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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