Question

Suppose a random sample of 50 college students is asked if they regularly eat breakfast. A 95% confidence interval for the proportion of all students that regularly eat breakfast is found to be 0.69 to 0.91. If a 90% confidence interval was calculated instead, how would it differ from the 95% confidence interval

Answer 1

The 90% confidence interval would be narrower than a 95% confidence interval

Explanation:

From the question we are told that

The 95% confidence interval is 0.69 to 0.91.

Generally the width of a confidence interval is dependent on the margin of error

Generally the margin of error is mathematically represented as

`E = Z_{(\alpha )/(2) } *  (\sigma )/(√(n) )`

Here we see that

`E \ \ \alpha \ \  Z_{(\alpha )/(2) }`

Here
`Z_{(\alpha )/(2) }` is the critical value of half of the level of significance .

This value increase as the value of confidence level increase and vise versa

So for 90% confidence interval , the confidence level is 90% , hence the value of
`Z_{(\alpha )/(2) }` will decrease which will in turn decrease the value of E , hence the width of the interval will reduce

So the 90% confidence interval is narrower than a 95% confidence interval when the sample proportion or sample mean is constant