Question

3. P343. 13. A sociologist is interested in surveying workers in computer-related jobs to estimate the proportion of such workers who has changed jobs within the past years. (a) In the absence of preliminary data, how large a sample must be taken to ensure that a 95% confidence interval will specify the proportion to within ±0.05? (b) In a sample of 100 workers, 20 of them had changed jobs within the past year. Find a 95% confidence interval for the proportion of workers who have changed jobs within the past year. (c) Based on the data in part (b), estimate the sample size needed so that the 95% confidence interval will specify the proportion to within ±0.05.

Answer 1

a) A sample of 385 must be taken.

b) The 95% confidence interval for the proportion of workers who have changed jobs within the past year is (0.1216, 0.2784).

c) A sample of 246 is needed.

Explanation:

Question a:

In a sample with a number n of people surveyed with a probability of a success of
`\pi`, and a confidence level of
`1-\alpha`, we have the following confidence interval of proportions.

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

The margin of error is:

`M = z\sqrt{(\pi(1-\pi))/(n)}`

Absence of preliminary data:

This means that we use
`\pi = 0.5`, which is when the largest sample will be needed.

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

How large a sample must be taken to ensure that a 95% confidence interval will specify the proportion to within ±0.05?

A sample of n must be taken.

n is found when
`M = 0.05`. So

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.05 = 1.96\sqrt{(0.5*0.5)/(n)}`

`0.05√(n) = 1.96*0.5`

Dividing both sides by 0.05

`√(n) = 1.96*10`

`(√(n))^2 = (19.6)^2`

`n = 384.16`

Rounding up,

A sample of 385 must be taken.

Question b:

In a sample of 100 workers, 20 of them had changed jobs within the past year.

This means that
`n = 100, \pi = (20)/(100) = 0.2`

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.2 - 1.96\sqrt{(0.2*0.8)/(100)} = 0.1216`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.2 + 1.96\sqrt{(0.2*0.8)/(100)} = 0.2784`

The 95% confidence interval for the proportion of workers who have changed jobs within the past year is (0.1216, 0.2784).

Question c:

Same procedure as question a, just with
`\pi = 0.2`

`M = z\sqrt{(\pi(1-\pi))/(n)}`

`0.05 = 1.96\sqrt{(0.2*0.8)/(n)}`

`0.05√(n) = 1.96√(0.2*0.8)`

`√(n) = (1.96√(0.2*0.8))/(0.05)`

`(√(n))^2 = ((1.96√(0.2*0.8))/(0.05))^2`

`n = 245.8`

Rounding up

A sample of 246 is needed.