DIUF3bOohzcihyLIAfmiDHseS&displayMode; Mathlass 12 webwork/fall201718-stat200-mc
ID: 3357492 • Letter: D
Question
DIUF3bOohzcihyLIAfmiDHseS&displayMode; Mathlass 12 webwork/fall201718-stat200-mccaa / stat 200_chapter7 /2 Stat 200 Chapter7: Problem 2 Previous Problem Problem List Next Problem (1 point) Starting salaries of 105 college graduates who have taken a statistics course have a mean of $43,491. The population standard deviation is known to be $9,137. Using 99% confidence, find both of the following: A. The margin of error: B. Confidence interval: Note: You can earn partial credit on this problem. Preview My Answers Submit Answers You have attempted this problem 0 times. You have unlimited attempts remaining. Email instructorExplanation / Answer
TRADITIONAL METHOD
given that,
standard deviation, =9137
sample mean, x =43491
population size (n)=105
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 9137/ sqrt ( 105) )
= 891.68
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
margin of error = 2.576 * 891.68
= 2296.967
III.
CI = x ± margin of error
confidence interval = [ 43491 ± 2296.967 ]
= [ 41194.033,45787.967 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
standard deviation, =9137
sample mean, x =43491
population size (n)=105
level of significance, = 0.01
from standard normal table, two tailed z /2 =2.576
since our test is two-tailed
value of z table is 2.576
we use CI = x ± Z a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
Za/2 = Z-table value
CI = confidence interval
confidence interval = [ 43491 ± Z a/2 ( 9137/ Sqrt ( 105) ) ]
= [ 43491 - 2.576 * (891.68) , 43491 + 2.576 * (891.68) ]
= [ 41194.033,45787.967 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 99% sure that the interval [41194.033 , 45787.967 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 43491
standard error =891.68
z table value = 2.576
a.
margin of error = 2296.967
b.
confidence interval = [ 41194.033 , 45787.967 ]
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.