that the trait is not distributed according to the
ratio 1:2:1.
10. p value. Since 13.71 7 10.597, the p value for the test is p 6 .005. ■