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

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  

Here we see that

    Here    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    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

unless you made a typo, the answers are both a and c.

step-by-step explanation:

1.5/3 is .5 whereas 3/9 is .3333333 not proportional

22/33 is .66666 and 4/6 is also .66666 proportional

2/4 is .5 whereas 142/213 is .66666 not proportional

14/7 is 2 and 7/3.5 is also 2 proportional

you could've also checked it by doing this simple division:

if the larger fraction divides with the smaller fraction to equal 1, it is proportional. so let me show you what i mean:

when i divide 22/33 by 4/6 i get 1, so we know it is proportional, and it is the same with 14/7 and 7/3.5. now if i divide 3/9 by 1.5/3 i get .66666 so therefore we know it is not proportional.

there are many ways to do this so i wanted to show you some possible ways.