Assuming that the population is normally distributed, construct a 90 %90% confid
ID: 3171590 • Letter: A
Question
Assuming that the population is normally distributed, construct a
90 %90%
confidence interval for the population mean, based on the following sample size of n equals 5.n=5.1, 2, 3,
44,
and
2929
Change the number
2929
to
55
and recalculate the confidence interval. Using these results, describe the effect of an outlier (that is, an extreme value) on the confidence interval.
Find a
90 %90%
confidence interval for the population mean, using the formula or technology.
nothing less than or equalsmuless than or equalsnothing
(Round to two decimal places as needed.)
Explanation / Answer
a.
Confidence Interval
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
Mean(x)=7.8
Standard deviation( sd )=11.9038
Sample Size(n)=5
Confidence Interval = [ 7.8 ± t a/2 ( 11.9038/ Sqrt ( 5) ) ]
= [ 7.8 - 2.132 * (5.324) , 7.8 + 2.132 * (5.324) ]
= [ -3.55,19.15 ]
b.
Mean(x)=3
Standard deviation( sd )=1.5811
Sample Size(n)=5
Confidence Interval = [ 3 ± t a/2 ( 1.5811/ Sqrt ( 5) ) ]
= [ 3 - 2.132 * (0.707) , 3 + 2.132 * (0.707) ]
= [ 1.492,4.508 ]