The following data represent the asking price of a simple random sample of homes

Question

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.

$266,900          $279,900              $219,900

$143,000          $205,800              $151,500

$459,000          $271,000              $187,500

$244,500          $147,800              $264,900

(PLEASE DISREGARD THE EMPTY TABLE...USE THE NUMBERS ABOVE IT)

(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.)

(c) Comment on the effect the outlier has on the confidence interval. (select one)

____ The outlier caused the width of the confidence interval to increase.

____ The outlier had no effect on the width of the confidence interval.

____ The outlier caused the width of the confidence interval to decrease

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 =    236808.3333          
t(alpha/2) = critical t for the confidence interval =    3.105806516          
s = sample standard deviation =    86248.91523          
n = sample size =    12          
df = n - 1 =    11          
Thus,              
              
Lower bound =    159480.2198          
Upper bound =    314136.4468          
              
Thus, the confidence interval is              
              
(   159480.2198   ,   314136.4468   ) [ANSWER]

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b)

Omitting 459000:

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 =    216609.0909          
t(alpha/2) = critical t for the confidence interval =    3.169272673          
s = sample standard deviation =    52887.03914          
n = sample size =    11          
df = n - 1 =    10          
Thus,              
              
Lower bound =    166071.7349          
Upper bound =    267146.447          
              
Thus, the confidence interval is              
              
(   166071.7349   ,   267146.447   ) [ANSWER]

********************

c)

As we can see, the one with the outlier is a wider interval, so

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

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