00:01
We have two tests, test 1 and test 2 for improving processing rates.
00:10
Corresponding observations are 334, 266, 240, 300, 324, 324, 324, 8, 369, 412, 353 ,000, 420, and 412, 353, for tests 2, corresponding results are 354, 295, 261, 270, 313, 3609, 317, 317, 3179, 369, 2 sample t test for that the null hypothesis is h not such that mu1 equal to mu 2 and h1 such that mu 1 less than mu 2 because the climb is to improve the processing rates here x1 bar equal to summation test 1 divided by n is 335 x2 bar is 355 0 .083.
01:39
Standard deviation of the x1 is root over.
01:45
Summation xi minus x bar whole square divided by n minus 1 is 66 .3229 and standard deviation of test is 59 .3747.
02:02
Full standard deviation is sp equal to n1 minus 1 into s .p 1 square plus n2 minus 1 into s 2 divided by n1 plus n2 root over that equal to 45 .049.
02:22
So now the t test statistic under the null hypothesis is x1 bar minus x2 bar divided by full standard deviation into 1 by n1 plus 1 by n1 equal to 335 minus 35 .083 divided by 55 .049 root over 1 by 12 plus 1 .12 equal minus 20 .0833 divided 18 .3913 equal to minus 1 .092 is the t test statistic.
03:05
Now corresponding p value with degrees of freedom n1 plus n2 minus 2 can be obtained using excel formula.
03:16
T dot ast, the first parameter is t test statistic value and the second parameter is degrees of freedom which is n1 plus n2 minus 2, that is 12 plus 12 minus 2 is 22 and the third parameter is true.
03:30
So the p value is 0 .1433 which is greater than 0 .05...