Question

Sleeping outlier: A simple random sample of nine college freshmen were asked how many hours of sleep they typically got per night. The results were

8.5 24 9 8.5 8 6.5 6 7.5 9.5  

Notice that one joker said that he sleeps 24 a day.

(a) The data contain an outlier that is clearly a mistake. Eliminate the outlier, then construct a 99.9% confidence interval for the mean amount of sleep from the remaining values. Round the answers to at least two decimal places.

(b) Leave the outlier in and construct the 99.9% confidence interval. Round the answers to at least two decimal places.

(c) Are the results noticeably different?

Answer 1

(a)
without outlier
calculated sample mean, x =7.9375
standard deviation, s =1.2082
sample size, n =8
level of significance, alpha = 0.001
from standard normal table, two tailed value of |t alpha/2| with n-1 = 7 d.f is 5.408
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 = [ 7.9375 ± t a/2 ( 1.2082/ Sqrt ( 8) ]
= [ 7.9375-(5.408 * 0.427) , 7.9375+(5.408 * 0.427) ]
= [ 5.627 , 10.248 ]
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(b)
with outlier data
given that,
sample mean, x =9.7222
standard deviation, s =5.4722
sample size, n =9
confidence interval = [ 9.7222 ± t a/2 ( 5.4722/ Sqrt ( 9) ]
= [ 9.7222-(5.041 * 1.824) , 9.7222+(5.041 * 1.824) ]
= [ 0.527 , 18.917 ]
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(c)
yes, signifincatly diffrent between both the intervals captured