1. [7pts The average loans they owed in a big city were $110,000 of a random sam
ID: 3386687 • Letter: 1
Question
1. [7pts The average loans they owed in a big city were $110,000 of a random sample of 1600 households. Given the standard deviation of household loan amount in this city sigma = $40,000. An approximate 95% confidence interval for the average amount of loans that all households owned is a)$110,000-$120,000 b)$30,000-$190,000 c)$108,000-$112,000 d)$70,000-$150,000 2. (7pts)Suppose that scores on the mathematics part of the National Assessment of Educational progress (NAEP) test for high school seniors follow a Normal distribution with standard deviation sigma = 30. You want to estimate the mean score within + 10 with 90% confidence. How large an SRS of scores must you choose? a)10 b)30 c)25 d) 24.4 3.[5pts] Determine whether a 95% confidence Internal involves a t-procedure, z-procedure, or neither. Sample data: n = 50, x = 984, s = 25. a) T-procedure B) Neither b) Z-procedureExplanation / Answer
1.
Note that
Margin of Error E = z(alpha/2) * s / sqrt(n)
Lower Bound = X - z(alpha/2) * s / sqrt(n)
Upper Bound = X + z(alpha/2) * s / sqrt(n)
where
alpha/2 = (1 - confidence level)/2 = 0.025
X = sample mean = 110000
z(alpha/2) = critical z for the confidence interval = 1.959963985
s = sample standard deviation = 40000
n = sample size = 1600
Thus,
Margin of Error E = 1959.963985
Lower bound = 108040.036
Upper bound = 111959.964
Thus, the confidence interval is
( 108040.036 , 111959.964 )
or
OPTION C: 108000-112000 [ANSWER]
*******************************************
Hi! Please submit the next part as a separate question. That way we can continue helping you! Please indicate which parts are not yet solved when you submit. Thanks!
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.