Question

Assuming that the population is normally​ distributed, construct a 95%

confidence interval for the population​ mean, based on the following sample size of n equals 7. ​1, 2,​ 3,

4, 5, 6, 7, and 23

In the given​ data, replace the value

23

with

7

and recalculate the confidence interval. Using these​ results, describe the effect of an outlier​ (that is, an extreme​ value) on the confidence​ interval, in general.

Find a 95%

confidence interval for the population​ mean, using the formula or technology.

Answer 1

sample mean, xbar = 6.29
sample standard deviation, s = 7.5656
sample size, n = 6
degrees of freedom, df = n - 1 = 5

Given CI level is 95%, hence α = 1 - 0.95 = 0.05
α/2 = 0.05/2 = 0.025, tc = t(α/2, df) = 2.571

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (6.29 - 2.571 * 7.5656/sqrt(6) , 6.29 + 2.571 * 7.5656/sqrt(6))
CI = (-1.65 , 14.23)

Replace 7 by 23
sample mean, xbar = 4
sample standard deviation, s = 2.1602

CI = (xbar - tc * s/sqrt(n) , xbar + tc * s/sqrt(n))
CI = (4 - 2.571 * 2.1602/sqrt(6) , 4 + 2.571 * 2.1602/sqrt(6))
CI = (1.73 , 6.27)

The CI widens because of the presence of outliers