Question

A genetic experiment with peas resulted in one sample of offspring that consisted of 415 green peas and 161 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.

< p < (Round to three decimal places as needed.)

b. Given that the percentage of offspring yellow peas is not 25%, do the results contradict expectations?

A. Yes, the confidence interval does not include 0.25, so the true percentage could not equal 25%

B. No, the confidence interval includes 0.25, so the true percentage could easily equal 25%

Answer #1

Answer)

N = 415

P = 161/415

First we need to check the conditions of normality that is if n*p and n*(1-p) both are greater than 5 or not

N*p = 161

N*(1-p) = 254

Both the conditions are met so we can use standard normal z table to estimate the interval

Critical value z from.z table for 90% confidence level is 1.645

Margin of error ( MOE) = Z*√{P*(1-P)}/√N = 0.03934808071

Interval is given by

(P-MOE, P+MOE)

(0.34860372651, 0.42729988793)

(34.86%, 42.73%)

B)

A. Yes, the confidence interval does not include 0.25, so the true percentage could not equal 25%

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