# Answer in Statistics and Probability for Goldie #250340

A budget head for SSS Sales Inc. would like to compare the daily travel expenses for the sales staff and the audit staff. He collected the following sample information.

Sales Staff

105

120

140

160

100

170

160

140

130

160

180

190

Audit Staff

130

50

90

80

70

125

180

80

110

At 0.05 significance level, can he conclude that the mean daily expenses of sales staff are greater than the audit staff?

Let “mu_1” be the sample mean for population 1 (sales staff) and “mu2” be the sample mean for population 2(audit staff) then, the hypothesis tested is,

“H_0:mu_1=mu_2”

“Against”

“H_1:mu_1gtmu_2”

To perform this hypothesis test we shall use the “p-value” technique the compare this value with the given “alpha =0.05” level of significance.

By applying the 2 sample t-test in “R” we can easily perform the test using the commands below,

x=c(105,120,140,160,100,170,160,140,130,160,180,190)

y=c(130,50,90,80,70,125,180,80,110)

t. test(x, y)

The output for these commands are as shown below.

Welch Two Sample t-test

data: x and y

t = 2.8812, df = 14.066, p-value = 0.01204

alternative hypothesis: true difference in means is not equal to 0

95 percent confidence interval:

11.40976 77.75691

sample estimates:

mean of x mean of y

146.2500 101.6667

From this output the “p-value” is divided by 2 since we are performing a one sided test thus, “p-value=0.01204/2=0.00602” and reject the null hypothesis if “p-valueltalpha”

Since “p-value=0.00602ltalpha=0.05,” we reject the null hypothesis and conclude that evidence exist to show that the mean daily expenses of sales staff are greater than the audit staff.

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