Question

A study suggests that about 80% of undergraduate students are working at a job. Researchers plan to do a new study and use the new data to make a z interval on of students who are working at a job. They plan on using a confidence level of 90% and they want the margin of error to be no more than 1%. Let p represent the proportion of students who are working at a job and assume p 0.80, what is the smallest sample size required to obtain the desired margin of error?

Answer 1

4330

Explanation:

Answer 2

The smallest sample size required to obtain the desired margin of error is 4330

Explanation:

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)}`

In this problem, we have that:

`\pi = 0.8`

90% confidence level

So
`\alpha = 0.1`, z is the value of Z that has a pvalue of
`1 - (0.1)/(2) = 0.95`, so
`Z = 1.645`.

What is the smallest sample size required to obtain the desired margin of error?

This is n when
`M = 0.01`

So

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

`0.01 = 1.645\sqrt{(0.8*0.2)/(n)}`

`0.01√(n) = 1.645√(0.8*0.2)`

`√(n) = (1.645√(0.8*0.2))/(0.01)`

`(√(n))^(2) = ((1.645√(0.8*0.2))/(0.01))^(2)`

`n = 4329.6`

Rounding up

The smallest sample size required to obtain the desired margin of error is 4330