Math Question, Need help pls

Question

A genetic experiment with peas resulted in one sample of offspring that consisted of

433

green peas and

154

yellow peas.

a. Construct a

90​%

confidence interval to estimate of the percentage of yellow peas.

b. It was expected that​ 25% of the offspring peas would be yellow. Given that the percentage of offspring yellow peas from the sample is not​ 25%, do the results contradict​ expectations?

Answer 1

Total peas: 433 green + 154 yellow = 587. Proportion yellow 154/587 = 0.26235 or ≈26.2%. This is your point estimate for the population proportion of yellow peas. p "hat" is 0.262; q "hat" is 1 - 0.262 = 0.738 . np"hat" is (587)(0.262) = 154; nq"hat" is (587)(0.738) = 433, so both results are ≥ 5 and the sample can be approximated by a normal distribution. For 90% confidence, z_c = 1.645.

The margin of error E is calculated by 1.645 x sqrt[(0.262)(0.738)/587)] = 0.030

The confidence limits are the point estimate ± the margin of error. 0.262 + 0.030 = 0.292 or 29.2%;

0.262 - 0.030 = 0.232, or 23.2%. You can say with 90% confidence that the proportion of yellow peas is between 23.2% and 29.2%, so the 25% estimate falls within the 90% confidence limits.