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?

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. = ~ N(0,1)

where, = proportion of companies engaging in such downsizing practices in a sample of 723 selected =

           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 < < 2.5758) = 0.99

P( < < ) = 0.99

P( < p < ) = 0.99

99% confidence interval for p =[ , ]

 = [ , ]

 = [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].

idk what the answer is but for the points

step-by-step explanation:

the first one and the second one that's the answer