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https://www.mathadl.com/Student/PlayerTest.aspx?testld-123933873&centerwin; yesåfromPlayerCheck-yes Joshua De OST UNIT 6-TEST-Ch. 9 T of s aomplite This Question: 1 pt This Test: 8pts 7 of 8 complete The following data represent the asking price of a simple random sample of homes for sale. Construct a 99% confidence interval with and without the outlier included. Comment on the effect the outlier has on the confidence interval. $252,500 143,000 $459,900 $214,900 $279,900 $205,800 $203,000 $147,800 $219,900 $295,900 $187,500 $264,900 Click the icon to view the table ofaras under the t-distribution. (a) Construct a 99% confidence interval with the outlier included. (Round to the nearest integer as needed.) (b) Construct a 99% confidence interval with the outlier removed. (Round to the nearest integer as needed.) (e) Comment on the effect the outlier has on the confidence interval. O The outlier caused the width of the confidence interval to decrease. O The outlier had no effect on the width of the confidence interval The outlier caused the width of the confidence interval to increase. Click to select your answerts) Previous Question Next question Submit Test

Explanation / Answer

a)

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    239583.3333          
t(alpha/2) = critical t for the confidence interval =    3.105806516          
s = sample standard deviation =    84253.81387          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
              
Lower bound =    164043.9661          
Upper bound =    315122.7006          
              
Thus, the confidence interval is              
              
(   164043.9661   ,   315122.7006   ) [ANSWER]

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b)

Removing the 459900 entry,

Note that              
              
Lower Bound = X - t(alpha/2) * s / sqrt(n)              
Upper Bound = X + t(alpha/2) * s / sqrt(n)              
              
where              
alpha/2 = (1 - confidence level)/2 =    0.005          
X = sample mean =    219554.5455          
t(alpha/2) = critical t for the confidence interval =    3.169272673          
s = sample standard deviation =    50133.46913          
n = sample size =    11          
df = n - 1 =    10          
Thus,              
              
Lower bound =    171648.4229          
Upper bound =    267460.668          
              
Thus, the confidence interval is              
              
(   171648.4229   ,   267460.668   ) [ANSWER]

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c)

As we can see, part a) is wider, as it has the outlier.

OPTION C: The outlier caused the width of the confidence interval to increase. [ANSWER]

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