This Question: 4 pts 18 of 35 (1 This Test: 105 pts possible rse Ho A real estat

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This Question: 4 pts 18 of 35 (1 This Test: 105 pts possible rse Ho A real estate agent wants to estimate the average age of those buying investment property in his area. He randomly selects 15 of his clients who purchased an investment property and obtains the data shown. Use this information to answer the following questions. 36 31 48 46 40 , the data shown, Use this 7 55 48 45 41 58 43 51 44 39 zzes & Click the icon to view a boxplot and a normal probability plot property (Use ascending order. Round to one decimal place as needed. ) s for S medi OB. There is a 90% chance that the true value of the mean will fall in the interval. hase fD. There is a 90% chance that the true value of the mean wil not fall in the interval. son T Use the normal probability plot and boxplot to assist in verifying the model requirements A. We are 90% confident that the interval actually does not contain the true value of the mean. O c. we are 90% confident that the interval actualy does contain the true value of the mean. ssion OA. The requirements are not met since all the data lie within the bounds of the normal probablity plot and the boxplot does not reveal any cullers C. The O D. The requirements are not m et ements are met since not all the data lie within the bounds of the normal probability plot and the boxplot does reveals outliers. requirements are met since all the data lie within the bounds of the normal probability plot and the boxplot does not reveal any outliers since not all the data lie within the bounds of the normal probability plot and the boxplot does reveals outliers. Click to select your answerfs). (17) 29

Explanation / Answer

TRADITIONAL METHOD
given that,
sample data:
(36 31 48 46 40 47 55 48 45 41 58 43 51 44 39)
From above sample data,
calucalted sample mean, x =44.8
calucalted standard deviation, s =7.002
sample size, n =15
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 7.002/ sqrt ( 15) )
= 1.81
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 14 d.f is 1.761
margin of error = 1.761 * 1.81
= 3.18
III.
CI = x ± margin of error
confidence interval = [ 44.8 ± 3.18 ]
= [ 41.62 , 47.98 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =44.8
standard deviation, s =7.002
sample size, n =15
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 14 d.f is 1.761
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 44.8 ± t a/2 ( 7.002/ Sqrt ( 15) ]
= [ 44.8-(1.761 * 1.81) , 44.8+(1.761 * 1.81) ]
= [ 41.62 , 47.98 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 41.62 , 47.98 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population mean

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