Please show work!!! NO hand written Answers, Only Typed Answers! of the mean tim
ID: 3355776 • Letter: P
Question
Please show work!!! NO hand written Answers, Only Typed Answers!
of the mean time takes al lilvidua confidence interval estimate income tax return? 10. Costs arerising for all kinds of medical care. The mean monthly rent at assed-iving act was reported to have increased 17% over the last five years to S3486 (The Wall sredj 8.2 Population Mean: Unknown 343 October 27, 2012). Assume this cost estimate is based on a sample of 120 facilities. From past studies, it can be assumed that the population standard deviation is $650. a. Develop a 90% confidence interval estimate of the population mean monthly rent. b. Develop a 95% confidence interval estimate of the population mean monthly rent. c. Develop a 99% confidence interval estimate of the population mean monthly rent. d. What happens to the width of the confidence interval as the confidence level is increased? Does this seem reasonable? ExplainExplanation / Answer
a. AT 90% CI
given that,
standard deviation, =650
sample mean, x =3486
population size (n)=120
level of significance, = 0.1
from standard normal table, two tailed z /2 =1.645
since our test is two-tailed
value of z table is 1.645
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 = [ 3486 ± Z a/2 ( 650/ Sqrt ( 120) ) ]
= [ 3486 - 1.645 * (59.337) , 3486 + 1.645 * (59.337) ]
= [ 3388.391,3583.609 ]
b.
AT 95% CI
confidence interval = [ 3486 ± Z a/2 ( 650/ Sqrt ( 120) ) ]
= [ 3486 - 1.96 * (59.337) , 3486 + 1.96 * (59.337) ]
= [ 3369.7,3602.3 ]
c.
AT 99% CI
confidence interval = [ 3486 ± Z a/2 ( 650/ Sqrt ( 120) ) ]
= [ 3486 - 2.576 * (59.337) , 3486 + 2.576 * (59.337) ]
= [ 3333.149,3638.851 ]
d.
When width of the CI increases the interval is wider