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Secure https//www mathxl.com/Student/PlayerHomework aspx?homeworkld 4640955108questionld-28cflushed falsecld 48286028centerwinyes BUS 352 TR700 Spring 2018 michael mccabe | 22718 7:08 AM Homework: Chapter 8 Homework Score: 0 of 1 pt 8.2.14-T HW Score: 66 67%, 4 of 6 pts Question Help * Asarmngthat te prulation is normaly distrtuted, oonstruct a 99% condenco rteralfor to population frean, based on the foliowrg sanie size6tn«6 I, 2, 3, 4, 5, and 20 In the given data, replace the value 20 with 6 and recalculate the confidence interval. Using these resuits, descnbe the offect of an outier (that is, an extreme value) on the confidence interval, in general Find a 99%cofidence rterval for the population mean, using the formulacrtechnology spsRund to two do ral places as needed .. I 1.1 |-fil:/i | 1. Iou) l Moro Enter your answer in the edit fields and then click Check Answer Clear Al Check Answe e to search End 8 2

Explanation / Answer

Sol:

done in R

Rcode is:

smpl <- c(1,2,3,4,5,20)
t.test(smpl,conf.level = 0.99)

One Sample t-test

data: smpl

t = 2.0174, df = 5, p-value = 0.0997

alternative hypothesis: true mean is not equal to 0

99 percent confidence interval:

-5.825844 17.492511

sample estimates:

mean of x

5.833333

99% confidence interval for mean with outlier is

99 percent confidence interval:
-5.825844 17.492511

-5.83 and 17.49

Now exclude outlier: and replace with 6:

Rocde is:

smpl2 <- c(1,2,3,4,5,6)
t.test(smpl2,conf.level = 0.99)

output:

One Sample t-test

data: smpl2
t = 4.5826, df = 5, p-value = 0.005934
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
0.4203999 6.5796001
sample estimates:
mean of x
3.5

ANSWER:

0.42 and 6.58

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