Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 420 green peas and 159 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? a. Construct a 90​% confidence interval. Express the percentages in decimal form. nothingless thanpless than nothing ​(Round to three decimal places as​ needed.)

Answer 1

a)

sample proportion, = 0.3786
sample size, n = 420
Standard error, SE = sqrt(pcap * (1 - pcap)/n)
SE = sqrt(0.3786 * (1 - 0.3786)/420) = 0.0237

Given CI level is 90%, hence α = 1 - 0.9 = 0.1
α/2 = 0.1/2 = 0.05, Zc = Z(α/2) = 1.64

Margin of Error, ME = zc * SE
ME = 1.64 * 0.0237
ME = 0.0389

CI = (pcap - z*SE, pcap + z*SE)
CI = (0.3786 - 1.64 * 0.0237 , 0.3786 + 1.64 * 0.0237)
CI = (0.34 , 0.417)
0.34 < p< 0.417

b)

Below are the null and alternative Hypothesis,
Null Hypothesis, H0: p = 0.25
Alternative Hypothesis, Ha: p ≠ 0.25

Rejection Region
This is two tailed test, for α = 0.1
Critical value of z are -1.64 and 1.64.
Hence reject H0 if z < -1.64 or z > 1.64

Test statistic,
z = (pcap - p)/sqrt(p*(1-p)/n)
z = (0.3786 - 0.25)/sqrt(0.25*(1-0.25)/420)
z = 6.09

P-value Approach
P-value = 0
As P-value < 0.1, reject the null hypothesis.