Howdy! I'm Professor Curtis of Aspire Mountain Academy here with more statistics homework help. Today we're going to learn how to perform correlation hypothesis testing. Here's our problem statement: Use the given data set to complete Parts A through C below. Use alpha equals 5%.
OK, Part A asks us to construct a scatterplot. To do this, we're going to use StatCrunch. And the first step to do that is putting the data inside StatCrunch. I’m going to resize this window so we can see better what's gonna go on.
OK, now inside StatCrunch, I could go to Graph and then select Scatter Plot. But I know from the problem I'm given here, I'm gonna have to do a regression analysis anyway, and the scatterplot comes as part of the regression analysis. So let's just do that. It's less buttons to push.
To get there, I'm going to go to Stat –> Regression –> Simple Linear. In the options window, I'm going to select my x-variable and my y-variable. And all the other default selections are good, so we're going to hit Compute! and in the results window, notice up here at the top it says 1 of 2. That means you're looking at page 1 of 2 pages total. The scatterplot is on the second page, which we can get to by hitting this arrow key down here at the bottom of the window. And voila!
So now we just match the one that looks the same. Notice how the axes go from 4 to 14 and 4 to about 13, so we're pretty close to the axes that are here. If I wanted to, I could select the little three bar icon in the lower left corner and then change my axes as appropriate. But we're close enough here that we get the general pattern. And we can see clearly that answer option A is the one we want. Well done!
Part B asks us to find the linear correlation coefficient. That was done here in our results window. If we go back to page 1, we can see our correlation coefficient here, in this case 0.81. We're asked to round to three decimal places, so I will do that! Good job!
Now this next part of Part B asks us to use the linear correlation coefficient to determine whether there is sufficient evidence to support the claim of linear correlation between the two variables. To do this, we need to compare the R-value we found from the problem statement and compare it with the critical R-values. So I'm going to click on this icon to get my table of critical values. Notice here — if I can get rid of this — we have 11 pairs in our sample. So we have 11 pairs, and we were asked earlier to use 5% for our alpha level. So 5% alpha matched with 11 pairs of data gives me a critical R-value of 0.602. The R-value that we obtained is 0.816. That's greater than the critical R-value. Therefore we have sufficient evidence to support a claim of linear correlation. I select the appropriate answer. Well done!
Now Part C asks us to identify the feature of the data that would be missed if Part B was completed without constructing the scatterplot. OK, let's go back and look at our scatterplot. And it looks like our scatterplot went away. Oh yeah, because I deleted it. OK, no matter! I'll just make it again. I want to do this.
So here's our scatterplot. What feature would be missed? Well, we've got a really good correlation coefficient, so it's not surprising that our data fit this line reasonably well. But clearly the scatterplot is showing us something that the correlation coefficient can't show us, and that is the presence of an outlier in our data set. If we didn't have this outlier, then this line would slope downward and fit the data that we have remaining much, much better. As it is this outlier is actually pulling that line up a little bit so that we're losing some of the fit. It's not that bad because we still have our value that's greater than the critical R-value. But it would be better if we didn't have that outlier in there, and this is what the scatterplot is showing us. So I'm going to select the answer option that best reflects that reasoning. Good job!
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11/23/2020 06:55:33 pm
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Frustrated with a particular MyStatLab/MyMathLab homework problem? No worries! I'm Professor Curtis, and I'm here to help.