Question 1204904: A genetic experiment involving peas yielded one sample of offspring consisting of 433 green peas and 166 yellow peas. Use a 0.01 significance level to test the claim that under the same circumstances, 26% of offspring peas will be yellow. Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim. Use the P-value method and the normal distribution as an approximation to the binomial distribution.
Answer by Theo(13342)   (Show Source):
You can put this solution on YOUR website! sample size = 433 + 166 = 599
green peas = 433
yellow peas = 166
p = .2771285476
q = 1-p = .7228714524
mean = p
standard error = sqrt(p*q/n) = .018287644
n = sample size
z = (x-m)/s is the formula used.
z is the critical z-score at 99% two tail confidence level.
x is the test mean of .26
m is the sample mean of p
s is the standard error
critical z-score at 99% two tail confidence interval = plus or minus 2.5758 rounded to 4 decimal places.
formula to find the test z-score is z = (.26 - .2771285476) / .018287644 = -.9366186037.
test z-score is less than critical z-score of -2.5758.
this indicates that the test value of .26 is within the 99% confidence interval limits of the sample mean of p.
this indicates that the percentage of yellow peas equal to 26% under the same circumstances is reasonable to assume.
the test p-value was .1745 which was higher than the critical p-value of .005 on the low end of the confidence interval (.01 / 2 = .005 on the low end of the confidence interval and .005 on the high end of the confidence interval)>
Identify the null hypothesis, alternative hypothesis, test statistic, P-value, conclusion about the null hypothesis, and final conclusion that addresses the original claim.
null hypothesis is that p = .26 is possible under the same circumstance.
alternate hypothesis is that p = .26 isn't.
test p-value = .1745
this is greater than critical p-value of .005 on the low end of the confidence interval.
conclusion is that p-value of .26 is entirely reasonable under the same conditions as it is well within the confidence interval.
to find the confidence interval limits, do the following.
on the low end of the confidence interval, you get -2.5758 = (x - .2771285476) / .018287644.
solve for x to get x = -2.5758 * .018187644 + .2771285476 = .2300232342.
on the high end of the confidence interval, you get 2.5758 = (x - .2771285476) / .018287644.
solve for x to get x = 2.5758 * .018187644 + .2771285476 = .324233861.
your 99% confidence interval is from .2300 to .3242 rounded to 4 decimal places.
.26 is well within those confidence interval limits.
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