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

Answer:

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

Step-by-step 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  

      $E = Z_{\frac{\alpha }{2} } *  \frac{\sigma }{\sqrt{n} }$

Here we see that

           $E \ \ \alpha \ \  Z_{\frac{\alpha }{2} }$

    Here  $Z_{\frac{\alpha }{2} }$  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  $Z_{\frac{\alpha }{2} }$  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