00:01
Now, for the first part of this question, we need to recall that our y is equal to 359 .2 -668 minus 5 .2 .272x.
00:15
Now, if we were to substitute the value of 30 for x, we would get a y value, which is equal to 2 .0 .90, which is approximately equal to 20 .1, which is approximately equal to 201.
00:30
Now moving on to the next part.
00:34
On the next part, we are looking at y bar plus or minus t at a certain significance level over 2 multiplied by the corresponding s at y bar.
00:48
This would be our confidence interval where we look at the nominal and we say plus or minus the margin of error which is represented by this right hand side.
00:58
So if we are to make the substitutions, we would be looking at 200 .95, which we have just determined here.
01:09
Into this, we add and subtract the margin of error, which would be 2 .62 multiplied by s at y bar, which would be 1 .4 .9.
01:20
This gives us an interval from 167 .95 right up to 2 .3 .3 .3 .000...