suppose a sample of 787 floppy disks is drawn. Of these disks 86 were defective. Use p=0.109 construct the 95% confidence interval for the population proportion of disks which are defective lower bound and upper bound round to 3 decimal places
Given an estimated proportion p = 0.109, a sample size n = 787 and a confidence interval of 95% we get:
p +/- z_crit x sqrt( p(1-p)/n )
z_crit can be found using:
N.inv ( p = 0.975, mean = 0, sd = 1 ) i.e. the 1-alpha/2 percentile of the unit normal = 1.96
So the lower and upper bound are given by:
0.109 +/- 1.96 x sqrt ( .109 x .891 / 787 )
= 0.109 +/- 0.0217
= 0.087 to 0.131
Hope this helped!
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