Answer to Question #295514 in Statistics and Probability for Ace

Question

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?

Answer 1

Sales staff(1)

$n_1=12\\\bar x_1={\sum x\over n_1}={1755\over12}=146.25$

$s_1^2={\sum x^2-{(\sum x)^2\over n_1}\over n_1-1}={265725-256668.75\over11}=823.2955$

Audit staff(2)

$n_2=9\\\bar x_2={\sum x\over n_2}={915\over9}=101.66667$

$s_2^2={\sum x^2-{(\sum x)^2\over n_2}\over n_2-1}={105325-93025\over8}=1537.5$

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

We test,

$H_0:\sigma_1^2=\sigma_2^2\\vs\\H_1:\sigma_1^2\not=\sigma_2^2$

The test statistic is,

$F_c={s_2^2\over s_1^2}={1537.5\over823.2955}=1.8675$

The table value is,

$F_{\alpha,n_2-1,n_1-1}=F_{0.05,8,11}= 2.94799$ and we reject the null hypothesis if $F_c\gt F_{\alpha,n_2-1,n_1-1}$

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

Now,

The hypothesis tested are,

$H_0:\mu_1=\mu_2\\vs\\H_1:\mu_1\gt\mu_2$ 

The test statistic is,

$t_c={(\bar x_1-\bar x_2)\over \sqrt{sp^2({1\over n_1}+{1\over n_2})}}$

where $sp^2$ is the pooled sample variance given as,

$sp^2={(n_1-1)s_1^2+(n_2-1)s_2^2\over n_1+n_2-2}={(8\times1537.5)+(11\times823.2955)\over19}={21356.2505\over19}=1124.01318$

Therefore,

$t_c={(146.25-101.66667)\over \sqrt{1124.01318({1\over 12}+{1\over 9})}}={44.583333\over14.7837}=3.0157$

$t_c$ is compared with the table value at $\alpha=0.05$ with $n_1+n_2-2=12+9-2=19$ degrees of freedom.

The table value is,

$t_{{0.05},19}=t_{0.05,19}= 1.729133$

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