According to one study, 50% of Americans admit to overeating sweet foods when st
ID: 3156915 • Letter: A
Question
According to one study, 50% of Americans admit to overeating sweet foods when stressed. Suppose that the 50% figure is correct and that a random sample of n = 100 Americans is selected
Does the distribution of , the sample proportion of Americans who relieve stress by overeating sweet foods, have an approximately normal distribution?
What is the mean?
What is the standard deviation?
What is the probability that the sample proportion, , exceeds 0.54?
What is the probability that lies within the interval 0.39 to 0.59?
Would it be unusual if the sample proportion were as small as 34%?
Explanation / Answer
According to one study, 50% of Americans admit to overeating sweet foods when stressed. Suppose that the 50% figure is correct and that a random sample of n = 100 Americans is selected
Does the distribution of , the sample proportion of Americans who relieve stress by overeating sweet foods, have an approximately normal distribution?
n=100
p=0.5
np=100*0.5=50, n(1-p)=50 > 10
we can use normal approximation to this binomial distribution.
What is the mean? p=0.5
What is the standard deviation?
Standard deviation = sqrt( 0.5*0.5/100) =0.05
What is the probability that the sample proportion, , exceeds 0.54?
Z value for 0.54, z=(0.54-0.5)/0.05 = 0.8
P(p >0.54) = P( z >0.8) = 0.2119
What is the probability that lies within the interval 0.39 to 0.59?
Z value for 0.39, z=(0.39-0.5)/0.05 = -2.2
Z value for 0.59, z=(0.59-0.5)/0.05 = 1.8
P( 0.39<p<0.59) = P( -2.2<z<1.8)
=P( z <1.8) –P( z <-2.2)
= 0.9641 - 0.0139
=0.9502
Would it be unusual if the sample proportion were as small as 34%?
Z value for 0.34, z=(0.34-0.5)/0.05 = -3.2
P( z < -3.2) = 0.0007 which is < 0.05
It is unusual if the sample proportion were as small as 34%.
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