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.
= sample mean salary of private sector.
=population mean salary of goverment sector
=population mean salary of private sector
= number of observation in goverment sector
=number of observation in private sector
hypothesis is to be tested
(average salaries of two sector are same)
against
(average salaries of two sector are different)
Test statistic:
it follows t distribution with 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 |
=72.7448= 93.4656
therefore test statistic is,
............. test statistic value
critical value:
=...............from table
decision criteria:
If t > then reject Ho at % level of significance.
here t < =2.064 accept Ho at 5% level of significance.
decision using p-value:
p-value=p[t >1.0767]=0.2923
if p value < then reject Ho
0.2923>0.05 therefore accpet Ho
Conclusion:
Average salaries of two sectors goverment and private are same.
2. Independent random samples of managers' yearly salaries (in $100) taken from governmental and private organizations...
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