Question

HW8#11

A genetic experiment involving peas yielded one sample of offspring consisting of 430 green peas and 168 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. Identify the null and alternative hypotheses.

B. What is the test statistic?

   z=

   Round two decimal places as needed.

C. What is the P-value?

D. Fail to reject OR Reject the null hypothesis?

E. There is or is not sufficient evidence?

Answer 1

Part A

Here, we have to use Z test for the population proportion.

The null and alternative hypotheses for this test are given as below:

Null hypothesis: H₀: The percentage of yellow offspring peas is 27%.

Alternative hypothesis: H_a: The percentage of yellow offspring peas is not 27%.

H₀: p = 0.27 versus H_a: p ≠ 0.27

This is a two tailed test.

Part B

The test statistic formula is given as below:

Z = (p̂ - p)/sqrt(pq/n)

Where, p̂ = Sample proportion, p is population proportion, q = 1 - p, and n is sample size

p̂ = x/n

n = 168 + 430 = 598

x = number of items of interest = 168

p̂ = 168/598 = 0.280936455

p = 0.27

q = 1 – p = 1 – 0.27 = 0.73

Z = (p̂ - p)/sqrt(pq/n)

Z = (0.280936455 – 0.27)/sqrt(0.27*0.73/598)

Z = 0.6024

Test statistic = Z = 0.60

Part C

P-value = 0.5485

(by using z-table)

Part D

P-value = 0.5485 > α = 0.01

So, we fail to reject the null hypothesis

Part E

There is sufficient evidence to conclude that 27% of offspring peas will be yellow.