Question 1132704: 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.
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nothing (Round to three decimal places as needed.)
Found 2 solutions by Boreal, rothauserc:
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 90% half-interval is z*sqrt(p*(1-p)/n) where z=1.645 and p is the point estimate of yellow or 159/579=0.275
half-interval is 1.645* sqrt (0.275*0.725/579)=0.0301
Therefore, the full interval is 0.275+/-1.645* sqrt (0.275*0.725/579)
the interval is (0.245, 0.305)
because 0.25 is within the confidence interval, the results to not contradict expectations.
Answer by rothauserc(4718)  (Show Source):
You can put this solution on YOUR website! a) sample proportion of yellow peas = 159/(420+159) = 0.275
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standard error = square root(0.275*(1=0.275)/(420+159)) = 0.019
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alpha(a) = 1 -(90/100) = 0.10
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critical probability(p*) = 1 -(a/2) = 0.95
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I assume the population is normally distributed and since the sample is > 30, a normal distribution is used to construct the 90% confidence interval
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The z-score associated with p* is 1.645, which is the critical value(CV)
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Margin of error = CV * SE = 1.645 * 0.019 = 0.031
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90% confidence interval is 0.275 + or - 0.031, (0.244, 0.306)
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b) The 90% confidence level means that we would expect 90% of the confidence interval estimates to include the population proportion for yellow peas.
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Ho: X = 0.25
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H1: X not = 0.25
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Ho is our null hypothesis and H1 implies a two-tailed test
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z = (0.275-0.25)/0.019 = 1.316
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since this is a two-tailed test, we reject Ho if 1.316 > 1.96 or if 1.316 < -1.96, since 1.316 < 1.96 we do not reject Ho
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