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# INTRODUCTION TO STATISTICS FOR ENGINEERS ST 314

OSU

GPA 3.9

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This 23 page Class Notes was uploaded by Shannon Krajcik Sr. on Monday October 19, 2015. The Class Notes belongs to ST 314 at Oregon State University taught by D. Dail in Fall. Since its upload, it has received 17 views. For similar materials see /class/224545/st-314-oregon-state-university in Statistics at Oregon State University.

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Date Created: 10/19/15

ST 314 HW 6 Solutions 1 Text p 235 exercise 439 a Parallel boxplots are Boxand Whisker Plot 1 8 2 07 1 13 16 19 22 25 VISCOSIT b Let H true average thickness for high viscosity and L true average thickness for low viscosity Then the appropriate test is 1 H0 pH L 0 Ha pH ML gt 0 looking to see ifhigher viscosity paint leads to thicker coatings 2 Use atest statistic oft yH yLsplnH lnLgt12 3 With at 05 and 11 nL 72 16 16 7 2 30 we nd t30 05 1697 so we reject H0 ift gt1697 4 Calculating gives yL 1348 st 1146 yH 1487 and SE2 2474 so s2p 15114615247430 0181 and sp 4254 Thus t yH yLsp1nH 1nL 2 1487 713484254116 11612 924 Since 924 is not gt 1697 we do not reject H0 5 At 0t 05 there is insuf cient evidence to conclude that the average thickness for high viscosity exceeds the average thickness for low velocity 0 With at 05 t30 025 2042 so a 95 con dence interval for the true mean diflierence in coating thickness between the high viscosity and low viscosity is yH YLit30025 sp1nH lnL1z148771348i 20424254116 116gt12 139 i 307 or 168 446 d We assumed the two samples were independent of one another We also assumed the two populations of viscosities have a common variance Normal probability plots of the two data sets would show some concern with the normality assumption in the thickness from the high viscosity e Fail to reject H0 at the ct 05 level Insuf cient evidence to conclude the higher viscosity leads to thicker coatings This appears to agree with the boxplots 2 Text p 237 exercise 442 a Parallel boxplots are U 0 3 1 D Boxand Whisker Plot 1 m 2 4393 4397 5391 5395 5399 6393 6397 MILES Let H1 true mean tread life of Brand 1 tires and u true mean tread life of Brand 2 tires Then the appropriate test is 1 H0 H1 u 0 Ha H1 u 7i 0 looking for any differences 2 Use atest statistic oft yl yzsplnH lnLgt12 3 With at 05 and n1nz 7 2 15 15 7 2 28 we nd I28 025 2048 so we reject H0 if ltl gt 2048 4 Calculating gives 1 520 s12 212857 z 56933 and s22 233524 so s2p l4212857 l423352428 223191 and SF 47243 Thus t 1 zsp1n1 ln112 520 7 5693347243115 115 2 286 Since 286 I 286 gt 2048 we reject H0 5 At 0c 05 there is suf cient evidence to conclude that the mean tread life of Brand 1 tires is different than that of Brand 2 tires With at 05 tzsy 025 2048 so a 95 con dence interval for the true difference between the mean tread life of Brand 1 tires and that of Brand 2 tires is yl h r us 025 sp1n1 lnz12 520 7 56933 r 204847243115 115 2 4933 r 3533 or 8466 1400 We assumed the two tread life samples were independent of one another We also assumed the two populations have a common variance Normal probability plots stemandleaf displays and summary statistics all support the normality assumption of the two populations Reject H0 at the ct 05 level Suf cient evidence to conclude a difference between Brand 1 mean tread life and that of Brand 2 This appears to agree with the boxplots 3 Text p 290 exercise 52 We are given 60 uo 5603907n3 and 7 LCL6 30 56 3 07 4388 d UCL 6 30 563 07 6812 Thenthe control chart is 0 A l J3 Control chart 52 7 5 E E 6 a o c 8 c 5 m a E 4 l l l l l l l l l l l l l l 1 6 11 16 Subgroup The process appears in control except for the last sample The process needs to be monitored 4 Text p 306 exercise 512 We are given m 20 subgroups in base Period RChart of ash content for 511 a From this base period we calculate R 044 Since n 4 D3 0 D4 2282 Then UCL D4 2282044 100 and we have LCL D3 0044 0 Our R chart is The process variation appears to be in control 1 6 11 16 AoAi39oo39I39U39Ioukn39oo39mA39A R 0 900009000 a pallet b For an X cha1t since 11 4 A2 0729 Since this process is in control over the base period we can use 0 m i 19388 which gives us UCL 0 41 19388 72944 1971 and LCL 0 7141 19388772944 1907 The J cha1t is Xbar Chart for 510 205 20 71 W 195 K W N 181 Average Ash Content 12 3 4 5 6 7 8 91011121314151617181920 Pallet The process mean location appears to be out of control An increasing trend eXists in the chart c The I cha1t assumes that the process is in control during the base period and the data follows a normal distribution The X chart is based on the sample means and assumes that the IT chart is in control that the subgroup means are in control during the base period and that the sample size is large enough to assume the subgroup means have a normal distribution by the CLT To check these assumptions we should look at a stem andleaf display of the data during the base period For these 80 observations SG gives us 2 18l33 7 18l44455 l4 18l6667777 20 18 888999 28 l9l00000111 l3 19 2222222333333 39 19 455 36 l9l666667777 27 19 888888999999 15 20l000001111 6 20l222 3 20l4 2 20l66 Thus the data does not appear to be normally distributed since there are multiple peaks The normality assumptions are very questionable ST 314 HYPOTHESIS TESTS AND CONFIDENCE INTERVALS Name of Test OneSample ztest OneSample ttest OneSample ztest for proportions of populations When used 1 continuous RV 1 continuous RV l binomial proportions 2 one population 2 one population 2 one population 3 039 known 3 O39unknown 3 npo 25and nqo 25 Assumptions 1 random sample taken 1 random sample taken 1 random sample taken 2 distribution of data 2 distribution of data 2 independent trials in pop is normal in pop is normal Hypothesis Does the sample of n subjects Same as Is the population proportion come from a population with to the left equal to some hypothesized LIX some specified value e value H03 Hxpo H03 Hxpo H03pp0 lt lt Teststatistic Zi hfi ZL 0 5 pomp0 5 Ja n Distribution of teststatistic normal tdistribution with n1 df Confidence Interval X i 2U X i tn15 The critical region was not given in any of the above as it depends on the alternative hypothesis normal if n is large enough i 2 Paired ttest or matched pairs 1 continuous RV 2 each subject eu receives both treatments or matched subjects each with a different treatment 1 random sampes taken 2 distribution of differences is norma Is the mean of the differences 0 H02 10 NOTE n of differences tdistribution with n1 df Two independent samples pooled twosample ttest 1 continuous RV 2 2 independent samples from 2 populations 3 to compare 2 pop means 4 0102 unknown 1 random sample taken from each population 2 distribution of data in each pop is normal if each n small 3 O391O392 see pp 229230 Do the two independent samples represent 2 populations with the same mean H03 1 2 OR H03 1 39 2 0 t lt139 2l391 2 quot1quot22 1 1 S 7 i p r11 r12 Where s n1391s12 n2391s p n1n22 tdistribution with n1n22 df 1 0 979 t n1 n2 nin22Sp refers to the confidence level alpha2 for Confidence Intervals ST 314 HW 1 Solutions 1 Text p 52 exercise 24 StemiandiLeaf Display for STRENGTH unlt 01 112 represents 12 1 014 3 0189 12 11011222344 29 1156666667777888899 41 21000112224444 22 215555666677788888889999 37 3100001112222223333344 17 31566777778 8 41244 5 41799 2 511 HI156 StatAdvlsor te represented by a of th a abe D0 more leadlng digits th On each row the lndividu put the Advisor output as your comments 39 row which contains the median Comments these appear to be wellbehaved data as they are roughly symmetrical with a single peak and have rapidly diminishing tails They compare quite well to the minimum acceptable breaking stress of 12 since only 6 out of the 100 values fall below this standard 2 Text p57 exercise 217 Ifyou do this in SG by creating two new columns in the data le called VISCil and VISC72 and then plotting stemandleaf displays for each column separately you end up with the following StemiandeLeaf Display for VISCil unlt 001 112 represents 012 memooooxlamww H w H StemiandeLeaf D1splay for VISC72 unlt 01 112 represents 12 0177 0189 11 2 4 4 4 11 8 1 1 4 4 4 4 4 5 55 4 11 4 1i 4 2 i 011 1 2 1 3 SG had taken it upon itself to use different units for the two displays Thus they are worthless for comparison purposes Doing it by hand with the leaves being the second decimal digit results in Low Viscosity High Viscosity 07 07 49 08 388 08 3 0 9 09 8 1 0 49 1 0 1 1 2 1 1 1 2 9 1 2 1 3 1 1 3 14 89 14 00266 15 9 15 119 1 6 25 1 6 1 7 16 1 7 1 8 3 1 8 1 9 19 20 20 5 21 21 27 2 2 2 2 23 23 6 Comments the two groups are distinctly different The Low Viscosity data have a small clump in the 08s and then are somewhat uniformly spread out from the 10s to the 18s The High Viscosity data have three distinct clumps one in the 0709 range one in the 14 and 15s and a third in the 2023 range Who knows why this is happening 3 Text p 67 exercise 223 Boxand Whisker Plot 0 1 2 3 4 5 6 STRENGTH Comments the data are centered around 27 with the bulk of the values between 18 and 32 Values range from 04 to 56 They are approximately symmetric With one mild outlier at 56 The boxplot immediately tells us if there are any outliers in the data unlike the stemandleaf display and can quickly tell us the range of the unquestionably good data 4 Text p 68 exercise 236 Boxand Whisker Plot GROUP 07 1 13 16 19 22 25 VISCOSIT Comments The Low Viscosity data Group 1 are centered at 14 with the bulk of values between 108 and 165 The High Viscosity data Group 2 are centered around 146 with most values between 120 and 182 Neither set has any apparent outliers 5 Text p 68 exercise 240 le Boxiandiwhlsker Plot PRESSURE by STARCH Analysis Su n Dependent variable Factor STARCH Don t Number of observatlons 24 Number of levels 3 want 01 need The StatAdvlsor Boxand Whisker Plot 6 9 l STARCH 2 15 18 21 24 PRESSURE Comments The Waxy Maize data Starch l is centered around 95 with most values between 7 and 10 The Native Corn data Starch 2 is also centered around 95 but most values are in the range of9 to 14 and there is one outlier at 225 The High Amylose Corn Starch 3 data are centered around 125 most values ranging between 10 and 15 and has no outliers The minimum injection pressure of Waxy Maize has the least variability and that of Native Corn has the most variability The center of the data for High Amylose Corn has the highest pressure while Waxy Maize has the lowest Since lower minimum injection pressure indicates easier processing Waxy Maize may be the easiest processable thermoplastic starch of the three ST 314 421 a 7 o HW 5 Solutions 1 H0 L 8 Ha L i 8 L true mean thickness ofmetal Wires 2 teststatistic t M s n 3 0c 05 olt2 025 49 degrees of freedom Reject if t gt 201 t497025 4 2y 3988 Eyz 31828 11 50 y Eyn7976 s2 2y2 Ey2n nl 31828 3988250 49 04023 s 2006 7976 8 85 2006 NE quot 5 85 lt 201 so we cannot reject H0 At 0c 05 there is insufficient evidence to conclude that the mean thickness of metal Wires differs from 8 microns y i 1042 5 MB 7976 i 2012006 M15 7976 i 057 So 7919 8033 is a 95 confidence interval for the one mean thickness of metal Wires y itm S11n 7976 i 201 20065150 7976i 407 So 757 838 is a 95 prediction interval for the thicknesses d A stemandleaf diagram andor normal probability plot can be constructed Stem Leaves No Depth 7 76 0 1 77 00000 5 6 E i 78 000000000 9 15 2 i 79 000000000 9 24 E t 80 000000000000 12 l 81 00 2 14 g I 82 000000 6 12 a i 83 0000 4 6 T 84 00 2 2 a g Of N 139 39 76 78 80 82 84 cknesses To do the analysis we assume that the distribution of metal Wire thicknesses is roughly normal ie single peaked approximately moundshaped tails die rapidly The stemandleaf diagram demonstrates that these assumptions While suspicious are not grossly violated Normal probability plot is roughly a straight line which looks good e 424 a 7 an The chipmanufacturing process appears to be producing metal wires with a target thickness of 8 microns at the 95 confidence level 1 H0 p 2125 Hz 11 i 2125 p average moisture content fixE0 2 tt tttiti es sa1s1c SHIT 1 3 0601 042 005 n 1 26 df reject H0 if1t1gt 2779 126005 2779 4 2y 5533 Eyz 1138385 y 553327 205 s2 1148385 5533227 26 0174 s 132 205 2125 295 132ij 5 Since 12951 gt 2779 we reject H0 At oc 01 there is sufficient evidence to conclude the true average moisture content differs from the target of 2125 y ital s A13 205 i 2779 1322 7 205 i 07 So 198 212 is a 99 confidence interval for the true mean moisture content y item sxll 1n 205 42779 132 2827 205 i 37 So 168 242 is a 99 prediction interval for the moisture content A stemandleaf diagram andor normal probability plot of the 27 observations can be constructed Stem Leaves No Depth u r 39 17 25 2 2 t 0 2 E 7 19 00488 5 7 z 1 20 002225669 9 g 5 21024555778 9 11 m o e g 22 29 2 2 6 i z a 397 4 0 g OI 7quot 17 18 19 20 21 22 23 moistllre1419 We assume that the distribution of moisture contents of polyol is approximately normal ie single peaked roughly moundshaped tails die rapidly Normal probability plot follows a straightline At the 99 confidence level evidence exists that the true moisture content has decreased from 2125 The confidence interval gives possible values between 198 and 212 This does indicate however that the shift has not been a large one 376 n P 1 2000 probability of underfilling 005 probability of not underfilling 995 Y number bottles underfilled u np 2000005 10 02 npq 2000005995 995 0 a b C 315 PrY 10 PrY 105 Pr 7pGS 95 2103151gtrZg 116 5636 PrY 215 PrY 2 145 PrY 215 Prmvt p o 2 145 2 10 315 PrZ 214319236 0764 PrY 2 20 PrY 2 195 PrY 2 20 Prmvtp o 2 195 710315PrZ 2 302 19987 0013 d The approximate probability of finding 20 or more underfilled is only l3 433 a If the consumer group finds 20 or more underfilled a rare event then we might conclude that the probability of a underfilled bottle is greater than 05 1 H0 p 10 Ha p gt 10 p true proportion of bores outside the specifications 2 Z P39130 VPOQon p010 q011090 n165 np0165 nq01485 is the test statistic 3 on 01 201 233 Reject H0 ifZ gt 233 218 10 7 41090 165 7505 Since 505 gt 233 we reject H0 4 36165218 z 5 At oc 01 there is sufficient evidence to conclude that the proportion of bores outside of the specifications exceeds 10 b 3 i z2 l an 218 i 257 1218782 165 218 i 0826 1354 3006 is a 99 confidence interval for the true proportion of bores outside the specifications 434 a 1 H0 p 05 Ha p gt 05 p true proportion of nonconforming bricks 19P 2 Z 0 is the test statistic VPOQO n p0 05 qo 95 n214 npo 107 nqo 2033 3 oc01 201 233 rejectHo ifZgt233 0841 05 4 A 182140841 z 229 O p 0595214 5 229 lt 233 so we cannot reject H0 There is insufficient evidence to conclude that the true proportion of nonconforming bricks is more than 5 b 3 iZm l an 0841 i258 08419159 214 0841 i 0489 0352 1330 is a 99 CI for the true proportion of nonconforming bricks 438 a 1 H0 p 1 Ha p gt 1 p true proportion of the homes in a nameless city inNC Florida that have radon levels higher than recommended by EPA 2 Z l k is the test statistic P0lt10 n p01 q09 n200 p020 nq0180 3 oc05 Z051645rejectH0ifZgt1645 1251 0 7 4 p 725200 125 z ix9W2 118 5 118 lt 1645 so we cannot reject H0 There is insufficient evidence to conclude that the true proportion of the homes in a nameless city in NC Florida with excessive levels of radon is higher than 10 b g izmxl an 125 i 196 125875200 1254 046 079 171 is a 95 confidence interval for the true proportion of hours in a nameless city in NC Florida that have radon level higher than recommended by EPA ST 314 HW 3 Solutions 1 Text p 133 exercise 340 The rv Y is distributed exponential with 7 001 PY lt 30 17 e quot30 7 1 7 60160 1 e 30 7 2592 PY gt151717 e M151 equot15 8607 PY 100 7 0 P50 g Y s 150 17 e quot159 7 1 7 e quotW1 e3915 e395 3834 909quot J Text p 141 exercise 347 Let Y thickness of bolts Then Y is normally distributed with mean u 100 mm and standard deviation 6 16 mm a P92 g Y s 108 7 P 921016 s Yu6 s 1081016 7 P 5 s z s 5 PZ lt 5 7 PZ lt 5 7 6915 73085 7 3830 b PY s 92 7 P Yp6 s 921016 PZ s 5 7 3085 3 Text p 142 exercise 352 Let Y weld ratio Then Y is normally distributed with u 083 and 6 0039 PY gt 09 PYuc gt 09 0830039 PZ gt 17949 7 1 7 PZ gt 179 179633 0367 note that using software we can nd PZ gt 17949 l 9637 00363 E b P08 lt Y lt 09 P080830039 lt Z lt 090830039 P0769 lt Z lt17949 P077 lt Z lt 179 PZ lt 179 7 PZ lt 077 9633 7 02206 07427 c We want to nd the smallest u so that PY gt 08 gt 095 But PY gt 08 PYp6 gt 08p6 PZ lt 08p0039 From Table 1 we see that PZgtl645 095 so 08p0039 1645 Solving for 1 gives 1 08 16450039 08641 4 Text p 152 exercise 356 Let Y concentricity ofan engine oil seal groove Then Y has mean u 56 and standard deviation 6 07 Sample size is n 3 Assuming the CLT applies Y is normally distributed with mean 56 and standard deviation 07N3 404 a Pprocess mean has shifted P Y lt 43 P Y gt 68 PZ lt 4356404 PZ gt 6856404 PZ lt 322 PZ gt 297 0 19985 0015 b Assumed the sample of n 3 was large enough to use the central limit theorem 5 Text p 153 exercise 359 Y has mean of l65and standard deviation of 5 Sample size is n 12 Then assuming the CLT applies Y is normally distributed with mean of 165 and standard deviation of 5712 71443 a P Y lt 162 PZ lt 1621651443 PZ lt 208 00188 b P 164lt Y lt 166 P 1641651443 lt Z lt 166 71651443 P069 lt Z lt 069 05098 Note Using software such as StatGraphics this can be computed more exactly as P6928 lt Z lt 6928 05116 We had to assume the sample of n 12 batches was large enough so that the distribution of Y is approximately normal by the CLT This could be checked by checking the summary statistics and a normal probability plot in Statgraphics O 6 Text p 153 exercise 360 Y has mean of 6500 and standard deviation of 250 Sample size is n 16 Then assuming the CLT applies Y is normally distributed with mean of 6500 and standard deviation of 250716 7 625 a P6400 lt Y lt 6550 P6400 7 6500625 lt Z lt 6550 7 6500625 P16 lt Z lt 08 PZ lt 08 7 PZ lt 16 078817 00548 07333 b We had to assume the sample of n 16 batches was large enough so that the distribution of Y is approximately normal by the CLT Note that this could be checked by checking the summary statistics and a normal probability plot in Statgraphics ST 314 HW 7 Solutions 1 Text p 333 exercise 535 We are given n 200 and m 25 subgroups in base period p 00478 Then we have LCL n 3npq 20000478 3 20000478l 00478 051 and UCL n 3anq 20000478 3 200004781 00478 1861 Then the control chart is npChart for Exercise 535 data of quotdeadquot locations Subgroup The production process appears in control 2 Text p 338 exercise 540 We are given In 20 subgroups in base period 5 130 Then we have the following LCL E 3JE 130 34130 218 and UCL E3JE 1303130 2382 Then the control chart is Control chart for 530 of flaws found 9 O111111111111111111111111111 11 1 6 11 16 21 26 Section of wire Control Chart for Exercise 540 data The process appears out of control since there are two out of control signals However the process does exhibit improvement over the last 10 subgroups 3 Text p 388 exercise 64 parts a c a A scatterplot is Plot of MODULUS VS PET 10 quot 39 I 8 in D 5 6 39 8 4 D E I 2 2 m 0 i E o 20 4o 60 so 100 PET b The analysis from StatGraphics is Simple Regression MODULUS v5 PET Regression Analysis Linear model Y 7 a bX Dependent variable MODULUS Independent variable39 PET Standard T Parameter Estimate Statistic PValue Intercept 822062 0351228 234054 00000 Slope 00601645 000435495 138152 00000 Analysis of Variance Source Sum of Squares Df Mean Square EiRatio PValue Model 289905 1 289905 19086 00000 Residual 0759469 5 0151894 Total Corr 2975 6 Correlation Coefficient 0987l53 Risquared 974472 percent Standard Error of Est 0389736 The StatAdvisor The output shows the results of fitting a linear model to describe the relationship between MODULUS and PET The equation of the fitted model is MODULUS 822062 00601645PET Since the Pivalue in the ANOVA table is less than 01 t 0 here is a statistically significant relationship between MODULUS and PET at the 99 confidence le h RiSquared statistic indicates that the model as fitted explains 974472 of the variability in MODU US The correlation coefficient indicating a relatively strong relationship between the variables The standard error of the estimate shows the standard deviation of the residuals to be 0389736 This value can be used to construct prediction limits for new observations by selecting the Forecasts option from the text menu Plot of Fitted Model I 8quot D I 5 Q 4 E O Z 2 2 0i 3 o s 5 e a g The prediction eguation is given above as MODULUS 822062 00601645PET c Analysis The complete StatGraphics analysis is given above R2 09745 is larger than 09 This model ts the data very well The Ftest value 19086 results in a pvalue of 00000 The pvalue for the tests on the individual coefficients is 00001 which is very small We can conclude that the modulus depends on the of PET in the blend 4 Text p 388 exercise 65 parts a b a The StatGraphics analysis is Regression Analysis Linear model Y a b Dependent variable VISCOSIT Independent variable TEMP Standard T Parameter Estimate Er r Statistic PValue Intercept 1 00468683 273428 00000 Slope 000875782 0000728359 l2024l 00000 Analysis of Variance Source Sum of Squares Df Mean Square EiRatio PValue Model 0325292 1 0325292 14458 00000 Residual 00134997 6 000224995 Total Corr 0338792 7 Correlation Coefficient 0979874 Risquared 960153 percent Standard Error of Est 00474336 The StatAdvisor The output shows the results of fitting a linear model to describe the relationship between VISCOSIT and TEMP The equation of the fitted model is VISCOSIT 7 128151 000875782TEMP Since the Pivalue in the ANOVA table is less than 001 statistically significant relationship between VISCOSIT the 99 confidence level t e s a and TEMP at e RSquared statistic indicates that 960153 of the variability in VISCOSIT The correlation coefficient equals 0979874 indicating a relatively strong relationship between The standard error of the estimate shows the standard deviation of the residuals to be 00474336 This value can be used to construct prediction limits for new observations by selecting the Forecasts option from the text menu the model as fitted explains Plot of Fitted Model 13 39 39 39 39 39 E 11 lt72 2 2 m S 07 05 39 39 24 44 64 84 104 TEMP The prediction eguation is given above as VISCOSIT 128151 7 000875782TEMP b Analysis From the full analysis above we see R2 09602 which is larger than 09 This model ts the data very well The Ftest value of 144578 results in a p value of 00000 indicating a high degree of overall model adequacy The pvalue on both individual coef cients is very small 00000 indicating that both are very likely to be nonzero We can conclude that the temperature does impact the viscosity of toluenetetralin blends 5 Text p 389 exercise 66 parts a b 6 Text p 406 exercise 618 The StatGraphics analysis is Multiple Regression Analysis Dependent variable SUREACEi Standard T Parameter Estimate Error Statistic Pi Value CONSTANT 124879 157121 794794 00000 COBALT 113369 489859 231432 00334 TEMP 01461 00315213 463498 00002 Analysis of Variance Source Sum of Squares Df Mean Square EiRatio Pe Value Model 119469 2 597343 1342 00003 Residual 75672 17 445129 Total Corr 195141 19 Risquared 612218 percent Risquared adjusted for df 566597 percent Standard Error of Est 210981 Mean absolute error 152035 DurbinelJatson statistic 140497 The StatAdvisor The output shows the results of fitting a multiple linear regression model to describe the relationship between SUREACEi and 2 independent variables The equation of the fitted model is SUREACEi 124879 113369COBALT 01461TEMP Since the Pivalue in the ANOVA table is less than 001 there is a statistically significant relationship between the variables at the 99 confidence level h ReSquared statistic indicates that the model as fitted explains 612218 of the variability in SU CEi T e adjusted Risquared statistic which is more suitable for comparing models with different numbers of independent variables is 566597 In determining whether the model can be simplified notice that the highest Pevalue on the independent variables is 00334 belonging to COBALT Since the Pivalue is less than 005 that term is statistically significant at the 95 confidence level Consequently you probably don39t want to remove any variables from the model Analysis From the StatGraphics results we see that R2 06122 and adjusted R2 05666 are both below 09 This model does not t the data well The Ftest value of 1342 results in a pvalue of 00003indicating a high degree of overall model adequacy The pvalues on the ttests of the individual coef cients are all below 005 indicating that both regressors are signi cant and neither should be dropped from the model We can conclude that the mole contents of cobalt and calcinations temperature both have negative effects on the surface area of an ironcobalt hydroxide catalyst 7 Text p 407 exercise 619 The analysis from StatGraphics is Multiple Regression Analysis Dependent variable ADSORPTI Standard T Parameter Estimate Error Statistic Pi Value CONSTANT 182509 207475 0879664 04019 PH 142082 2663 0533542 06066 AL 0398339 0118439 336324 00083 IRON 0112162 00308319 363785 00054 Analysis of Variance Source Sum of Squares Df Mean Square EeRatio PValue Model 353578 3 117859 5706 00000 Residual 185909 9 206566 Total Corr 372169 12 Risquared 950047 percent Risquared adjusted for df 933396 percent Standard Error of Est 454495 Mean absolute error 29911 DurbinelJatson statistic 26522 The StatAdvisor The output shows the results of fitting a multiple linear regression model to describe the relationship between ADSORPTI and 3 independent variables The equation of the fitted model is ADSORPTI l82509 142082PH 0398339AL 0112162IRON Since the Pivalue in the ANOVA table is less than 001 there is a statistically significant relationship between the variables at the 99 confidence level Th RSquared statistic indicates that the model as fitted explains 950047 of the variability in ADSORPTI The adjusted Risquared statistic which is more suitable for comparing models with different numbers of independent variables is 933396 In determining whether the model can be simplified notice that the highest Pevalue on the independent variables is 06066 belonging to P Since the Pevalue is greater or equal to 010 that term is not statistically significant at the 90 or higher confidence level Consequently you should consider removing PH from the model Analysis From the StatGraphics ouput we see that R2 09500 and the adjusted R2 09334 are both larger than 09 This model ts the data very well The Ftest value of 5706 results in a pvalue of 00000 indicating a high degree of overall model adequacy The ttests on the individual coef cients indicate that AL and IRON are signi cant regressors but that PH is not As indicated by StatGraphics we should consider removing PH from the model and regressing the response on only AL and IRON which have signi cant positive effect on the soils adsorption of phosphate

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