Question

The sales manager for a hardware wholesaler finds that 229 of the previous 500 calls to hardware store owners resulted in new product placements. Assuming that the 500 calls represent a random sample, find a 95% confidence interval for the long-run proportion of new product placements

Answer 1

$95\%CI=(p-z_{0.025}\sqrt{\frac{p(1-p)}{n}},p+z_{0.025}\sqrt{\frac{p(1-p)}{n}})=$

$=(\frac{229}{500}-1.96\sqrt{\frac{\frac{229}{500}(1-\frac{229}{500})}{500}},\frac{229}{500}+1.96\sqrt{\frac{\frac{229}{500}(1-\frac{229}{500})}{500}})=$

$=(0.4143,0.5017).$