A genetic experiment involving peas yielded one sample of offspring consisting of 408 green peas and 154 yellow peas. Use a 0.05 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. H0: p=0.27 H1: p>0.27 B. H0: p=0.27 H1: p≠0.27 C. H0: p=0.27 H1: p<0.27 D. H0: p≠0.27 H1: p<0.27 E. H0: p≠0.27 H1: p=0.27 F. H0: p≠0.27 H1: p>0.27 What is the test statistic? z=enter your response here (Round to two decimal places as needed.) What is the P-value? P-value=enter your response here (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 to the significance level, α. B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α. C. Fail to reject the null hypothesis because the P-value is greater than the significance level, α. D. Reject the null hypothesis because the P-value is greater than the significance level, α. What is the final conclusion? A. There is not sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. B. There is not sufficient evidence to support the claim that less than 27% of offspring peas will be yellow. C. There is sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. D. There is sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.

A genetic experiment involving peas yielded one sample of offspring consisting of 408 green peas and 154 yellow peas. Use a 0.05 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. H0: p=0.27 H1: p>0.27 B. H0: p=0.27 H1: p≠0.27 C. H0: p=0.27 H1: p<0.27 D. H0: p≠0.27 H1: p<0.27 E. H0: p≠0.27 H1: p=0.27 F. H0: p≠0.27 H1: p>0.27 What is the test statistic? z=enter your response here (Round to two decimal places as needed.) What is the P-value? P-value=enter your response here (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 to the significance level, α. B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, α. C. Fail to reject the null hypothesis because the P-value is greater than the significance level, α. D. Reject the null hypothesis because the P-value is greater than the significance level, α. What is the final conclusion? A. There is not sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. B. There is not sufficient evidence to support the claim that less than 27% of offspring peas will be yellow. C. There is sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow. D. There is sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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Related questions

Question

A genetic experiment involving peas yielded one sample of offspring consisting of

408

green peas and

154

yellow peas. Use a

0.05

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?

H0: p=0.27

H1: p>0.27

H0: p=0.27

H1: p≠0.27

H0: p=0.27

H1: p<0.27

H0: p≠0.27

H1: p<0.27

H0: p≠0.27

H1: p=0.27

H0: p≠0.27

H1: p>0.27

What is the test statistic?

z=enter your response here

(Round to two decimal places as needed.)

What is the P-value?

P-value=enter your response here

(Round to four decimal places as needed.)

What is the conclusion about the null hypothesis?

Reject

the null hypothesis because the P-value is

less than or equal to

the significance level,

α.

Fail to reject

the null hypothesis because the P-value is

less than or equal to

the significance level,

α.

Fail to reject

the null hypothesis because the P-value is

greater than

the significance level,

α.

Reject

the null hypothesis because the P-value is

greater than

the significance level,

α.

What is the final conclusion?

There

is not

sufficient evidence to warrant rejection of the claim that

27%

of offspring peas will be yellow.

There

is not

sufficient evidence to support the claim that less than

27%

of offspring peas will be yellow.

There

sufficient evidence to warrant rejection of the claim that

27%

of offspring peas will be yellow.

There

sufficient evidence to support the claim that less than

27%

of offspring peas will be yellow.

A genetic experiment involving peas yielded one sample of offspring consisting of 408 green peas and 154 yellow peas. Use a 0.05 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?
В. Но: р30.27
А. Но р3D0.27
H1:p>0.27
H1:p#0.27
С. Но р30.27
D. Ho: p+ 0.27
H1:p<0.27
H1:p<0.27
F. Hо: р#0.27
H1:p>0.27
E. Но: р#0.27
H4: p= 0.27
What is the test statistic?
z =
(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 to the significance level, a.
B. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, oa.
C. Fail to reject the null hypothesis because the P-value is greater than the significance level, a.
D. Reject the null hypothesis because the P-value is greater than the significance level, a.
What is the final conclusion?
A. There is not sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow.
B. There is not sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.
C. There is sufficient evidence to warrant rejection of the claim that 27% of offspring peas will be yellow.
D. There is sufficient evidence to support the claim that less than 27% of offspring peas will be yellow.
O O O

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

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