Hi,
I just getting confused by these two questions below.
2. Refer to Exercise 8.11. Would a 99% confidence interval be wider or
narrower than the one that you found in that exercise? Verify your results
by computing the interval.
For question 1, I came up with the interval [.16 to .20] work shown below
p-hat = 192/1050= .18
Margin of error = (1.960)√((.18 (1-.18))/1050)= .02
CI 95% = .18 +.02 = .16 and .20
For question 2, I came up with the interval [.15 to .21] work shown below
p-hat = 192/1050= .18
Margin of error = (2.576)√((.18 (1-.18))/1050)= .03
CI 95% = .18 +.03 = .15 and .21
But for the Question 2, the book provides an answer of [0.3157 to .3843 ].
Did I do something wrong
Please Help and Thank You
I just getting confused by these two questions below.
- In the survey described in the previous exercise, there were 1050 borrowers whose total debt was $10,000 or more. Of these, 192 left school without completing a degree. Consider the population to be borrowers whose total debt was $10,000 or more. Find a 95% confidence interval for the proportion of borrowers who left school without completing a degree in this population.
2. Refer to Exercise 8.11. Would a 99% confidence interval be wider or
narrower than the one that you found in that exercise? Verify your results
by computing the interval.
For question 1, I came up with the interval [.16 to .20] work shown below
p-hat = 192/1050= .18
Margin of error = (1.960)√((.18 (1-.18))/1050)= .02
CI 95% = .18 +.02 = .16 and .20
For question 2, I came up with the interval [.15 to .21] work shown below
p-hat = 192/1050= .18
Margin of error = (2.576)√((.18 (1-.18))/1050)= .03
CI 95% = .18 +.03 = .15 and .21
But for the Question 2, the book provides an answer of [0.3157 to .3843 ].
Did I do something wrong
Please Help and Thank You