How large a sample is needed in problem 9 if we wish to be 9

How large a sample is needed in problem 9 if we wish to be 95% confident that the sample percentage of those equating success with personal satisfaction is within 1% of the population percentage? Hint, use p=0.52 as a preliminary estimate. (answer should be: 9589)

Problem 9. The national study of the changing workforce conducted an extensive survey of 2958 wage and salaried workers on issues ranging from relationships with their bosses to household chores. The data were gathered throught hour-long- telephone interviews with a nationally representative sample. In response to the question, \"What does success mean to you?\" 1538 responded. Let p be the population proportion of all wage and salaried workers who would respond the same way to the stated question. Find a 90% confidence interval for p.

Please show ALL work.

Solution

p = 0.52...standard error = sqrt( 0.52 * 0.48 / 2958 ) =0.009185931.......
90% confidence interval = ( 0.52 - 1.644854*0.009185931 , 0.52 + 1.644854*0.009185931) =( 0.5049 ,0.5351)...

Now, diffrence in poltn mean and sample mean = 0.01..we have to find the sample size...z for 95% confidence= 1.96..

so, z = (poltn mean- smaple mean) / (std. error)
z= 1.96 = 0.01 / sqrt( 0.52*.48 / n ).....where n= reqd sample size....
or, n = 9588.6336 = 9589 (rounded because,sample size should be an integer)