A random sample of n 500 observations from a binomial popul
A random sample of n = 500 observations from a binomial population produced x = 420 successes.
(a) Find a point estimate for p.
 p =  
 
 Find the 95% margin of error for your estimator. (Round your answer to three decimal places.)
 
 
 (b) Find a 90% confidence interval for p. (Round your answers to three decimal places.)
 to  
Solution
a)
 point estimate for p = 0.84
 b)
 Margin of Error = Z a/2 Sqrt(p*(1-p)/n))
 x = Mean
 n = Sample Size
 a = 1 - (Confidence Level/100)
 Za/2 = Z-table value
 CI = Confidence Interval
 Mean(x)=420
 Sample Size(n)=500
 Sample proportion =0.84
 Margin of Error = Z a/2 * ( Sqrt ( (0.84*0.16) /500) )
 = 1.96* Sqrt(0)
 =0.032
 c)
 Confidence Interval For Proportion
 CI = p ± Z a/2 Sqrt(p*(1-p)/n)))
 x = Mean
 n = Sample Size
 a = 1 - (Confidence Level/100)
 Za/2 = Z-table value
 CI = Confidence Interval
 Mean(x)=420
 Sample Size(n)=500
 Sample proportion = x/n =0.84
 Confidence Interval = [ 0.84 ±Z a/2 ( Sqrt ( 0.84*0.16) /500)]
 = [ 0.84 - 1.645* Sqrt(0) , 0.84 + 1.65* Sqrt(0) ]
 = [ 0.813,0.867]

