252solngr1-071 2/13/07 (Open in ‘Print Layout’ format. Graded Assignment 1 Please show your work! Neatness and whether the papers are stapled may affect your grade. 1. (Keller & Warrack) A random sample of 11 young adult men was asked how many minutes of sports they watched daily. Results are below. 50 48 65 74 66 30 40 60 60 60 50 Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 66}. 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 99% confidence interval for the mean. Does the mean from your interval differ significantly from 50 minutes? Why? What about 40 minutes? 2. If there are 0.9 million young adult men in the metropolitan area , do a confidence interval for the total amount of time they watch in a day. (See problem 8.50.) It might help to convert your result to hours. 3. Show how would these results change if the 11 individuals were a random sample taken from a dormitory with 100 male residents? 4. Assume that the population standard deviation is 10 (and that the sample of 11 is taken from a very large population). Find z .0075 using the Normal table (If you have several values of z that you can use, pick the average of the extreme ones.) and use it to compute a 98.5% confidence interval. Does the mean differ significantly from 50 now? From 40? Why? Extra Credit Problems 5. a. Use the data above to compute a 98% confidence interval for the population standard deviation. b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a. c. Fool around with the method for getting a confidence interval for a median and try to come close to a 99% confidence interval for the median. 6. 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 is 20%. You can do the same for the F and chisquared tables in your book of tables. A good answer will explain what you did and contain the command dialog and graphs. 252solngr1-071 2/8/07 1. (Keller & Warrack) A random sample of 11 young adult men was asked how many minutes of sports they watched daily. Results are below. 50 48 65 74 66 30 40 60 60 60 50 Personalize the data as follows: add the digits of your student number to the last six numbers. Example: Ima Badrisk has the student number 123456; so the last six numbers become {31, 42, 63, 64, 65, 56}. 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 99% confidence interval for the mean. Does the mean from your interval differ significantly from 50 minutes? Why? What about 40 minutes? Solution: Two data sets for computations of the variance are shown here. They will be referred to as solution 1 and solution 2. The first represents a student number of 000000 and the second 999999. x12 x1 index 1 2 3 4 5 6 7 8 9 10 11 50 48 65 74 66 30 40 60 60 60 50 603 x 22 x2 2500 2304 4225 5476 4356 900 1600 3600 3600 3600 2500 34661 50 48 65 74 66 39 49 69 69 69 59 657 2500 2304 4225 5476 4356 1521 2401 4761 4761 4761 3481 40547 From Table 3 x tn 1 s x is the formula for a two sided confidence interval when the population 2 standard deviation is unknown. x n1 n 2 15 , 1 The means are x1 s12 s x1 s 22 s x2 x12 s1 nx12 n 1 2 2 nx 22 n 1 s2 x 1 n x 2 1 34661 , x 2 657 and 603 54 .8182 and x 2 11 34661 1154 .8182 10 2 x n 2 x 2 2 40547 . 657 59 .7273 . 11 1605 .61 160 .56 s1 160 .56 12 .671 10 s12 160 .56 14 .5965 3.821 n 11 n x 603 , 40547 1159 .7273 2 1306 .15 130 .61 s 2 130 .61 11 .429 10 10 s 22 130 .61 11 .874 3.446 n 11 n 1 .99 .01 2 .005 10 t n1 t.005 3.169 2 1 x1 t n1 s x1 54.8182 3.1693.821 54.818 12.107 or 42.711 to 66.925. 2 2 x 2 t n1 s x2 59.7173 3.1693.446 59.717 10.920 or 48.797 to 70.637. 2 Since 50 minutes is included in these intervals and 40 is not, our results are significantly different from 50 but not 40. 2 252solngr1-071 2/8/07 2. If there are 0.9 million young adult men in the metropolitan area , do a confidence interval for the total amount of time they watch in a day. (See problem 8.50.) It might help to convert your result to hours. Solution 1: 42.711 to 66.925 when multiplied by 0.9 million gives us 38.440 to 60.233 million minutes or 0.641 to 1.004 million hours. Solution 2: 48.797 to 70.637 when multiplied by 0.9 million gives us 43.917 to 63.573 million minutes or 0.732 to 1.060 million hours. 3. Show how would these results change if the 11 individuals were a random sample taken from a dormitory with 100 male residents? If you interpreted this as a re-do of Problem 2, multiply solution below by 0.9 million. Solution 1: If N 100 , the sample of 11 is more than 5% of the population, so use a finite population correction. Recall that x1 54 .8182 and s x1 s1 n 3.821 . Now s x1 s1 n N n 100 11 3.821 N 1 100 1 3.821 0.8990 3.8210.94815 3.623 . 10 Recall that .01 , 2 .005 , t n1 t .005 3.169 and that x tn 1 s x is the formula for a two 2 2 sided interval. 1 x1 t n1 s x1 54.8182 3.1693.623 54.818 11.481 or 43.337 to 66.299. The 2 interval is smaller, but it doesn’t change anything – the mean is still significantly different from 40 (but not 50). Solution 2: If N 100 , the sample of 11 is more than 5% of the population, so use a finite population correction. Recall that x 2 59 .7273 and s x2 s2 3.446 . Now n N n 100 11 3.446 . 3.446 0.8990 3.446 0.94815 3.267 . N 1 100 1 n 10 Recall that .01 , 2 .005 , t n1 t .005 3.169 and that x tn 1 s x is the formula for a two s x2 s2 2 2 sided interval. 2 x 2 t n1 s x2 59.7273 3.1693.267 59.7273 10.348 or 43.3390 to 70.075. The 2 interval is smaller, but it doesn’t change anything – the mean is still significantly different from 40 (but not 50). 4. Assume that the population standard deviation is 10 (and that the sample of 11 is taken from a very large population). Find z .0075 using the Normal table (If you have several values of z that you can use, pick the average of the extreme ones.) and use it to compute a 98.5% confidence interval. Does the mean differ significantly from 50 now? From 40? Why? Solution: Make a diagram! The diagram for z will be a Normal curve centered at zero and will show one point, z .0075 , which has 0.75% above it (and 99.25% below it!) and is above zero because zero has 50% below it. Since zero has 50% above it, the diagram will show 49.25% between zero and z .0005 . From the diagram, we want one point z .0025 so that Pz z .0075 .0075 or P0 z z .0025 .4925 . On the interior of the Normal table we can find to .4925 exactly. In fact, it says P0 z z 0 .4925 for 2.43. This means that we will say z .0075 2.43 . If the confidence level is 1 .9850 .0150 and x 10 2 .0075 . 10 , so 10 2 9.0909 3.0151 . 11 n 11 Solution 1: 1 x1 z x1 54.8182 2.433.0151 54.82 7.33 or 47.50 to 62.15 Solution 2: 2 x 2 z x2 59.7273 2.433.0151 59.73 7.33 or 52.40 to 67.06 3 252solngr1-071 2/8/07 First Extra Credit Problem 5. a. Use the data above to compute a 98% confidence interval for the population standard deviation. b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a. c. Fool around with the method for getting a confidence interval for a median and try to come close to a 99% confidence interval for the median. Solution: a. The problem says that .02 and n 1s 2 2 2 2 n 1s 2 1 2 2 .01 . From the supplement pg 1 (or Table 3), n1 n 1 10 10 . We use 2 .01 23.2093 and 21 .99 2.5582 . 2 2 2 Solution 1: s12 160 .56 s1 160 .56 12 .671 The formula becomes 10 160 .56 2 10 160 .56 23 .2093 2.5582 or 69 .1792 2 627 .6288 . If we take square roots, we get 8.3174 25.0525 . b. We will repeat a) with n 45 . Now DF n 1 44 . From the supplement pg 2 (or Table 3), the formula for large samples is s 2 DF z 2 DF 2 s 2 DF z 2 DF . Since the 2 table has no values for 44 2 degrees of freedom, we must use the large sample formula. We use z.01 2.327 and 2 DF 2(44 ) 88 9.3808 . Solution 1: s1 160 .56 12 .671 . The formula becomes 12 .671 9.3808 2.327 9.3808 12 .671 9.3808 or 2.327 9.3808 118 .8641 118 .8641 or 10.15 16.8511 . 11 .7078 7.0538 c. We fool around with the method for getting a confidence interval for a median and try to come close to a 99% confidence interval for the median. The numbers in order are x3 x5 x6 x7 x8 x9 x10 Solution 1: x1 x11 x2 x4 30 40 48 50 50 60 60 60 65 66 74 It says on the outline that, if we use the k th numbers from the end, 2Px k 1 . We want to be 1% or lower which means Px k 1 .005 . There are two ways to do this. If we take the easy way out and n 1 z .2 n 11 1 2.576 11 12 8.5436 1.7282 . This seems to 2 2 2 be telling us to use the numbers that are 1st from each end or x1 52 and x11 74. (To be conservative, round the result down.) This really looks sloppy, so the next solution is preferred. To be more precise, use the Binomial table with n 15 . Possible intervals are x1 to x11 , x 2 to x10 etc. Let’s try a few intervals. use a Normal approximation k 4 252solngr1-071 2/8/07 The binomial table for n 11 is copied here. We need only look at the last p .5 column. 11 0 1 2 3 4 5 6 7 8 9 10 11 0.89534 0.99482 0.99984 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.56880 0.89811 0.98476 0.99845 0.99989 0.99999 1.00000 1.00000 1.00000 1.00000 1.00000 1.00000 0.31381 0.69736 0.91044 0.98147 0.99725 0.99970 0.99998 1.00000 1.00000 1.00000 1.00000 1.00000 0.16734 0.49219 0.77881 0.93056 0.98411 0.99734 0.99968 0.99997 1.00000 1.00000 1.00000 1.00000 0.08590 0.32212 0.61740 0.83886 0.94959 0.98835 0.99803 0.99976 0.99998 1.00000 1.00000 1.00000 0.04224 0.19710 0.45520 0.71330 0.88537 0.96567 0.99244 0.99881 0.99987 0.99999 1.00000 1.00000 0.01977 0.11299 0.31274 0.56956 0.78970 0.92178 0.97838 0.99571 0.99942 0.99995 1.00000 1.00000 0.00875 0.06058 0.20013 0.42555 0.66831 0.85132 0.94986 0.98776 0.99796 0.99979 0.99999 1.00000 0.00363 0.03023 0.11892 0.29628 0.53277 0.75350 0.90065 0.97072 0.99408 0.99927 0.99996 1.00000 0.00139 0.01393 0.06522 0.19112 0.39714 0.63312 0.82620 0.93904 0.98520 0.99779 0.99985 1.00000 0.00049 0.00586 0.03271 0.11328 0.27441 0.50000 0.72559 0.88672 0.96729 0.99414 0.99951 1.00000 2Px k 1 Interval k x1 to x11 or 30 to 74 x 2 to x10 or 40 to 66 1 2Px 0 2.00049 .00098 2 2Px 1 2.00586 .01172 x3 to x9 or 48 to 65 3 2Px 2 2.03271 .06542 4 2Px 3 2.11328 .22656 x 4 to x8 or 50 to 60 Notice that we could have answered the question by finding the largest value of k with Px k 1 .005 . Since the smallest interval with a significance level below 1% is 30 to 74, this is the best that we can do. We can check our results using the Normal distribution. The outline says, using a continuity correction, k 1 1 2 np k .5 .5n P x k 1 1 2 P z P z 2 npq . 5 n . 1 1 .5 11.5 0.5 5.5 k 1 Px 0.5 P z Pz 3.02 .5 .4987 .0013 P z 1.65831 11 . 5 . 5 2 1 .5 11.5 1.5 5.5 k 2 Px 1.5 P z Pz 2.41 .5 .4920 .0080 Pz 1.65831 11.5.5 Since we need Px k 1 .005 , k 1 was correct. 5 252solngr1-071 2/8/07 Extra Credit Minitab Problem 5. Check some numbers in the Normal, t, Chi-Squared or F tables using the new 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 6.16.4 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 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. ————— 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 MTB > %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 ...working... t Curve Area 6 252solngr1-071 2/8/07 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 ...working... Normal Curve Area 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 7 252solngr1-071 2/8/07 ...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) l Please answer Yes or No. y Enter the value for which you want the area to the left. DATA> 4.8650 ...working... Chi Squared Curve Area 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 8 252solngr1-071 2/8/07 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 ...working... F Curve Area Data Display mode 0.818182 std dev 0.968246 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 9 252solngr1-071 2/8/07 ...working... F Curve Area Data Display mode 0.818182 std dev 0.968246 10