A genetic experiment involving peas yielded one sample of offspring consisting of 455 green peas and 176 yellow peas. Use a 0.05 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.

Question
Asked Oct 30, 2019

A genetic experiment involving peas yielded one sample of offspring consisting of 455 green peas and 176 yellow peas. Use a 0.05 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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Expert Answer

Step 1

State the appropriate hypothesis:

Let p be the population proportion of yellow offspring peas.

The given claim is that, the percentage of yellow offspring peas is equal to 24%.

That is, p = 0.24.

The researcher’s objective is to test, whether or not the proportion of yellow offspring peas is equal to 0.24.

Step 2

The hypotheses are given below:

Null hypothesis:

H0 : p = 0.24

That is, the proportion of yellow offspring peas is equal to 0.24.

Alternative hypothesis:

Ha : p ≠ 0.24 (Two tailed test)

That is, the proportion of yellow offspring peas is not equal to 0.24.

Obtain the sample proportion:

It is given that among a sample of 631 (=455 + 176) offspring peas, 176 offspring peas are yellow in color.

The sample proportion of yellow offspring peas is obtained as 0.2789 (=176/631).

Obtain the test statistic value:

The test statistic value is obtained as follows:

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Step 3

p-value:

In this case, the test is two tailed. ...

p-value P(-2.29)+P( 22.29)
=P(s-2.29)+(1-P(<2.29))
(Using standard normal table)
=0.0110 1-0.9890
=0.0220
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p-value P(-2.29)+P( 22.29) =P(s-2.29)+(1-P(<2.29)) (Using standard normal table) =0.0110 1-0.9890 =0.0220

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