Question

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%

Answer 1

a) 0.239 < p < 0.311.

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

Explanation:

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

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

For this problem, we have that:

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

So
`n = 592, \pi = (163)/(592) = 0.275`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.275 - 1.96\sqrt{(0.275*0.725)/(592)} = 0.239`

The upper limit of this interval is:

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

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%