Question

A university dean is interested in determining the proportion of students who receive some sort of financial aid. Rather than examine the records for all students, the dean randomly selects 200 students and finds that 118 of them are receiving financial aid.

a) Use a 90% confidence interval to estimate the true proportion of students who receive financial aid. Explain/show how you obtain your answer.

b) If the dean wanted to estimate the proportion of all students receiving financial aid to within 3% with 99% reliability, how many students would need to be sampled? Explain/show how you obtain your answer.

Answer 1

a

The 90% confidence interval that estimate the true proportion of students who receive financial aid is

`0.533  <  p <  0.64`

b

`n = 1789`

Explanation:

Considering question a

From the question we are told that

The sample size is n = 200

The number of student that receives financial aid is
`k = 118`

Generally the sample proportion is

`\^ p = (k)/(n)`

=>
`\^ p = (118)/(200)`

=>
`\^ p = 0.59`

From the question we are told the confidence level is 90% , hence the level of significance is

`\alpha = (100 - 90 ) \%`

=>
`\alpha = 0.10`

Generally from the normal distribution table the critical value of
`(\alpha )/(2)` is

`Z_{(\alpha )/(2) } =  1.645`

Generally the margin of error is mathematically represented as

`E =  Z_{(\alpha )/(2) } * \sqrt{(\^ p (1- \^ p))/(n) }`

=>
`E =  1.645 * \sqrt{(0.59 (1- 0.59))/(200) }`

=>
`E = 0.057`

Generally 90% confidence interval is mathematically represented as

`\^ p -E <  p <  \^ p +E`

=>
`0.533  <  p <  0.64`

Considering question b

From the question we are told that

The margin of error is E = 0.03

From the question we are told the confidence level is 99% , hence the level of significance is

`\alpha = (100 - 99 ) \%`

=>
`\alpha = 0.01`

Generally from the normal distribution table the critical value of is

`Z_{(\alpha )/(2) } = 2.58`

Generally the sample size is mathematically represented as

`[\frac{Z_{(\alpha )/(2) }}{E} ]^2 * \^ p (1 - \^ p )`

=>
`n = [(2.58)/(0.03) ]^2 * 0.59 (1 - 0.59 )`

=>
`n = 1789`