252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Name: Class days and time: Student number: Please include this on what you hand in! Graded Assignment 1 Please show your work! Write on one side of your paper. Neatness and whether the papers are stapled may affect your grade. 1. (Duane and Seward) A random sample of rents (to the nearest dollar amount) paid by students living off campus gives the data below. We assume that rents are approximately Normally distributed. 910 820 780 870 860 800 820 810 900 730 920 650 Personalize the data as follows: add the third-to-last digit of your student number to the third-to-last rent above; add the second to last digit to the second-to-last number above, change the last digit the same way. Make sure that your student number is clearly visible on your paper. Example: Ima Badrisk has student number 123456; so the sample of rents that she uses is below. 910 820 780 870 860 800 820 810 900 734 925 656 Compute the sample standard deviation using the computational formula. (If you don’t know what that means, find out!). Use this sample standard deviation to compute a 90% confidence interval for the mean rent. The Off-Campus Housing Office says that the average amount spent on rents is $870/month. Using the concept of significant difference, does it appear that the office is stating the wrong rent? Why? (This was not an opinion question! If you don’t know what a significant difference is, find out!) Is the mean significantly different from $880? 2. The university wants to publish a document showing its economic impact on the community. If university students rent 965 units, do a confidence interval for the total amount of rent being paid. (See problem 8.50 in the text.) 3. Show how your results in 1) would change if the 12 rents were a random sample chosen from a population of only 100 rentals. 4. Assume that the population standard deviation is 77 (and that the sample of 12 is taken from a very large population). Find z .04 (the 96th percentile of z ) using the Normal table and use it and the mean that you found in 1) to compute a 92% confidence interval for the average rent. Does the mean differ significantly from $870 ounces now? From $860? Why? Extra Credit: (You have to do extra credit now – not after you receive an unacceptable final grade.) 5. a. Use the data above to compute a 90% confidence interval for the population standard deviation. b. Assume that you got the sample standard deviation that you got above from a sample of 50, repeat a. c. Fool around with the method for getting a confidence interval for a median and try to come close to a 90% confidence interval for the median. 6. Use the computer labs to access Minitab. Click on the command part of the display (the blank upper sheet). Use editor and ‘enable commands’ to start. Check some numbers in the Normal, t, Chi-Squared or F tables using the set of Minitab routines that I have prepared. To use the new set of routines, follow the instructions inAreadoc1. There are several things that you can do. For the Normal distribution use the computer to check the answers to Examples 6.1-6.4 on pp 198-200 in the text. For the t-table pick a number of degrees of freedom and show that for that number of degrees of freedom, the probability above, say, t .20 on the t-table is 20%. You can do the same for the F and chi-squared tables in your book of tables. A good answer will explain what you did and contain the command dialog and graphs. 1. (Duane and Seward) A random sample of rents (to the nearest dollar amount) paid by students living off campus gives the data below. We assume that rents are approximately Normally distributed. 910 820 780 870 860 800 820 810 900 730 920 650 Personalize the data as follows: add the third-to-last digit of your student number to the third-tolast rent above; add the second to last digit to the second-to-last number above, change the last digit the same way. Make sure that your student number is clearly visible on your paper. Example: Ima Badrisk has student number 123456; so the sample of rents that she uses is below. 910 820 780 870 860 800 820 810 900 734 925 656 Compute the sample standard deviation using the computational formula. (If you don’t know what that means, find out!). Use this sample standard deviation to compute a 90% confidence interval for the 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) mean rent. The Off-Campus Housing Office says that the average amount spent on rents is $870/month. Using the concept of significant difference, does it appear that the office is stating the wrong rent? Why? (This was not an opinion question! If you don’t know what a significant difference is, find out!) Is the mean significantly different from $860? Solution: Three of the many possible solutions are shown. Row 1 2 3 4 5 6 7 8 9 10 11 12 x12 x1 x 22 x2 910 828100 820 672400 780 608400 870 756900 860 739600 800 640000 820 672400 810 656100 900 810000 730 532900 920 846400 650 422500 9870 8185700 910 828100 820 672400 780 608400 870 756900 860 739600 800 640000 820 672400 810 656100 900 810000 734 538756 925 855625 656 430336 9885 8208617 x 32 x3 910 828100 820 672400 780 608400 870 756900 860 739600 800 640000 820 672400 810 656100 900 810000 739 546121 929 863041 659 434281 9897 8227343 Computation of Sample Variances using the computational formula. n1 n 2 n3 12 , and x x x 2 1 8185700 , x 2 9885 , x 2 2 8208617 , x 3 9897 8227343 . 2 3 The means are x1 x3 9870 , 1 x 2 1 nx12 s12 x s 22 x 2 2 x 2 3 1 n 9870 822 .50 , x 2 12 x n 2 9885 823 .75 and 12 9897 824 .75 12 3 n x n 1 nx 22 n 1 67625 .00 8185700 12 822 .50 2 6147 .73 s1 6147 .73 78 .407 11 11 65848 .25 8208617 12 823 .75 2 5986 .20 s 2 5986 .20 77 .371 11 11 64792 .25 8227343 12 824 .75 2 5890 .20 s 3 5890 .20 76 .748 11 n 1 11 If you used another formula, you should lose some credit but get the same answer as I did. s 32 nx32 Computation of Standard Errors. We will use the formula x tn1s x , where s x s 2 s x1 s1 s x3 s3 s s12 6147 .73 512 .31 22 .634 s x2 2 n 12 n s 32 5890 .20 490 .85 22 .155 n 12 n n . s 22 5986 .20 498 .85 22 .335 n 12 n Finding t The significance level is given as 90%. n1 n 2 n3 12 . 1 .90 .10 2 .05 t n1 t 11 1.796 {ttable}. Please repeat this rhyme after me “Sigma means 2 .05 z and s means use t (unless you have a very high freedom degree).” 2 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Putting it all together Remember x1 822 .50 , x 2 823 .75 and x 3 824 .75 . The formula for a 2-sided interval is x t n1 s 822.50 1.79622.634 822.50 40.65 or 781.85 to 863.15. 1 1 2 x1 2 x 2 t n1 s x2 823.75 1.79622.335 823.75 40.11 or 783.64 to 863.86. 2 3 x3 tn1 s x3 824.75 1.79622.155 824.75 39.79 or 784.96 to 864.54. 2 The Off-Campus Housing Office says that the average amount spent on rents is $870/month. Using the concept of significant difference it appears that the office is stating the wrong rent? In all of these cases it appears that the mean is significantly different from 870 since 870 is not on any of the confidence intervals. The same is not true of 860, since it falls within all of these intervals. Minitab output follows. The instruction is ‘onet’ followed by the appropriate column name. The instruction is modified by the subcommand ‘conf 90.’ This indicates a 90% confidence level. MTB > onet c1; SUBC> conf 90. One-Sample T: Rents1 Variable N Mean Rents1 12 822.5 StDev 78.4 SE Mean 22.6 90% CI (781.9, 863.1) MTB > onet c2; SUBC> conf 90. One-Sample T: Rents2 Variable N Mean Rents2 12 823.8 StDev 77.4 SE Mean 22.3 90% CI (783.6, 863.9) MTB > onet c3; SUBC> conf 90. One-Sample T: Rents3 Variable N Mean Rents3 12 824.8 StDev 76.7 SE Mean 22.2 90% CI (785.0, 864.5) 2. The university wants to publish a document showing its economic impact on the community. If university students rent 965 units, do a confidence interval for the total amount of rent being paid. (See problem 8.50 in the text.) Solution 1: 781.85 to 863.15 when multiplied by 965 gives us $754,485 to $832,939. Solution 2: 783.64 to 863.86 when multiplied by 965 gives us $756,212 to $833,625. Solution 3: 784.96 to 864.54 when multiplied by 965 gives us $757,486 to $834,281. 3. Show how your results in 1) would change if the 12 rents were a random sample chosen from a population of only 100 rentals. Solution: Recall that n n n 12 t n1 t 11 1.796 , x 822 .50 , x 823 .75 , x 824 .75 , 1 2 3 2 .05 1 2 3 6147 .73 , 5986 .20 and 5890 .20 . Solution 1: If N 100 , the sample of 12 is more than 5% of the population, so use a finite population s12 s 22 s32 s2 N n 6147 .73 100 12 N n 1 455 .387 21 .34 .. The . 12 100 1 n N 1 n N 1 interval becomes 1 x1 t n1 s x1 822.50 1.79621.34 822.50 38.33 or 784.17 to 860.83. correction. Now s x1 s1 2 Solution 2: If N 100 , the sample of 12 is more than 5% of the population, so use a finite population s2 N n 5986 .20 100 12 N n 1 443 .422 21 .06 .. The . 12 100 1 n N 1 n N 1 interval becomes 2 x 2 t n1 s x2 823.75 1.79621.06 823.75 37.82 or 785.93 to 861.57. correction. Now s x1 s1 2 3 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Solution 3: If N 100 , the sample of 12 is more than 5% of the population, so use a finite population s2 N n 5890 .20 100 12 N n 1 436 .311 20 .89 .. The . 12 100 1 N 1 n N 1 n interval becomes 3 x3 t n1 s x3 824.75 1.79620.89 824.75 37.51 or 787.24 to 862.26. correction. Now s x1 s1 2 It would have been easier for me to multiply the original standard errors ( s x1 22.634 , s x2 22.335 and 100 12 .8888 0.9428 , but the results would be the same. It is still true that 100 1 the results are significantly different from 870, but not 860. s x3 22.155 ) by 4. Assume that the population standard deviation is 77 (and that the sample of 12 is taken from a very large population). Find z .04 (the 96th percentile of z ) using the Normal table and use it and the mean that you found in 1) to compute a 92% confidence interval for the average rent. Does the mean differ significantly from $870 ounces now? From $880? Why? Solution: Recall that n1 n 2 n3 12 , x1 822 .50 , x 2 823 .75 and x 3 824 .75 . Our new data is 77 . We will use the formula x z x , where x 2 n 77 12 77 2 494 .0833 12 = 22.228. We must now find z .04 . Make a diagram! The diagram for z will be a Normal curve centered at zero and will show one point, z .04 , which has 4% above it (and 96% below it!) and is above zero because zero has 50% below it. Since zero has 50% above it, the diagram will show 46% between zero and z .04 . From the diagram, we want one point z .04 so that Pz z.04 .0400 or P0 z z.04 .4600 . On the interior of the Normal table{norm} we cannot find .4600 exactly. If you look at the 1.7 row you will find 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 z 1.7 0.4554 0.4564 0.4573 0.4582 0.4591 0.4599 0.4608 0.4616 0.4625 0.4633 0.0 0.0000 0.0040 0.0080 0.0120 0.0160 0.0199 0.0239 0.0279 0.0319 0.0359 This means that the best we could do are P0 z 1.75 .4599 or P0 z 1.76 .4608 . 1.75 is definitely better than 1.76 and something like 1.751 might be even better. Any of these are acceptable. I will use z.04 1.75 . Note that the first line of the table reads as below. 0.00 0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08 0.09 z This means that P0 z 0.04 .0160 . It does not give us any information about z .04 . 1 x1 z 2 x 822 .50 1.75 22 .228 822 .50 38.90 or 783.60 to 861.40. 2 x 2 z 2 x 823 .75 1.75 22.228 823 .75 38.90 or 784.85 to 862.65. 3 x3 z 2 x 824 .75 1.75 22.227 824 .75 38 .90 or 785.85 to 863.65. It is still true that the results are significantly different from 870, but not 860. We really ought to check our value of z .04 . We could try Pz 1.75 Pz 0 P0 z 1.75 .5 .4599 .0401 .04 , or we could show that 1.75 is the 96th percentile. Pz 1.75 Pz 0 P0 z 1.75 .5 .4599 .9599 .96 . 4 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Extra Credit Problem (You have to do extra credit now – not after you receive an unacceptable final grade.) 5. a. Use the data above to compute a 90% confidence interval for the population standard deviation. b. Assume that you got the sample standard deviation that you got above from a sample of 50, repeat a. c. Fool around with the method for getting a confidence interval for a median and try to come close to a 90% confidence interval for the median. Solution: a) n1 n 2 n3 12 , s12 6147 .73 , s 22 5986 .20 and s32 5890 .20 . The problem says that .10 and 2 .05 . From the outline (the supplement pg 1 or Table 3), n 1 21 2 n 1s 2 2 2 n 1s 2 2 .29511 2 1 n1 . We use 2 .20511 19.6751 and 2 2 4.5748 {chiSq}. 11 6147 .73 2 11 6147 .73 or 3442 .34 2 14782 .07 . If we 19 .6451 4.5748 take square roots, we get 58.67 121 .58 . The other solutions are very similar. Solution 1: The formula becomes b. We will repeat a) with n 50 . Now DF n 1 49 . From the supplement pg 2 (or Table 3), the formula for large samples is s 2 DF z 2 DF s 2 DF z 2 DF 2 . Since the 2 table has no values for 49 degrees 2 of freedom, we must use the large sample formula. We use z .05 1.645 {ttable} and 2 DF 2(49 ) 98 9.8995 . Solution 1: s x1 22.634 , s x2 22.335 and s x3 22.155 . The formula becomes 22 .634 9.8995 1.645 9.8995 22 .634 9.8995 1.645 9.8995 or 224 .065 224 .065 or 19.4087 17.1446 . 11 .5452 8.2545 c. We are now trying to find an approximate 99% confidence interval for the median. This was one many people blew because they didn’t read the question! Some people tried to use the formula for the confidence interval for the mean on every other parameter. Though that can work in some unusual cases, it is a bad idea. We must put them in order as below. Rank 1 2 3 4 5 6 7 8 9 10 11 12 x1 x2 650 730 780 800 810 820 820 860 870 900 910 920 656 734 780 800 810 820 820 860 870 900 910 925 x3 659 739 780 800 810 820 820 860 870 900 910 929 5 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) We could, instead, just list their ranks. Row r1 x1 x 2 r2 1 2 3 4 5 6 7 8 9 10 11 12 910 820 780 870 860 800 820 810 900 730 920 650 11.0 6.5 3.0 9.0 8.0 4.0 6.5 5.0 10.0 2.0 12.0 1.0 910 820 780 870 860 800 820 810 900 734 925 656 11.0 6.5 3.0 9.0 8.0 4.0 6.5 5.0 10.0 2.0 12.0 1.0 x3 r3 910 820 780 870 860 800 820 810 900 739 929 659 11.0 6.5 3.0 9.0 8.0 4.0 6.5 5.0 10.0 2.0 12.0 1.0 It says on the outline that, if we use the k th numbers from the end, 2Px k 1 . We want to be 10% or lower which means Px k 1 .05 . There are two ways to do this. If we take the easy way out and use a Normal approximation, z .05 1.645 so that n 1 z .2 n 12 1 1.645 12 13 5.69845 3.6507 . This seems to be telling us to use the 2 2 2 numbers that are 3rd from each end or x1,3 780 and x1,10 900. Actually, these are the same for all three k columns shown here. (To be conservative, round the value of k down.) This looks sloppy for such a low sample size, so the next solution is preferred. To be more precise, use the Binomial table with n 12 . Possible intervals are x i ,1 to x i ,12 , x i , 2 to x i ,11 etc. (Here i is the column number and the second subscript is the rank of the number in the column.) The part of the Binomial table for n 12 {bin} is shown below. We need only look at the last p .5 column. n x 12 0 1 2 3 4 5 6 7 8 9 10 11 12 .01 0.88638 0.99383 0.99979 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 .05 0.54036 0.88164 0.98043 0.99776 0.99982 0.99999 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 .10 0.28243 0.65900 0.88913 0.97436 0.99567 0.99946 0.99995 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 .15 0.14224 0.44346 0.73582 0.90779 0.97608 0.99536 0.99933 0.99993 0.99999 1.00000 1.00000 1.00000 1.00000 .20 0.06872 0.27488 0.55835 0.79457 0.92744 0.98059 0.99610 0.99942 0.99994 1.00000 1.00000 1.00000 1.00000 .25 0.03168 0.15838 0.39068 0.64878 0.84236 0.94560 0.98575 0.99722 0.99961 0.99996 1.00000 1.00000 1.00000 Let us list the possible intervals for k Interval 2Px k 1 x i ,1 to x i ,12 1 2Px 0 2.00024 .00048 x i , 2 to x i ,11 2 x i ,3 to x i ,10 3 x i , 4 to x i ,9 4 2Px 1 2.00317 .00634 2Px 2 2.01929 .03858 2Px 3 2.07300 .14600 .30 0.01384 0.08503 0.25282 0.49252 0.72366 0.88215 0.96140 0.99051 0.99831 0.99979 0.99998 1.00000 1.00000 .35 0.00569 0.04244 0.15129 0.34665 0.58335 0.78726 0.91537 0.97449 0.99439 0.99915 0.99992 1.00000 1.00000 .40 0.00218 0.01959 0.08344 0.22534 0.43818 0.66521 0.84179 0.94269 0.98473 0.99719 0.99968 0.99998 1.00000 .45 0.00077 0.00829 0.04214 0.13447 0.30443 0.52693 0.73931 0.88826 0.96443 0.99212 0.99892 0.99993 1.00000 .50 0.00024 0.00317 0.01929 0.07300 0.19385 0.38721 0.61279 0.80615 0.92700 0.98071 0.99683 0.99976 1.00000 Significance 1 .9995 .9937 .9614 .8540 If we are being conservative we will take the smallest interval with a significance level above 99%. This will be as before x1,3 780 and x1,10 900 and, again, will be the same for the other two columns. 6 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Extra Credit Minitab Problem 6. Use the computer labs to access Minitab. Click on the command part of the display (the blank upper sheet). Use editor and ‘enable commands’ to start. Check some numbers in the Normal, t, Chi-Squared or F tables using the set of Minitab routines that I have prepared. To use the new set of routines, follow the instructions in Areadoc1. There are several things that you can do. For the Normal distribution use the computer to check the answers to Examples 6.1-6.4 on pp 198-200 in the text. For the t-table pick a number of degrees of freedom and show that for that number of degrees of freedom, the probability above, say, t .20 on the t-table is 20%. You can do the same for the F and chi-squared tables in your book of tables. A good answer will explain what you did and contain the command dialog and graphs. 10 10 10 Results: I looked at the tables and found t.10 1.372 , z.10 1.282 , 2 .10 15.9872 , 2.90 4.8650 , 10,10 2.32 and F 10,10 1 F.10 0.431 . For the numbers with .10 as a subscript, I checked that the .90 2.32 probability above them was .10. For the numbers with .90 as a subscript, I checked that the probability below them was .10. This was run in 2005, but there is no reason to believe that the new version of Minitab will give different results. ————— 9/19/2005 5:33:43 PM ———————————————————— Welcome to Minitab, press F1 for help. MTB > WOpen "C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW". Retrieving worksheet from file: 'C:\Documents and Settings\rbove\My Documents\Minitab\notmuch.MTW' Worksheet was saved on Thu Apr 14 2005 Results for: notmuch.MTW #This is a dummy worksheet. MTB > %tarea6a # It is used to locate %tarea6a Executing from file: tarea6a.MAC Graphic display of t curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K120) Enter the degrees of freedom. DATA> 10 Do you want the area to the left of a value? (Y or N) n Do you want the area to the right of a value? (Y or N) y Enter the value for which you want the area to the right. DATA> 1.372 #Verify that 1.372 is a 10% value of t. ...working... t Curve Area 7 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Data Display mode median 0 0 MTB > %normarea6a Executing from file: normarea6a.MAC Graphic display of normal curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K116) Enter the mean and standard deviation of the normal curve. DATA> 0 DATA> 1 Do you want the area to the left of a value? (Y or N) n Do you want the area to the right of a value? (Y or N) y Enter the value for which you want the area to the right. DATA> 1.282 #Verify that 1.282 is the 10% value of z. ...working... Normal Curve Area 8 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) MTB > %chiarea6a Executing from file: chiarea6a.MAC Graphic display of chi square curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K120) Enter the degrees of freedom. DATA> 10 Do you want the area to the left of a value? (Y or N) n Do you want the area to the right of a value? (Y or N) y Enter the value for which you want the area to the right. DATA> 15.9872 #Verify that 15.9872 is a 10% value of chi-squared. ...working... ChiSquare Curve Area Data Display std_dev mode median 4.47214 8.00000 9.33333 MTB > %chiarea6a Executing from file: chiarea6a.MAC Graphic display of chi square curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K120) Enter the degrees of freedom. DATA> 10 Do you want the area to the left of a value? (Y or N) y Enter the value for which you want the area to the left. DATA> 4.8650 #Verify that 4.865 is a 90% value of chi-squared. ...working... Chi Squared Curve Area 9 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) Data Display std_dev mode median 4.47214 8.00000 9.33333 MTB > %farea6a Executing from file: farea6a.MAC Graphic display of F curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K120) Enter the degrees of freedom.DF2 must be above 4. DATA> 10 DATA> 10 Do you want the area to the left of a value? (Y or N) n Do you want the area to the right of a value? (Y or N) y Enter the value for which you want the area to the right. DATA> 2.32 #Verify that 2.32 is a 10% value of F. ...working... F Curve Area Data Display mode 0.818182 std dev 0.968246 10 252solngr1 1/24/08 (Open this document in 'Page Layout' view!) MTB > %farea6a Executing from file: farea6a.MAC Graphic display of F curve areas Finds and displays areas to the left or right of a given value or between two values. (This macro uses C100-C116 and K100-K120) Enter the degrees of freedom.DF2 must be above 4. DATA> 10 DATA> 10 Do you want the area to the left of a value? (Y or N) y Enter the value for which you want the area to the left. DATA> .431 # Verify that 0.431 is a 90% value of F. # 0.431 is 1 divided by 2.31. # It does not appear on the F table. # See material on F distribution for explanation. ...working... F Curve Area Data Display mode 0.818182 std dev 0.968246 11