A Genetic Experiment With Peas
nothing (Round to iii decimal places every bit needed.)
Found two solutions past Boreal, rothauserc:
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90% half-interval is z*sqrt(p*(i-p)/n) where z=1.645 and p is the bespeak guess of yellow or 159/579=0.275
half-interval is 1.645* sqrt (0.275*0.725/579)=0.0301
Therefore, the total interval is 0.275+/-1.645* sqrt (0.275*0.725/579)
the interval is (0.245, 0.305)
considering 0.25 is within the confidence interval, the results to not contradict expectations.
Answer by rothauserc(4718)
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a) sample proportion of yellow peas = 159/(420+159) = 0.275
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standard error = square root(0.275*(ane=0.275)/(420+159)) = 0.019
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alpha(a) = one -(ninety/100) = 0.x
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disquisitional probability(p*) = one -(a/2) = 0.95
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I assume the population is commonly distributed and since the sample is > xxx, a normal distribution is used to construct the ninety% confidence interval
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The z-score associated with p* is i.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 xanthous 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 nil hypothesis and H1 implies a two-tailed test
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z = (0.275-0.25)/0.019 = one.316
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since this is a two-tailed exam, nosotros reject Ho if 1.316 > 1.96 or if 1.316 < -i.96, since 1.316 < ane.96 we do non reject Ho
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A Genetic Experiment With Peas,
Source: https://www.algebra.com/algebra/homework/Probability-and-statistics/Probability-and-statistics.faq.question.1132704.html
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