Question

In economic​ downturns, companies attempt to downsize their workforces by offering early retirement incentives to older employees. A survey of 723 randomly selected companies found that 195 engage in such downsizing practices. What is the​ 99% confidence interval for the proportion of companies that downsize their workforces by offering early retirement​ incentives?

Answer 1

Answer:

99% confidence interval for the proportion of companies that downsize their work forces by offering early retirement​ incentives is [0.227 , 0.312].

Step-by-step explanation:

We are given that in economic​ downturns, companies attempt to downsize their work forces by offering early retirement incentives to older employees. A survey of 723 randomly selected companies found that 195 engage in such downsizing practices.

Firstly, the pivotal quantity for 99% confidence interval for the population proportion of companies that downsize their work forces by offering early retirement​ incentives is given by;

     P.Q. = $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ~ N(0,1)

where, $\hat p$ = proportion of companies engaging in such downsizing practices in a sample of 723 selected = $\frac{195}{723}$

           n = sample of companies = 723

           p = population proportion of companies

Here for constructing 99% confidence interval we have used One-sample z proportion statistics.

So, 99% confidence interval for the population proportion, p is ;

P(-2.5758 < N(0,1) < 2.5758) = 0.99  {As the critical value of z at 0.5%

                                                 significance level are -2.5758 & 2.5758}

P(-2.5758 < $\frac{\hat p-p}{\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < 2.5758) = 0.99

P( $-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < ${\hat p-p}$ < $2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

P( $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ < p < $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$ ) = 0.99

99% confidence interval for p =[ $\hat p-2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$, $\hat p+2.5758 \times {\sqrt{\frac{\hat p(1-\hat p)}{n} } }$]

 = [ $\frac{195}{723} -2.5758 \times {\sqrt{\frac{\frac{195}{723} (1-\frac{195}{723} )}{723} } }$ , $\frac{195}{723} +2.5758 \times {\sqrt{\frac{\frac{195}{723} (1-\frac{195}{723} )}{723} } }$ ]

 = [0.227 , 0.312]

Therefore, 99% confidence interval for the proportion of companies that downsize their work forces by offering early retirement​ incentives is [0.227 , 0.312].