Question

George Mendel conducted hybridization of peas. One experiment resulted in 580 peas with 26.2% having yellow...

  1. George Mendel conducted hybridization of peas. One experiment resulted in 580 peas with 26.2% having yellow pods. According to the theory, 25% of the peas should have yellow pods. Using a = 0.05, to test the claim that the proportion of peas with yellow pods is equal to 25%. (two tails)

    1. H0: p = ________

    2. H1: p ≠ ________

    3. z-score________

    4. P=_______

    5. Reject or fail to reject________

Homework Answers

Answer #1

Solution :

This is the two tailed test .

The null and alternative hypothesis is

H0 : p = 0.25

Ha : p 0.25

= 0.262

P0 = 0.25

1 - P0 = 1 - 0.25 = 0.75

Test statistic = z

= - P0 / [P0 * (1 - P0 ) / n]

= 0.262 - 0.25 / [(0.25 * 0.75) / 580 ]

Test statistic = z = 0.67

P(z > 0.67) = 1 - P(z < 0.67) = 1 - 0.7486 = 0.2514

P-value = 2 * 0.2514 = 0.5028

= 0.05    

P-value >

Fail to reject the null hypothesis .

Know the answer?
Your Answer:

Post as a guest

Your Name:

What's your source?

Earn Coins

Coins can be redeemed for fabulous gifts.

Not the answer you're looking for?
Ask your own homework help question
Similar Questions
3. When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in...
3. When Gregor Mendel conducted his famous hybridization experiments with peas, one such experiment resulted in 580 offspring peas, with 26.2% of them having yellow pods. According to Mendel’s theory, 25% of them should have yellow pods. Use a .05 significance level to test the claim that the proportion of peas with yellow pods is 25%. a. Check your assumptions (SINS).    b. Set up the null and alternative hypotheses. ?0:______________________ ?1:______________________ c. You will run the following type of...
A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed...
A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed peas in such a way that​ 25% (or 143​) of the 572 offspring peas were expected to have yellow pods. Instead of getting 143 peas with yellow​ pods, he obtained 148. Assume that the rate of​ 25% is correct. a. Find the probability that among the 572 offspring​ peas, exactly 148 have yellow pods. b. Find the probability that among the 572 offspring​ peas,...
A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed...
A scientist conducted a hybridization experiment using peas with green pods and yellow pods. He crossed peas in such a way that​ 25% (or 142142​) of the 568568 offspring peas were expected to have yellow pods. Instead of getting 142142 peas with yellow​ pods, he obtained 144144. Assume that the rate of​ 25% is correct. a. Find the probability that among the 568568 offspring​ peas, exactly 144144 have yellow pods. b. Find the probability that among the 568568 offspring​ peas,...
When Gregor Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of...
When Gregor Mendel conducted his famous genetics experiments with peas, one sample of offspring consisted of 428 green peas and 152 yellow peas. a) Find a 95% confidence interval estimate of the proportion of yellow peas. b) Based on his theory of genetics, Mendel expected that 25% of the offspring peas would be yellow. Do the results contradict Mendel’s theory? Why or why not?
A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and...
A genetic experiment involving peas yielded one sample of offspring consisting of 420 green peas and 149 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. What are...
HW8#11 A genetic experiment involving peas yielded one sample of offspring consisting of 448 green peas...
HW8#11 A genetic experiment involving peas yielded one sample of offspring consisting of 448 green peas and 174 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. A. what...
A genetic experiment involving peas yielded one sample of offspring consisting of 447 green peas and...
A genetic experiment involving peas yielded one sample of offspring consisting of 447 green peas and 178 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. A. What are...
A genetic experiment involving peas yielded one sample of offspring consisting of 438 green peas and...
A genetic experiment involving peas yielded one sample of offspring consisting of 438 green peas and 161 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. A. Upper H...
For one binomial experiment, n1 = 75 binomial trials produced r1 = 30 successes. For a...
For one binomial experiment, n1 = 75 binomial trials produced r1 = 30 successes. For a second independent binomial experiment, n2 = 100 binomial trials produced r2 = 50 successes. At the 5% level of significance, test the claim that the probabilities of success for the two binomial experiments differ. (a) Compute the pooled probability of success for the two experiments. (Round your answer to three decimal places.) (b) Check Requirements: What distribution does the sample test statistic follow? Explain....
For one binomial experiment, n1 = 75 binomial trials produced r1 = 45 successes. For a...
For one binomial experiment, n1 = 75 binomial trials produced r1 = 45 successes. For a second independent binomial experiment, n2 = 100 binomial trials produced r2 = 65 successes. At the 5% level of significance, test the claim that the probabilities of success for the two binomial experiments differ. (a) Compute the pooled probability of success for the two experiments. (Round your answer to three decimal places.) (b) Check Requirements: What distribution does the sample test statistic follow? Explain....
ADVERTISEMENT
Need Online Homework Help?

Get Answers For Free
Most questions answered within 1 hours.

Ask a Question
ADVERTISEMENT