Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 431 green peas and 151 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 is not​ 25%, do the results contradict​ expectations?

in detail plz

Answer 1

Total peas = 431 + 151 = 582

Proportion of yellow peas = 151 / 582 = 0.259

a)

Z critical value from Z table at 0.10 significance level = 1.645

90% confidence interval for p is

- Z * sqrt( ( 1 - ) / n) < p < + Z * sqrt( ( 1 - ) / n)

0.259 - 1.645 * sqrt( 0.259 * 0.741 / 582 ) < p < 0.259 + 1.645 * sqrt( 0.259 * 0.741 / 582 )

0.229 < p < 0.289

90% CI is ( 0.229 , 0.289 )

b)

Since 0.25 (or 25%) contained in confidence interval, the result does not contradict expectations.