A 98% confidence interval estimate for a population mean was computed to be (45.
ID: 2957471 • Letter: A
Question
A 98% confidence interval estimate for a population mean was computed to be (45.9, 60.5). Determine the mean of the sample, which was used to determine the interval estimate (show all work).A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children. A sample of 145 was taken, and the mean amount spent was $187.75. Assuming a standard deviation equal to $39.83, find the 90% confidence interval for m, the mean for all such families (show all work).
A confidence interval estimate for the population mean is given to be (41.16, 49.85). If the standard deviation is 16.290 and the sample size is 54, answer each of the following (show all work):
(A) Determine the maximum error of the estimate, E.
(B) Determine the confidence level used for the given confidence interval.
Explanation / Answer
A 98% confidence interval estimate for a population mean was computed to be (45.9, 60.5).
mean = (lower limit + upper limit of confidence interval)/2
=(45.9 + 60.5)/2 = 53.2
A study was conducted to estimate the mean amount spent on birthday gifts for a typical family having two children.
sample size, n= 145
mean , m = $187.75.
standard deviation, s = $39.83
standard error, e = s/n = 39.83/145 = 3.31
at 90% confidence , for a two tail distribution, z = 1.65 (from z table)
hence confidence interval for m, the mean for all such families = m ± (z*e)
= 187.75 ± (1.65*3.31)
=187.75 ± 5.5
= (182.25, 193.25)
A confidence interval estimate for the population mean is given to be (41.16, 49.85). If the standard deviation, s = 16.290 and the sample size, n= 54, :
hence mean, m = (lower limit +upper limit)/2 = (41.16+49.85)/2 =45.505
(A) the maximum error of the estimate, E = one half the width of the confidence interval
= (49.85-41.16)/2 =4.345
(B)standard error, e = s/n = 16.290/54 = 2.217
hence, maximum error of estimate, E = z * e
hence, 4.345 = z * 2.217
hence, z = 1.96
referring to the z table we find that for a two tail distribution, the value of z - 1.96 corresponds to 95% confidence level
Related Questions
drjack9650@gmail.com
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.