Assuming that the population is normally distributed, construct a 90% confidence
ID: 3319246 • Letter: A
Question
Assuming that the population is normally distributed, construct a 90% confidence interval or the population mean, based on the following sample size of n-6 1, 2,3, 4, and 23 In the given data, replace the value 23 with 5 and recalculate the confidence interval. Using these results, describe the effect of an outlier (that is, an extreme value) on the confidence interval ingeneral Find a 90% confidence interval for the population mean, using the formula or technology. Os « In the given data replace the value 23 with 5 Find a 90% confidence interval or he population mean using he or nu a or technology. | | (Round to two decimal places as needed.) s (Round to two decimal places as needed.) Using the results from the previous two steps, what is the effect of an outlier (that is, an extreme value) on the confidence interval, in general? A he presence o an outlier in he original data increases he value o the sample mean and greatly decreases e sample standard de ation, narrowing econ ence ter a O B. The presence of an outlier in the original data increases the value of the sample mean and greatly inflates the sample standard deviation, widening the confidence interval. C The presence of an outlier in the original data decreases the value of the sample mean and great y inflates the sample standard deviation, widening the confidence interval. D. The presence of an outlier in the original data decreases the value of he sample mean and great y decreases the sample standard deviation, narrowing the confidence intervalExplanation / Answer
TRADITIONAL METHOD
given that,
sample mean, x =3
standard deviation, s =1.5811
sample size, n =5
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 1.5811/ sqrt ( 5) )
= 0.707
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 = 4 d.f is 2.132
margin of error = 2.132 * 0.707
= 1.508
III.
CI = x ± margin of error
confidence interval = [ 3 ± 1.508 ]
= [ 1.492 , 4.508 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =3
standard deviation, s =1.5811
sample size, n =5
level of significance, = 0.1
from standard normal table, two tailed value of |t /2| with n-1 = 4 d.f is 2.132
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 = [ 3 ± t a/2 ( 1.5811/ Sqrt ( 5) ]
= [ 3-(2.132 * 0.707) , 3+(2.132 * 0.707) ]
= [ 1.492 , 4.508 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 90% sure that the interval [ 1.492 , 4.508 ] 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