Question

2. Independent random samples of managers' yearly salaries (in $100) taken from governmental and private organizations are provided in the data sheet. At 95% confidence, test to determine if there is a significant difference between the average salaries of the managers in the two sectors. The standard deviations of both populations is unknown. a. State the null and alternative hypotheses. b. Reject or Fail to Reject the null hypothesis based on the level of significance and the p-value.

Answer 1

Here we have check whether the average salaries of two sector (goverment and private) are different or not.

here population standard deviation unknown so we use t -test

= sample mean salary of goverment sector.

X1

$\bar{X}_{2}$= sample mean salary of private sector.

$\mu_{1}$=population mean salary of goverment sector

=population mean salary of private sector

112

= number of observation in goverment sector

11

$n_{2}$=number of observation in private sector

hypothesis is to be tested

$H_{0}:$  $\mu_{1}=\mu_{2}$ (average salaries of two sector are same)

against

$H_{1}:$  (average salaries of two sector are different)

Zylin

Test statistic:

$t=\frac{\bar{X}_{1}-\bar{X}_{2}}{S*\sqrt(\frac{1}{n1}+\frac{1}{n2})}$

it follows t distribution with $n_{1}+n_{2}-2$ degrees of freedom.

	Government(X)	Private(Y)	(X-Xbar)^2	(Y-Ybar)^2
	540	474	208.184498	225
	432	380	8755.6069	11881
	528	463	5.89809796	676
	574	612	2345.3293	15129
	448	420	6017.3221	4761
	502	526	555.610898	1369
	480	430	2076.7525	3481
	499	459	706.039298	900
	610	615	7128.1885	15876
	572	541	2155.6149	2704
	390	335	18379.6045	23716
	593	613	4546.6161	15376
	651		15732.3337
	539		180.327298
Total	7358	5868	68793.4286	96094

$\bar{X}_{1}= \frac{\sum X}{n_{1}}=\frac{7358}{14}=525.5714$

$\bar{Y}_{1}= \frac{\sum Y}{n_{2}}=\frac{5868}{12}=489$

$S_{1}=\sqrt{\frac{\sum (X-\bar{X})^{2}}{n_{1}-1}}=\sqrt\frac{68793.4286}{13}=\sqrt{5291.8022}$=72.7448$S_{2}=\sqrt{\frac{\sum (Y-\bar{Y})^{2}}{n_{2}-1}}=\sqrt\frac{96094}{11}=\sqrt{8735.818}$= 93.4656

$S^{2}=\frac{n_{1}*S_{1}^{2}+n_{2}*S_{2}^{2}}{n_{1}+n_{2}-2}$

$S^{2}=\frac{14*5291.8022 +12*8735.8182 }{14+12-2}$

$=\frac{74085.2308+104829.818}{24}$

$=\frac{178912.049}{24}$

$=7454.6687$

$S=\sqrt{7454.6687}$$=86.3404$

therefore test statistic is,

$t=\frac{525.5714-489}{86.3404*\sqrt{\frac{1}{14}+\frac{1}{12}}}$

$t=\frac{36.5714}{86.3404*\sqrt {0.1548}}$

$t=\frac{36.5714}{33.9663}$

$t=1.0767$............. test statistic value

critical value:

$t_{n_{1}+n_{2}-2,\alpha/2}$ =$t_{24,0.025}=2.064$...............from table

decision criteria:

If t >$t_{n_{1}+n_{2}-2,\alpha/2}$ then reject Ho at $\alpha$ % level of significance.

here t <  $t_{n_{1}+n_{2}-2,\alpha/2}$=2.064 accept Ho at 5% level of significance.

decision using p-value:

p-value=p[t >1.0767]=0.2923  

if p value < $\alpha$ then reject Ho

0.2923>0.05 therefore accpet Ho

Conclusion:

Average salaries of two sectors goverment and private are same.