You want to rent an unfurnished one-bedroom apartment in Boston next year. The m

Question

You want to rent an unfurnished one-bedroom apartment in Boston next year. The mean monthly rent for a random sample of 11 apartments advertised in the local newspaper is $1200. Assume that the standard deviation is $.250. Find a 95% confidence interval for the mean monthly rent for unfurnished one-bedroom apartments available for rent in this community. (Round your answers to two decimal places.) will the 95% confidence interval include approximately 95% of the rents of all unfurnished one-bedroom apartments in this area? Explain why or why not. No, this is a range of values for the mean rent, not for individual rents. O Yes, this range implies that it will cover 95% of all of the rents in this area It is impossible to know because we would have to find all of the rents for unfurnished one-bedroom apartments in this area. O No, this confidence interval would include 95% of all individual rents in the area Yes, this is a range where 95% of the means and individual costs will be covered by the interval

Explanation / Answer

TRADITIONAL METHOD
given that,
standard deviation, =250
sample mean, x =1200
population size (n)=11
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 250/ sqrt ( 11) )
= 75.378
II.
margin of error = Z a/2 * (stanadard error)
where,
Za/2 = Z-table value
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
margin of error = 1.96 * 75.378
= 147.741
III.
CI = x ± margin of error
confidence interval = [ 1200 ± 147.741 ]
= [ 1052.259,1347.741 ]
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DIRECT METHOD
given that,
standard deviation, =250
sample mean, x =1200
population size (n)=11
level of significance, = 0.05
from standard normal table, two tailed z /2 =1.96
since our test is two-tailed
value of z table is 1.96
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 = [ 1200 ± Z a/2 ( 250/ Sqrt ( 11) ) ]
= [ 1200 - 1.96 * (75.378) , 1200 + 1.96 * (75.378) ]
= [ 1052.259,1347.741 ]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [1052.259 , 1347.741 ] contains the true population mean
2. if a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population mean
[ANSWERS]
best point of estimate = mean = 1200
standard error =75.378
z table value = 1.96
margin of error = 147.741
confidence interval = [ $1052.259 , $1347.741 ]
yes, this range implies that will cover 95% of all the rents covered in this area

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