Can you help me with this question

Question

A genetic experiment involving peas yielded one sample of offspring consisting of 433 green peas and 166 yellow peas. Use a 0.01 significance level to test the claim that under the same​ circumstances, 26​% 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.

Answer 1

Hi Dylan,

Not sure how you would use the normal approximation to binomial here, since that requires a hypothesized x-value, not a hypothesized proportion. We can test this, though, using a Normal test statistic, since we have at least ten successes (yellow peas) and failures (non-yellow). First things first, the hypotheses:

H₀: p=0.26 True population proportion is approximately 0.26

H_A: p is not equal to 0.26; true population proportion differs significantly from 0.26.

Now, for the z-test statistic, the formula is:

z=(p-hat - p)/SE

SE=sqrt([p(1-p)/n])

p-hat=probability from sample of yellow peas

p-hat=166/599=0.277

p=hypothesized population probability=0.26

n=sample size=(433+166)=599

SE=standard error

Thus:

SE=sqrt((0.277*0.723)/599)

SE=0.0183

Now, for our z-test statistic:

z=(p-hat - p)/SE

z=(0.277 - 0.26)/0.0183

z=0.93

This is our test statistic. Now, from the z-table:

P(z<0.93)=0.8238

Take 1-0.8238=0.1762

Multiply that value by 2--remember we had a two-sided alternative, not a greater or less than--

P=2(0.1762)=0.3524

This result is not significant, so we fail to reject H₀ and conclude that 0.26 is a plausible value for the true population proportion of yellow pea offspring. I hope this helps. Sorry if there is a normal approximation for probability that I am not familiar with.