A genetic experiment with peas resulted in one sample of offspring that consisted of 429 green peas and 163 yellow peas.

A) construct a 95% confidence interval to estimate of the percentage of yellow peas.

B)it was expected that 25% of the offspring peas would be yellow. given the percentage ofoffspring yellow peas is not 25% do the results contradict expectation?

A)< p < (round the three decimal places as needed.)

B) given that the percentage of offspring yellow peas is not 25% do the results contradict expectation?

yes, the confidence interval does not include 0.25, so the true percentage could not equal 25%
no, the confidence interval include 0.25, so the true percentage could easly equal 25%

In a sample with a number n of people surveyed with a probability of a success of [tex]\pi[/tex], and a confidence level of [tex]1-\alpha[/tex], we have the following confidence interval of proportions.

[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]

In which

z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].

For this problem, we have that:

163 yellow peas out of 163 + 429 = 592 total peas

So [tex]n = 592, \pi = \frac{163}{592} = 0.275[/tex]

95% confidence level

So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]Z = 1.96[/tex].

The lower limit of this interval is:

[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.275 - 1.96\sqrt{\frac{0.275*0.725}{592}} = 0.239[/tex]

The upper limit of this interval is:

[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.275 + 1.96\sqrt{\frac{0.275*0.725}{592}} = 0.311[/tex]

So

A) 0.239 < p < 0.311.

B)

25% is part of the confidence interval.

So the answer is:

no, the confidence interval include 0.25, so the true percentage could easly equal 25%