Question 1202873: A genetic experiment with peas resulted in one sample of offspring that consisted of 429 green peas and 156 yellow peas.
a. Construct a 90% confidence interval to estimate of the percentage of yellow peas.
b. Based on the confidence interval, do the results of the experiment appear to contradict the expectation that 25% of the offspring peas would be yellow?
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
n = sample size
n = 429 green + 156 yellow
n = 585 total
At 90% confidence, the critical z value is roughly z = 1.645
Use a table or stats calculator to determine this.
p = population proportion of yellow peas
phat = sample proportion of yellow peas
phat's job is to estimate p.
phat = (# of yellow peas in sample)/(sample size)
phat = (156 yellow peas)/(585 total)
phat = 156/585
phat = 4/15
phat = 0.2666667 approximately
E = margin of error
E = z*sqrt(phat*(1-phat)/n)
E = 1.645*sqrt(0.2666667*(1-0.2666667)/585)
E = 0.0300762070348
E = 0.0300762
L = lower boundary of confidence interval
L = phat - E
L = 0.2666667 - 0.0300762
L = 0.2365905
L = 0.237
U = upper boundary of confidence interval
U = phat + E
U = 0.2666667 + 0.0300762
U = 0.2967429
U = 0.297
The 90% confidence interval in the format L < p < U is roughly 0.237 < p < 0.297
That can be condensed to the format (L, U) so we get (0.237, 0.297).
We're 90% confident the population proportion p is somewhere between 0.237 and 0.297
This translates to the statement "We're 90% confident the percentage of yellow peas is somewhere between 23.7% and 29.7%".
Those values are approximate.
Based on the confidence interval, it appears that p = 0.25 = 25% is possible.
This is because 0.237 < 0.25 < 0.297 is a true statement.
In other words, p = 0.25 is a solution to 0.237 < p < 0.297
The results of the experiment do not contradict the expectation that 25% of the offspring peas would be yellow.
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