Question #295514

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?


1
Expert's answer
2022-02-10T04:01:11-0500

Sales staff(1)

n1=12xˉ1=xn1=175512=146.25n_1=12\\\bar x_1={\sum x\over n_1}={1755\over12}=146.25

s12=x2(x)2n1n11=265725256668.7511=823.2955s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={265725-256668.75\over11}=823.2955

Audit staff(2)

n2=9xˉ2=xn2=9159=101.66667n_2=9\\\bar x_2={\sum x\over n_2}={915\over9}=101.66667

s22=x2(x)2n2n21=105325930258=1537.5s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={105325-93025\over8}=1537.5

Before we go to test the means first we have to test their variability using F-test.

We test,

H0:σ12=σ22vsH1:σ12σ22H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2

The test statistic is,

Fc=s22s12=1537.5823.2955=1.8675F_c={s_2^2\over s_1^2}={1537.5\over823.2955}=1.8675

The table value is,

Fα,n21,n11=F0.05,8,11=2.94799F_{\alpha,n_2-1,n_1-1}=F_{0.05,8,11}= 2.94799 and we reject the null hypothesis if Fc>Fα,n21,n11F_c\gt F_{\alpha,n_2-1,n_1-1}

Since Fc=1.8675<F0.05,8,11=2.94799F_c=1.8675\lt F_{0.05,8,11}=2.94799, we accept the null hypothesis that the population variances for sales staff and audit staff are equal.


Now,

The hypothesis tested are,

H0:μ1=μ2vsH1:μ1>μ2H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\gt\mu_2 

The test statistic is,

tc=(xˉ1xˉ2)sp2(1n1+1n2)t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}

where sp2sp^2 is the pooled sample variance given as,

sp2=(n11)s12+(n21)s22n1+n22=(8×1537.5)+(11×823.2955)19=21356.250519=1124.01318sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(8\times1537.5)+(11\times823.2955)\over19}={21356.2505\over19}=1124.01318

Therefore,

tc=(146.25101.66667)1124.01318(112+19)=44.58333314.7837=3.0157t_c={(146.25-101.66667)\over \sqrt{1124.01318({1\over 12}+{1\over 9})}}={44.583333\over14.7837}=3.0157

tct_c is compared with the table value at α=0.05\alpha=0.05 with n1+n22=12+92=19n_1+n_2-2=12+9-2=19 degrees of freedom.

The table value is,

t0.05,19=t0.05,19=1.729133t_{{0.05},19}=t_{0.05,19}= 1.729133

The null hypothesis is rejected since tc=3.0157>t0.05,19=1.729133t_c=3.0157\gt t_{0.05,19}= 1.729133 Therefore, we conclude that there is sufficient evidence to show that the mean daily expenses of sales staff are greater than the audit staff.


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