Question

A genetic experiment involving peas yielded one sample of offspring consisting of 442 green peas and 123 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 26% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.26

H_{a} : p
0.26

= x / n = 123 / 442 = 0.2783

P_{0} = 0.26

1 - P_{0} = 0.74

Test statistic = z

=
- P_{0} / [P_{0
*} (1 - P_{0} ) / n]

= 0.2783 - 0.26 / [(0.26 * 0.74) / 442]

= 0.876

P(z > 0.876) = 1 - P(z < 0.876) = 0.1905

P-value = 0.381

= 0.01

P-value >

Fail to reject the null hypothesis .

There is not sufficient evidence that the 0.01 significance level to test the claim that under the same circumstances, 26% of offspring peas will be yellow .

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null hypothesis, alternative hypothesis, test statistic,
P-value, conclusion about the null hypothesis, and final
conclusion that addresses the original claim. Use the P-value
method and the normal distribution as an approximation to the
binomial distribution.

A genetic experiment involving peas yielded one sample of
offspring consisting of 404 green peas and 134 yellow peas. Use a
0.01 significance level to test the claim that under the same
circumstances, 24% of offspring peas will be yellow. Identify the
null hypothesis, alternative hypothesis, test statistic,
P-value, conclusion about the null hypothesis, and final
conclusion that addresses the original claim. Use the P-value
method and the normal distribution as an approximation to the
binomial distribution.
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addresses the original claim use the P Value method and the normal
distribution as an approximation to the binomial

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binomial distribution.
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