Question

A study of​ full-time workers in a certain field found that only about 66​% leave their jobs in order to retire. Assume that the true proportion of all​ full-time workers in the field who leave their jobs in order to retire is p=0.06. In a random sample of 850 full-time workers in the​ field, let ModifyingAbove p with caret p^ represent the proportion who leave their jobs in order to retire. Complete parts a through c below.

a. Describe the properties of the sampling distribution of ModifyingAbove p with caret p^.

The mean of the sampling distribution of ModifyingAbove p with caret p^ is ___.

The standard deviation of the sampling distribution of ModifyingAbove p with caret p^ is ___.

​(Round to four decimal places as​ needed.)

b. Compute P(p^<0.08).

The probability that ModifyingAbove p with caret p^is less than 0.08 is ___.

​(Round to four decimal places as​ needed.)

Interpret this result.

The probability that ___(the population of all workers/a random sample of 850 workers/any random sample of workers)___ in the field has a ___(proportion/mean/standard deviation/variance)___ of workers that leave the job in order to retire ___(less than/greater than/equal to) ___ 0.08 is ___(greater than/equal to/less than)___ the calculated probability.

c. Compute P(p^>0.046).

The probability that ModifyingAbove p with caret p^ is greater than 0.046 is ___.

​(Round to four decimal places as​ needed.)

Interpret this result.  

The probability that ___(any random sample of workers/ a random sample of 850 workers/ the entire population of workers)___ in the field has a ___(mean/ proportion/ standard deviation)___ of workers that leave the job in order to retire ___(less than/ equal to/ greater than 0.046 is ___(equal to/ greater than/ less than)___ the calculated probability.

Answer 1

a) The mean of the sampling distribution of sample proportion = p(population proportion )= 0.06

the standard deviation of the sampling distribution of sample proportion

= = = 0.0081

b) By central limit theorem , the sampling distribution of sample proportion follow Normal

that is

then

= P( z < 2.47)

= 0.9932 ( from z table)

The probability that a random sample of 850 workers has a proportion of workers that leave the job in order to retire less than 0.08 equal to 0.9932

c)

= P( z > -1.73 )

= 0.9582 ( from z table)

The probability that a random sample of 850 workers has a proportion of workers that leave the job in order to retire greater than  0.08 equal to 0.9582