In a population, 68% of all tax returns lead to a refund. A random sample of...
In a population, 68% of all tax returns lead to a refund. A random sample of 1000 tax returns is taken.
(a) Describe the distribution of the number of returns leading to refunds in the sample.
(b) What is the expected number of returns leading to refunds in the sample? What is its variance?
(c) What is the mean of the sample proportion of returns leading to refunds?
(d) What is the variance of the sample proportion?
(e) What is the standard error of the sample proportion?
(f) What is the probability that the sample proportion exceeds 0.65? (assume CLT holds)
Homework Answers
Ans:
a)number of returns leading to refunds has binomial distribution with n=1000 and p=0.68
b)
expected number of returns leading to refunds=np=1000*0.68=680
Variance=np(1-p)=1000*0.68*(1-0.68)=217.6
c)mean of the sample proportion=p=0.68
d)Variance of the sample proportion =0.68*(1-0.68)/1000=0.000218
e) standard error of the sample proportion=sqrt(0.000218)=0.01475
f)
z=(0.65-0.68)/0.01475
z=-2.034
P(z>-2.034)=0.9790
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