Question

A genetic experiment involving peas yielded one sample of offspring consisting of 411 green peas and 140 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 27% 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.

What are the null and alternative hypotheses?

A.

Upper H 0 : p not equals 0.27

Upper H 1 : p equals 0.27

B.

Upper H 0 : p not equals 0.27

Upper H 1 : p greater than 0.27

C.

Upper H 0 : p not equals 0.27

Upper H 1 : p less than 0.27

D.

Upper H 0 : p equals 0.27

Upper H 1 : p less than 0.27

E.

Upper H 0 : p equals 0.27

Upper H 1 : p not equals 0.27

F.

Upper H 0 : p equals 0.27

Upper H 1 : p greater than 0.27

What is the test statistic?

zequals

nothing

(Round to two decimal places as needed.)

What is the P-value?

P-value=_____

(Round to four decimal places as needed.)

What is the conclusion about the null hypothesis?

A.

Reject the null hypothesis because the P-value is less than or equal tothe significance level, alpha.

B.Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, alpha.

C.Reject the null hypothesis because the P-value is greater thanthe significance level, alpha.

D.Fail to reject the null hypothesis because the P-value is greater than the significance level, alpha.

What is the final conclusion?

A.There is sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.

B.There is sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow.

C.There is not sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.

D.There is not sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow.

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H_{0} : p = 0.27

H_{a} : p
0.27

= x / n = 140 / 411 = 0.3406

P_{0} = 0.27

1 - P_{0} = 0.73

Test statistic = z

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

= 0.3406 - 0.27 / [(0.27 * 0.73) / 411]

= 3.225

= 3.23

P(z > 3.225) = 1 - P(z < 3.225) = 0.0006

P-value = 2 * 0.0006 = 0.0012

= 0.01

P-value <

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