Question

**A genetic experiment involving peas yielded one sample
of offspring consisting of 416 green peas and 165 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 less than 0.27

B.

Upper H 0 : p equals 0.27

Upper H 1 : p not equals 0.27

C.

Upper H 0 : p not equals 0.27

Upper H 1 : p greater than 0.27

D.

Upper H 0 : p equals 0.27

Upper H 1 : p less than 0.27

E.

Upper H 0 : p not equals 0.27

Upper H 1 : p equals 0.27

F.

Upper H 0 : p equals 0.27

Upper H 1 : p greater than 0.27

**What is the test statistic? (Round to two decimal
places as needed.)**

**What is the P-value? (Round to four decimal places as
needed.)**

**What is the conclusion about the null
hypothesis?**

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

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

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

D. 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 warrant rejection of the claim that 24% of offspring peas will be yellow.

B. There is not sufficient evidence to support the claim that less than 24% of offspring peas will be yellow.

C. There is not sufficient evidence to warrant rejection of the claim that 24% of offspring peas will be yellow.

D. There is sufficient evidence to support the claim that less than 24% 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

n = 416

x = 165

= x / n = 165 / 416 = 0.3966

P_{0} = 0.27

1 - P_{0} = 1 - 0.27 = 0.73

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

= 0.3966 - 0.27 / [(0.27 * 0.73) / 416]

= 5.818

Test statistic = 5.82

P(z > 5.82) = 1 - P(z < 5.82) = 1 - 1 = 0

P-value = 2 * 0 = 0

= 0.01

P-value <

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

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