00:01
Hello everyone in this problem we have given normal bone density and here we have osteophenia here we have osteoporosis so here we have 1024 -4 851 837 768 6 47, 682, 837, 775, 640, 686.
00:49
And in the austropania, 864, 936, 600, 684, 936, 600, 684, 599, 599, 686 434 598 and in the osteoporosis 76759 948 766 344 308 940 940 7708 778 378 341 298 298 and our null hypothesis for this, mu 1 is equal to mu 2 is equal to mu 3.
01:47
And alternate hypothesis that means are not equal at alpha is equal to 0 .10.
01:59
So degree of freedom is k minus 1 and degree of freedom 2 is n minus k.
02:08
So, degree of freedom 1 is 3 minus 1 that is 2 and degree of freedom 2 is 30 minus 3, that is 27.
02:21
So our critical value is 3 .68 and decision rule follows that we have to reject the null hypothesis if our value of f is greater than equal.
02:40
Equal to 3 .68.
02:42
So here we have normal bone density.
02:49
Normal bone density.
02:53
Here we have osteopenia and here osteoporosis.
03:02
Here we have n1 is 10 and 2 is 10 and 3 is 10.
03:12
Here, me.
03:13
Mean of 1 is 774 .7.
03:18
Mean of 2 is 643 .4 and mean of 3 is 604 .9.
03:27
Now we get overall mean is 674 .3.
03:35
So ssb is 10 into 774 .7.
03:45
Plus 10 into 643 .4 plus 6704 .3 square plus 10 into 604 .9 minus 674 .3 whole square.
04:05
So we have 100801 .6 plus 9548 .1 .1 plus 48 .1 plus 481 plus 481 .6 plus 481 .6.
04:17
And by adding all these we get 158 513 .3 and our formula for s s e is summation x minus x j square so now here we have normal normal bone density and here we have x minus x1 square.
04:53
Here we have 1 .04 -851 -837.
05:03
Next, 768 -648, 647, 6 -82, 837, 775640, 6 -86.
05:21
Now, here in this column we have 6 .2 .250 .4.
05:32
Here we have 5821 .389 3881 .29 .49 .4881 .49.
05:47
46307.
05:49
829 .859.
05:51
859 .859.
05:51
859 .859.
05:51
859 .9.
05:51
8559 3 .293 .29 -9 .8559 .8 .8 59 3 .2 .2 .2 .9 .9 3 .2 .2 9 .22222 9 .8 .8 .8 .8 3881 .29 .0 .09 .1818144 .0 .79 .7867 .69...