Question

A genetic experiment involving peas yielded one sample of offspring consisting of 434 green peas and 137 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 25% 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? OA. O B. Ho: p 0.25 H: p>0.25 D. Ho:p#0.25 H: p 0.25 F. Ho :p#0.25 H: p<0.25 Ho:p:0.25 H1 : p #0.25 H: p<0.25 H:p>0.25 ° C. H0: p = 0.25 What is the test statistic? (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? O A. Fail to reject the null hypothesis because the P-value is less than or equal to the significance level, a B. Reject the null hypothesis because the P-value is less than or equal to the significance level, α. O C. Reject the null hypothesis because the P-value is greater than the significance level, a. D. Fail to reject the null hypothesis because the P-value is greater than the significance level, «. What is the final conclusion? ○ A. There is not sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. There is sufficient evidence to warrant rejection of the claim that 25% of offspring peas will be yellow. There is sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. There is not sufficient evidence to support the claim that less than 25% of offspring peas will be yellow. B. c. D.

Answer 1

p : proportion of yellow peas

Null hypothesis H₀ : p = 0.25

Alternate hypothesis : H₁ ; p $\small \neq$ 0.25

Two tailed test

Ans : A

Given,

Hypothesized proportion : p₀ = 0.25

Number of green peas yielded = 434

Number of yellow pears yielded = 137

Total number peas yielded : n = 434+137 = 571

Sample proportion of yellow peas yielded : $\small \widehat{p}$ = 137/571 = 0.2399

P-Po Test Statistic: Z= 1 - Po)/n

0.2399 0.25 =-0.5574 0.25(1 - 0.25)/571 0.0181

Value of the test statistic = -0.5574

p-value for two tailed test :

$\small P-Value = P(Z<-|Z_{stat}|) + P(|Z>|Z_{stat}|)= 2\times P(Z<-0.5574)$

P(Z <-0.5574) 0.2886

P-Value-2 x 0.2886 0.5772

p-value = 0.5772

As P-Value i.e. is greater than Level of significance i.e (P-value:0.5772 > 0.01:Level of significance); Fail to Reject Null Hypothesis

Conclusion about the hypothesis:

D. Fail to reject the null hypothesis because the p-value is greater than the level of significance level $\small \alpha$

Final Conclusion:

There is not sufficient evidence to warrant rejection of the claim that 25% of the offspring peas will be yellow