Question: A social researcher found in a sample of 3,600 households in the state of Arkansas that the average household income is $49,500 with a population standard deviation of $4,500. Estimate the population income at a 95% and a 99% confidence interval.  Edit

Answer: Ans. sample mean, x bar= 49500 , population sd(sigma)=4500 and sample size, n=3600
margin of error,e= {sigma/sqrt(n)}*z_c
z_c= critical z-value at c% confidence interval.
for 95% z_c=1.96 so, e= {4500/sqrt(3600)}*1.96=147
for 99% z_c=2.58 so, e= {4500/sqrt(3600)}*2.58=193.5
95% confidence interval=(x bar - e , xbar + e)=(49500-147, 49500+147)=(49353, 49647)(Ans.)

99% confidence interval=(x bar - e , xbar + e)=(49500-193.5, 49500+193.5)=(49306.5, 49693.5)(Ans.)

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