Stat 1000: Tips for Web Assign HW 05
Published: Wed, 03/02/11
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General Tips for Web Assign and JMP
When working with Web Assign, always enter the answer to one specific box and then click "Submit Answer" to confirm that is correct before you move on to another box. Do not enter several answers all at once in several boxes before you click "Submit Answer". You risk being marked wrong due to some typo or something.
For some strange reason, JMP 8 occasionally computes wrong answers even if you have copied and pasted your data correctly. I suggest that, if it is feasible, type the given data into your calculator (in Stat mode as shown in Appendix D of my book), and have your calculator compute the sample mean. Compare that answer with JMP's answer for the sample mean. If they are the same, everything is fine. If they are not the same, close JMP 8 and restart it, recopy and paste the data, and check again. Sometimes you have to do this 2 or 3 times before JMP finally works. If it is not feasible to use your calculator to compute the sample mean, have JMP do the question 2 or 3 times, being sure to restart JMP and recopy the data each time, and confirm that JMP gives you the same answer each time before risking entering the results into Web Assign.
If you are taking the course by distance/online (Section D01) click here to see your tips for HW 05.
Study Lesson 5 in Volume 2 of my book, if you have it, to prepare for this assignment.
Question 1:
When you are asked to list sample spaces, generally you will use a two-way table to visualize all the possible outcomes. If you are asked to count the number of things (number of coins, number of successes, etc.), list the sample space from lowest possible number up. Don't forget that, when counting the possible number of things, there is always the chance that there could be 0 things. Some of you are asked about coins. Note, they are not asking you what denomination the coins are (pennies, nickels, dimes, etc.). That is irrelevant. They don't care what kind of coins you might have, just how many and how much money are they worth.
When they ask you, "Are the outcomes equally likely?", think carefully. Remember, to be equally likely means that the first outcome you have listed in your sample space, has the same probability of happening as the second outcome, etc. Don't think that because the outcomes are equally easy to list, that makes them equally likely.
For example, if I ask you to the list the sample space for the possible medals an Olympic athlete might win, I could say the sample space is {Gold, Silver, Bronze, No medal at all}. Even though there are four possible outcomes, they are not equally likely! There is not a 25% chance of winning a Gold medal, for example. It depends what athlete we are talking about, what event, etc. The mere fact I have to say "it depends" tells me the outcomes are not equally likely. Even if the actual medal someone won was randomly determined and we were assuming every person has an equal chance of winning, if there were 20 people competing for the medal, there would be only a 1/20 chance they would win gold (since only one person can win gold), and a 17/20 chance they win no medal at all. Of course, some athletes may be the favourite to win gold and they would then have a much higher chance than 1/20 of winning.
Of course, some outcomes are equally likely. For example, if you are flipping a fair coin, or rolling a fair die, you have every right to say either side of the coin or die are equally likely to come up.
Question 2
Of course, you can use the formulas I show you to work out P(A and B) for when A and B are mutually exclusive (disjoint) and when A and B are independent. You can also use the formula for P(A or B). I recommend that you draw a Venn diagram to help you solve the rest of this question. Note, if A and B are mutually exclusive (disjoint), they would be represented on a Venn diagram by two completely separate circles. Obviously, A is everything in the A circle, while the complement of A, AC, is everything outside the A circle. Similarly for B and BC.
Questions 3, 4, and 5 are quite similar to my questions in Lesson 5 where you can used sample spaces and probability distributions to solve the problems.
Study Lesson 3 in my book, if you have it, to prepare for this assignment.
Question 1:
To compute the correlation coefficient by hand, follow my example in Lesson 3, question 1, part (c). Note, you are not given the means and standard deviations for x and y already, so you are certainly allowed to use the Linear Regression Stat Mode on your calculator to tell you the means and standard deviations of both x and y. Put your calculator in Linear Regression Stat Mode (see Appendix D of my book). After you enter all the data, you can ask it for the mean and standard deviation of the x values and the mean and standard deviation of the y values. For example, Sharps use "RCL 4" to get x-bar and "RCL 7" to get y-bar. "RCL 5" gives you Sx and "RCL 8" gives you Sy.
Note, only those of you who have my blue or green study book have an example that follows exactly the headings given in this question. Even though they tell you to do everything to three decimal places, don't do that. Record every single decimal place your calculator gives you for each calculation, or else your answers won't be accurate enough. I suggest you do everything on paper first, then you can type in the results, rounding all of your numbers off to 3 decimal places at that time (even though you actually did the calculations using all the decimal places). Of course, your calculator actually tells you the value of r, so you can use that as a check.
Question 2 is just an algebra question. They give you three of x, y, a, and b and want you to figure out the missing one. Sub the givens into the appropriate places of
y = a + bx and solve what is missing.
Question 3 is a good run through of the formulas I show you in Lesson 3.
Question 4 uses JMP.
Here is how to use JMP for linear regression. First copy and paste the data into a New Data Table the usual way (see my previous homework tips if you are not sure how to paste the data). If you have to type the data in manually, simply double-click the space to the right of "Column 1" to create "Column 2". Enter the X data down column 1 and the Y data down column 2. Be sure to double-click each column to give it an appropriate name and to ensure the Data Type is Numeric and the Modeling Type is Continuous.
Select Analyze, then Fit Y By X. Highlight the column you have determined should be X, and click the X, Factor button. Highlight the column you have determined should be Y and click the Y, Response button. Click OK.
You should now see a scatterplot. Click the red triangle above the scatterplot and select Fit Line and JMP will draw in the least-squares regression line. Note, it shows you the regression equation directly below the scatterplot. JMP also shows you the value of r-squared (the coefficient of determination), rather than r, the correlation coefficient. Remember, the coefficient of determination is the percentage of y's variation explained by the regression equation. You can always square root this number to get r, the correlation coefficient, but use your scatterplot to help you decide if r is negative or positive because your calculator can't tell you that.
If you want to get rid of anything, click the red triangle and deselect anything you don't want to see. Note, if you click the blue triangle next to something, that will make part of the output disappear as well, if you wish. Just click the blue triangle again to make it reappear.