00:01
Here we are given sst equal to 6724 .125, n equal to 10 and sse is equal to 507 .75, then sb1 equal to 0 .0813 and sb2 equal to 0 .0567.
00:24
So the regression line is y hat equal to 29 .127 plus 0 .5906x1 plus 0 .4980x2.
00:37
So from this we can have b1 is 0 .5906 and b2 is 0 .4980 that is the coefficient.
00:46
So now we will calculate ssr which is sst minus sse which is 6724 .125 minus 507 .25 which is 6216 .375.
01:04
Now r square equal to ssr divided by sst which is equal to 6216 .375 divided by 6724 .125 which is 0 .924.
01:20
Now in the second part we have adjusted r square is equal to 1 minus sse divided by n minus 3 divided by sst divided by n minus 1 which is 1 minus 507 .75 divided by 7 divided by 6724 .125 divided by 9 which is 1 minus 0 .097 which is 0 .903.
01:46
Next we have to find msr which is equal to ssr divided by k minus 1 which is 6216 .375 divided by 2 which is 3108 .18.
02:04
Next we have to find mse which is equal to sse divided by n minus k which is 507 .75 divided by 10 minus 3 which is 507 .75 divided by 7 which is 72 .536.
02:23
From this we can have f equal to msr divided by mse which is equal to 3108 .1875 divided by 72 .536 which is 42 .85 and the f critical value at 0 .05 level of significance comma 27 degrees of freedom is 4 .74.
02:48
So here the calculated f is greater than the f critical value...