A random sample of 130 mortgages in the state of Mississippi was randomly select
ID: 3181684 • Letter: A
Question
A random sample of 130 mortgages in the state of Mississippi was randomly selected. From this sample, 14 were found to be delinquent on their current payment. The margin of error for an 80% confidence interval for the proportion based on this sample is ________.
0.0447
0.0560
0.0622
0.0737
Steve's average golf score at his local course is 92.9 from a random sample of 60 rounds of golf. Assume that the population standard deviation for his golf score is 4.2. The 99% confidence interval around this sample mean is ________.
(87.6, 98.2)
(90.7, 95.1)
(89.5, 96.3)
(92.4, 94.2)
(87.6, 98.2)
(90.7, 95.1)
(89.5, 96.3)
(92.4, 94.2)
Explanation / Answer
p = 14/130 = 0.107 , q = 1- p = .893
z value at 805 confidence interval = 1.28
margin of error = z * sqrt(pq / n)
= 1.28 * sqrt ( 0.107 * .893/ 130)
= .0347
2)
mean =92.9 , std. deviation = 4.2 , n =60
z value at 99% confidence interval = 2.57
CI = mean +/- z * SE
= 92.9 + / - 2.57 *(4.2 / sqrt(60))
= (91.5 , 94.2)
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.