Question

A simple random sample of college students was performed recently in which the students were asked to self identify as to whether they are binge drinkers or not. In a 1995 study it was found that about 44% of students were self identified as such and the researcher believed that this number has increased. In this study 7851 of the 17592 students interviewed self identified as binge drinkers. A 90% confidence interval of the estimated binge drinking rate is

Answer 1

The 90% confidence interval is
`0.445< p < 0.456`

Explanation:

From the question we are told that

The sample size is
`n = 17592`

The number of binge drinkers is
`k = 7851`

Given that the confidence level is 90% then the level of significance is mathematically represented as

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

`\alpha = 0.10`

The critical value of
`(\alpha )/(2)` from the normal distribution table is
`Z_{(\alpha )/(2) } = 1.645`

The sample proportion is mathematically represented as

`\r p = ( 7851)/( 17592)`

`\r p = 0.45`

Generally the margin of error is mathematically represented as

`E = Z_{ (x)/(y) } * \sqrt{(\r p(1 - \r p ))/(n) }`

`E =1.645 * \sqrt{( 0.45 (1 - 0.45 ))/(17592) }`

`E =0.00617`

The 90% confidence interval is mathematically represented as

`\r p -E < p < \r p +E`

`0.45 -0.00617 < p < 0.45 + 0.00617`

`0.445< p < 0.456`