00:01
For this problem, we are asked to consider the following pairs of observations, which i've also copied into my excel table here.
00:07
Part a, we are asked to construct a scatter plot of the data.
00:10
So we go to insert, charts, scatter.
00:13
So we have our scatter plot.
00:16
In part b, we are asked to use the method of leased squares to fit a straight line to the seven data points in the table.
00:24
Now, one way we can do this is actually using the lin -sht command in excel.
00:28
So we select the known y's, then the known x's.
00:32
We put true for the constant argument so that we calculate a b value.
00:37
Then false, we don't need any additional regression statistics.
00:41
So we can see that, now i'll note, the output values are m first and then b.
00:48
So i'll move this over just to get a little bit of space.
00:51
And i'll put in, i'll write it explicitly as y hat, the predicted y values for each.
00:56
So we'll have that y hat is going to be equal to b plus m times the x value.
01:06
So we can plot out our y hat values.
01:11
Actually, i need to make an adjustment here.
01:25
There we go.
01:27
And now what i'll do is add into the chart and another series.
01:35
Let's say series name is y hat.
01:38
Series x values are the same.
01:43
And the series y values are shown there.
01:54
And go to trendline linear for y hat.
01:58
Of course, y hat is explicitly linear, so that would be displaying exactly what we'd expect.
02:03
So that gives us the equation of our straight line.
02:07
It would be, and actually i'll, you can display that directly on screen here.
02:12
Display equation on chart.
02:14
0 .619 plus 0 .5444.
02:18
In part d, we are asked to specify the null and alternative hypotheses.
02:25
You would use to test whether the data provided or provide sufficient evidence to indicate that x contributes information for the linear prediction of y.
02:34
So the null hypothesis here is that beta 1 equals 0.
02:40
The alternative hypothesis is that beta 1 does not equal 0, which i'll write as exclamation mark, equals 0.
02:51
For part e, we are asked, what is the test statistic that should be used in conducting the hypothesis test of part b, and to specify the number of degrees of freedom associated with the test statistic.
03:03
In this case, the test statistic will be t, and the degrees of freedom will be, let's see here, what's the count? 7.
03:14
So the degrees of freedom equals the count minus 2, so we have 5 degrees of freedom.
03:20
In part f, we are asked to conduct the hypothesis test of part d using alpha equals 0 .05.
03:27
So at that 0 .05 confidence level and 5 degrees of freedom, we'll have that the critical t value will be equal to, i'm just cross -referencing a table off screen, t -c equals 2 .571.
03:43
Now we need to actually calculate our t value...