Hi, I'm back!
Can someone have a look at the 2010 Exam 1? I'm wondering about the following questions:
1. Question 3. Can someone explain how to do that please?
2. Question 8. The correct answer is B, but I was pretty darn sure it was C. Isn't the y-intercept at 167? Am I reading it wrong?
3. Question 9. I got the answer right, but just generally, when doing a question like that, am I just supposed to sort of see a similar pattern between the actual data and the residual plots? Is that how it works? Coz that's what I've been doing....but is there a better way?
Thanks!
Question 3The 68% 95% 99.7% rule tells us that in a normally distributed data set:
68% of the data is within 1 standard deviation
95% of the data is within 2 standard deviations
99.7% of the data is within 3 standard deviations.
Because the question is asking you to use this rule, we say that the data is approximately normally distributed. Now using the boxplot, we can see that the median is 180 degrees. In a normally distributed data set, the mean=median. So the mean is 180 degrees. To find the standard deviation, we can find the interval which 68% of the data values lie. Approximately 38% of the data lie between 179 and 181 so 1 degree is incorrect. Keep doing this, and once we hit 68% at about 2 standard deviations making B the right answer.
Question 8You are making a very common mistake! Look carefully at your x and y axis, and we can see that the origin is not located at (0, 0) and its located at (20, 160). This means that the y intercept will be a lot less than 167. The best way to do this is to find the gradient, and use the point (20, 167) to find the real y intercept.
Question 9You can continue to do it that way, but the easiest way to do this question is to just rule out the incorrect answers by looking at different features of the graph. For example, we can see that there needs to be two values below the x axis, when x = 23.5. This rules out A and C. Next, we can see that four numbers have to be above the line y = 0 and four have to be below, because that is how it is with the original graph. This rules out D and E also