A family is relocating from St. Louis, Missouri, to California. Due to an increa
ID: 2922474 • Letter: A
Question
A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. Her realtor friend informs them that the last 23 houses that sold in their neighborhood took an average time of 180 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 62 days. Use Table 1.
What assumption regarding the population is necessary for making an interval estimate for the population mean?
Construct the 95% confidence interval for the mean sale time for all homes in the neighborhood. (Round intermediate calculations to 4 decimal places. Round "z-value" to 3 decimal places and final answer to 2 decimal places.)
A family is relocating from St. Louis, Missouri, to California. Due to an increasing inventory of houses in St. Louis, it is taking longer than before to sell a house. The wife is concerned and wants to know when it is optimal to put their house on the market. Her realtor friend informs them that the last 23 houses that sold in their neighborhood took an average time of 180 days to sell. The realtor also tells them that based on her prior experience, the population standard deviation is 62 days. Use Table 1.
Explanation / Answer
TRADITIONAL METHOD
given that,
standard deviation, =62
sample mean, x =180
population size (n)=23
I.
stanadard error = sd/ sqrt(n)
where,
sd = population standard deviation
n = population size
stanadard error = ( 62/ sqrt ( 23) )
= 12.928
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 * 12.928
= 25.339
III.
CI = x ± margin of error
confidence interval = [ 180 ± 25.339 ]
= [ 154.661,205.339 ]
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DIRECT METHOD
given that,
standard deviation, =62
sample mean, x =180
population size (n)=23
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 = [ 180 ± Z a/2 ( 62/ Sqrt ( 23) ) ]
= [ 180 - 1.96 * (12.928) , 180 + 1.96 * (12.928) ]
= [ 154.661,205.339 ]
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interpretations:
1. we are 95% sure that the interval [154.661 , 205.339 ] 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 = 180
standard error =12.928
z table value = 1.96
margin of error = 25.339
confidence interval = [ 154.661 , 205.339 ]
[ANSWERS]
a. Assume that the population has a normal distribution
b. [ 154.661,205.339 ]
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