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252solngr1 2/05/04 (Open this document in 'Page Layout' view!) Graded Assignment 1 Please show your work! Neatness and whether the papers are stapled may affect your grade. 1. A manufacturer is concerned that a machine is not adequately filling a 200 oz container. 15 measurements are taken. Results are below. 198 204 189 182 205 195 188 201 200 203 201 193 194 196 190 Compute the sample standard deviation using the computational formula. Use this sample standard deviation to compute a 90% confidence interval for the mean. Does the mean differ significantly from 200 oz? Why? 2. How would your results change if the sample of 15 had been taken from a population of 100? 3. Assume that the population standard deviation is 7.00 (and that the sample of 15 is taken from a very large population). Find z .04 and use it to compute a 92% confidence interval. Does the mean differ significantly from 200 oz? Why? Solution: 1) index x x 2939 , x 576471 , x 2939 195.9333 x x2 2 1 198 39204 2 204 41616 3 189 35721 4 182 33124 5 205 42025 6 195 38025 7 188 35344 8 201 40401 9 200 40000 10 203 41209 11 201 40401 12 193 37249 13 194 37636 14 196 38416 15 190 36100 sum 2939 576471 sx sx n n 15 x nx 2 2 s x2 n 1 n 15 576471 15195.9333 14 2 623.12928 44.5093 14 s x 44.5093 6.67152 . s x2 44.5993 2.9673 1.7226 n 15 The formula is in table 20 of the Supplement. .10 2 .05 Confidence interval: 14 tn2 1 t.05 1.761 x tn1 s x is the formula for a two sided interval. 2 x tn1 s x 195.9333 1.7611.7227 195.9333 3.0337 2 we ask if the mean is significantly different from 200, our null hypothesis is or 192.900 to 198.967. If H 0 : 200 and since 200 is not between the top and the bottom of the confidence interval, reject H 0 and say that the mean is significantly different from 200. If we ask if the mean is significantly different from 198, the null hypothesis is H 0 : 198 and 198 is between the top and the bottom of the confidence interval. So we reject H 0 or we can say that the mean is significantly different from 198. 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) N 100 , the sample of 15 is more than 5% of the population, so use s N n 100 15 sx x 1.7226 1.7226 0.8586 1.72260.92660 1.5962 . 100 1 n N 1 n 1 14 Recall that x 195.9333 , .10 , 2 .05 and t t.05 1.761 . 2 2) If Confidence interval: x tn1 s x is the formula for a two sided interval. 2 x tn1 s x 195.9333 1.7611.5962 195.9333 2.8109 2 or 193.122 to 198.744. The interval is smaller, but it doesn’t change anything – the mean is still significantly different from 200 but not 198. 3) a) Find z.04 . Please don’t tell me that because P0 z 0.04 .0160 , that z.04 is .0160. 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 . z.04 so that Pz z.04 .04 or P0 z z.04 .4600 . From the interior of the Normal table the closest we can come to .4600 is P0 z 1.75 .4599 . This means that z.04 1.75 . Check: Pz 1.75 Pz 0 P0 z 1.75 .5 .4599 .0401 .04. From the diagram, we want one point Normal Curve with Mean 0 and Standard Deviation 1 The Area to the Right of 1.75 is 0.0401 0.4 Density 0.3 0.2 0.1 0.0 -5 b) We know that -4 -3 -2 -1 0 Data A xis 1 x 195.9333 , n 15 and 7 . So x 2 n 3 7 15 =1.8074. The 92% confidence interval has 1 .92 or .08 , so z z.04 2 4 72 3.2667 15 1.75 . The confidence interval is x z x 2 195.9333 1.751.8074 195.93 3.16 or 192.77 to 199.09. If we test the null hypothesis H 0 : 200 against the alternative hypothesis H 0 : 200 , since 200 is not on the confidence interval, we reject the null hypothesis. The result is not significantly different from 198 because 198 is within the interval. Extra Credit: 4) a. Use the data above to compute a 90% confidence interval for the population standard deviation. 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) n 1s 2 Solution: From the supplement pg 1, 2 2 s x 6.67152 , n 15 , .10 and 2 2 1 2 n 1s 2 .05 . We use n 1 21 .9514 6.5706 . The formula becomes . We know s x 44.5093 , 2 2 2 n 1 2 14 .05 23.6848 and 1444.5093 2 1444.5093 23.6848 6.5706 2 26.3083 94.8361. If we take square roots, we get 5.129 9.738 2 or b. Assume that you got the sample standard deviation that you got above from a sample of 45, repeat a. Solution: From the supplement pg 2, s 2 DF s 2 DF . We now have z 2 DF z 2 DF s x 6.67152 , n 45 , .10 and 2 2 2 .05 . We use z.05 1.645 and 2DF 2(44) 88 9.3808 . The formula becomes 6.671529.3808 6.671529.3808 or 5.681 8.090 . 1.645 9.3808 1.645 9.3808 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. The numbers in order are x1 x 2 x3 x 4 x5 x6 x7 x8 x 9 x10 x11 x12 x13 x14 x15 182 188 189 190 193 194 195 196 198 200 201 201 203 204 205 It says on the outline that 2Px k 1 . If we check a Binomial table with n 15 and p .50 , we find that the first quantity below 5% is Px 3 .01758 . So if k 4 , k 1 3 and 2.01758 .03516 and P190 201 1 .03516 96.48% . If you are willing to live dangerously use k 5 , k 1 4 , 2.05923 .11846 and P193 201 1 .11846 88.15% . If we use the formula k n 1 z . 2 n 2 15 1 1.645 15 4.81 , 2 it tells us to use the 4th number from the end, since, if we want to be conservative we round the answer down. How I got these results ‘MTB >’ is the Minitab prompt. The retrieval is done using the ‘file’ pull-down menu and the ‘open worksheet’ command followed by finding where I put the data. Other instructions were typed in the ‘session’ window. I put the data in column 1 in Minitab and used the ‘Gsummary’ command to get the mean and standard deviation. ————— 2/3/2005 8:31:48 PM ———————————————————— Welcome to Minitab, press F1 for help Results for: 2gr1-051.MTW MTB > GSummary c1; SUBC> Confidence 90.0. Summary for C1 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) Summary for C1 A nderson-Darling N ormality Test 184 188 192 196 200 A -S quared P -V alue 0.23 0.763 M ean S tDev V ariance S kew ness Kurtosis N 195.93 6.67 44.50 -0.505497 -0.417569 15 M inimum 1st Q uartile M edian 3rd Q uartile M aximum 204 182.00 190.00 196.00 201.00 205.00 90% C onfidence Interv al for M ean 192.90 198.97 90% C onfidence Interv al for M edian 192.72 201.00 90% C onfidence Interv al for S tD ev 9 0 % C onfidence Inter vals 5.13 9.74 Mean Median 192 194 MTB > let c2=c1*c1 MTB > Print c1 c2 196 198 200 202 I computed the square of C2 in c1 and got the sums for computing the variance. Data Display Row 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 C1 198 204 189 182 205 195 188 201 200 203 201 193 194 196 190 C2 39204 41616 35721 33124 42025 38025 35344 40401 40000 41209 40401 37249 37636 38416 36100 MTB > sum c1 Sum of C1 Sum of C1 = 2939 MTB > sum c2 Sum of C2 Sum of C2 = 576471 MTB > describe c1 Descriptive Statistics: C1 Variable N N* Mean SE Mean StDev Minimum Q1 Median Q3 252solngr1 2/05/04 C1 Variable C1 (Open this document in 'Page Layout' view!) 15 0 195.93 Maximum 205.00 1.72 6.67 182.00 190.00 196.00 201.00 MTB > Onet c1; This does a 90% confidence interval and a test for a mean of 200 using s. SUBC> Test 200; SUBC> Confidence 90. One-Sample T: C1 Test of mu = 200 vs not = 200 Variable N Mean StDev SE Mean C1 15 195.933 6.670 1.722 90% CI (192.900, 198.967) T -2.36 P 0.033 MTB > OneZ c1; This does a 92% confidence interval and a test for a mean of 200 using sigma. SUBC> Sigma 7; SUBC> Test 200; SUBC> Confidence 92. One-Sample Z: C1 Test of mu = 200 vs not = 200 The assumed standard deviation = 7 Variable N Mean StDev SE Mean C1 15 195.933 6.670 1.807 92% CI (192.769, 199.098) Z -2.25 P 0.024 MTB > %normarea5a This does the graph shown above. Executing from file: normarea5a.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> 1.75 ...working... Normal Curve Area MTB > let c5=c1 MTB > Sort c5 c5; SUBC> By c5. MTB > print c5 This sorts c1, which I moved to c5. Data Display C5 182 204 188 205 189 190 193 194 195 196 198 200 201 201 203 Extra Credit: 5. Check some numbers in the t, Chi-Squared or F tables using the new set of Minitab routines that I have prepared. To use the new set of routines, set up a file to hold your work. Then go to http://courses.wcupa.edu/rbove , open the Minitab folder and download any of the following: Normal Distribution Area programs: NormArea5A.txt 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) or NormArea5C.txt and NormArea5.txt. t Distribution Area programs: tAreaA.txt or tAreaC.txt and tArea.txt Chi-squared Distribution Area programs: ChiAreaA.txt or ChiAreaC.txt and ChiArea.txt F Distribution area programs: FAreaA.txt or FAreaC.txt and FArea.txt Use Notepad (under ‘tools’ in Minitab’) to convert their extensions from .txt back to .mac. To see how they are used, look at http://courses.wcupa.edu/rbove/Minitab/Area.doc. Routines like tAreaA are self prompting. To use routines like tAreaC, you need to set up your data in advance. If you want to use one of the worksheets that are mentioned in http://courses.wcupa.edu/rbove/Minitab/Area.doc, click on ‘File’ and then ‘Open Worksheet.’ Copy a URL like the ones below into File Name.’ http://courses.wcupa.edu/rbove/Minitab/252PrA1d-f.MTW http://courses.wcupa.edu/rbove/Minitab/tEx1.MTW http://courses.wcupa.edu/rbove/Minitab/ChiEx1.MTW http://courses.wcupa.edu/rbove/Minitab/FEx1.MTW 10 Results: I looked at the tables and found t .10 1.372 , 10 2 .90 4.8650 , F.1010,10 2.32 and 10,10 F.90 1 10 z.10 1.282 , 2 .10 23.2093 , 2.32 0.431 . For the numbers with .10 as a subscript, I checked that the probability above them was .10, for the numbers with .90 as a subscript, I checked that the probability below them was .10. MTB > %tareaA Executing from file: tareaA.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 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) t Curve with 10 Degrees of Freedom and Standard Deviation 1.11803 The Area to the Right of 1.372 is 0.1000 0.4 Density 0.3 0.2 0.1 0.0 -5.0 -2.5 0.0 Data A xis 2.5 5.0 Data Display mode median 0 0 MTB > %normarea5a Executing from file: normarea5a.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 Normal Curve with Mean 0 and Standard Deviation 1 The Area to the Right of 1.282 is 0.0999 0.4 Density 0.3 0.2 0.1 0.0 -5 -4 -3 -2 -1 0 Data A xis 1 2 3 4 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) MTB > %ChiareaA Executing from file: ChiareaA.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> 1.282 ...working... ChiSquare Curve Area ChiSquare Curve with 10 Degrees of Freedom and Standard Deviation 4.47214 The Area to the Right of 1.282 is 0.9995 0.10 Density 0.08 0.06 0.04 0.02 0.00 0 10 20 Data A xis 30 Data Display mode median 8.00000 9.33333 MTB > %chiareaA Executing from file: chiareaA.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 ...working... Chi Squared Curve Area 40 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) ChiSquare Curve with 10 Degrees of Freedom and Standard Deviation 4.47214 The Area to the Left of 4.865 is 0.1000 0.10 Density 0.08 0.06 0.04 0.02 0.00 0 10 20 Data A xis 30 40 Data Display mode median 8.00000 9.33333 MTB > %fareaA Executing from file: fareaA.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 ...working... F Curve Area F Curve with numerator DF of 10 and Denominator DF of 10 The Area to the Right of 2.32 is 0.1003 0.8 0.7 Density 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 Data Display mode 0.818182 2 4 6 8 Data A xis 10 12 14 16 252solngr1 2/05/04 (Open this document in 'Page Layout' view!) Data Display std dev 0.968246 MTB > %fareaA Executing from file: fareaA.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 ...working... F Curve Area F Curve with Numerator DF of 10 Denominator DF of 10 The Area to the Left of 0.431 is 0.1003 0.8 0.7 Density 0.6 0.5 0.4 0.3 0.2 0.1 0.0 0 Data Display mode 0.818182 Data Display std dev 0.968246 2 4 6 8 Data A xis 10 12 14 16