Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 416 green peas and 156 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?

Answer 1

Solution :

Given that,

n = 416 + 156 = 572

x = 156

a) Point estimate = sample proportion = = x / n = 156 / 572 = 0.273

1 - = 1 - 0.273 = 0.727

At 90% confidence level the z is,

= 1 - 90%

= 1 - 0.90 = 0.10

/2 = 0.05

Z_/2 = Z _0.05 = 1.645

Margin of error = E = Z _{/ 2} * (( * (1 - )) / n)

= 1.645 * (((0.273 * 0.727) / 572)

= 0.031

A 90% confidence interval for population proportion p is ,

- E < p < + E

0.273 - 0.031 < p < 0.273 + 0.031

0.242 < p < 0.304

The 90% confidence interval for the population proportion p is :24.2% < p < 30.4%

b) No, the confidence interval includes 0.25, so the true percentage could easily equal 25%