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

answer: the answer is (b) amie should have multiplied

step-by-step explanation: given a sphere and a cylinder with same radius and height.

let, 'r' be the radius of the cylinder and sphere and 'h' be the height of the cylinder. according to the question,

r = h.

now, the volume of the sphere will be

and the volume of the cylinder will be

therefore, we have

thus, ammie should have multiplied the volume of the sphere