STAT METH RESEARCH 1
STAT METH RESEARCH 1 STA 6166
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This 10 page Class Notes was uploaded by Golden Bernhard on Friday September 18, 2015. The Class Notes belongs to STA 6166 at University of Florida taught by Lawrence Winner in Fall. Since its upload, it has received 19 views. For similar materials see /class/206563/sta-6166-university-of-florida in Statistics at University of Florida.
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Date Created: 09/18/15
Chapter 1 Statistics and the Scienti c Method ET 9 n L 957 EL 57 n L 957 L 579 0 CL The population of interest is the weight of shrimp maintained on the speci c diet for a period of 6 months The sample is the 100 shrimp selected from the pond and maintained on the speci c diet for a period of 6 mont s The weight gain of the shrimp over 6 months Since the sample is only a small proportion of the whole population it is necessary to evaluate what the mean weight may be for any other randomly selected 100 shrimpsl The amount of radioactivity at all points in the suspect area The 200 randomly selected points in the suspect area The level of radioactivity in the suspect area We want to relate the level of radioactivity of the 200 points in the sample to the level in whole suspect area Thus we need to know how accurate a protrayal of the population is provided by the 200 points in the sample All households in the city that receive welfare support The 400 households selected from the city welfare rolls The number of children per household for those households in the city which receive welfare In order to evaluate how closely the sample of 400 households matches the number of children in all households in the city receiving welfarel All football helmets produced by the ve companies over a given period of time The 540 helmets selected from the output of the ve companies The amount of shock transmitted to the neck when the helmets face mask is twistedl The neck strength of players is extremely variable for high school playersl Hence the amount of damage to the neck varies considerably from player to player for exactly the same amount of shock transmitted by the helmet The 35000 students enrolled in the university The 500 students selected by the faculty First a list of all 35000 students would be constructed Using a computer program generate 500 random numbers between 1 and 35000 The students corresponding to these 500 numbers on the list will be the students to whom questionaires will be sentl The questions should determine what impact this change may have on the students7 professional development if the change would encourage students to work harder when they knew their grade was on the margin between two grades and similar questions concerning whether or not the students felt the change would have a true impact on students learning experience Chapter 8 The Completely Randomized Design n L 39D no m EL Yes the mean for Device A is considerably relative to the standard deviations smaller than the mean for Device H0 MA MB Mo MD versus Ha Difference in Ms Reject H0 if F 2 F05320 310 SSW 517672 l20912 1532 2492 08026 y 00826 e SSB 670l1605 7 08262 009477 08262 012277 08262 02735 7 0826 05838 a 5838 3 F 302520 485 gt 310 e Reject H0 and conclude there is signi cant difference among the mean difference in pH readings for the four devices pvalue PF320 2 485 pvalue 00107 The data must be independently selected random samples from normal populations having the same value for a Suppose the devices are more accurate at higher levels of pH in the soil and if by chance all soil samples with high levels of pH are assigned to a particular device then that device may be evaluated as more accurate based just on the chance selection of soil samples and not on a true comparison with the other devices Yes because the box for the LowTar Brand is completely below all the other boxes Yes because the Ftest has an extremely low pvalue less than 0001 pvalue lt 0001 We would be stating that the brands have different average tar content when in fact they have the same level of tar contentl Thus the manufacturer s claim about pro ducing a brand having a lower average tar content would be false Consumers may change brands and not be receiving the bene t of a lower tar content The box plot and normal probability plot are given here Normal Probablllty Plot Residuals from Fitted Model Residual u uizmxn2n2nxmizumwm 2 4 a 1 2 Ouanmes 07 standard Normal Based on the box plots and normal probability plots it appears that the distributions of several of the varities have a nonnormal distribution The Levine7s test yields L 2381 With pvalue 074 Thus there is not a signi cant difference in the variety variances b The AOV table is given here Source df SS MS F pvalue Variety 4 1096i 2742 373 0014 Error 30 22054 735 Total 34 33021 Reject H0 if F 2 269 Since F 373 gt 269 reject H0 and conclude there is a signi cant difference in the mean yield of ve varietiesl c The Kruskal Wallis yields H 1001 With pvalue 0040 Thus reject H0 and conclude there is a signi cant difference in the distributions for the yields of the ve varietiesl d The only difference in the conclusions is that the pvalue for the Ftest is somewhat smaller than the pvalue for the Kruskal Wallis test 84 The Ftest is testing for a difference in the mean yields of the ve varieties Whereas the KruskalWallis test is testing Whether there is a difference in the distribution of yields for the ve varietiesl 85 a The KruskalWallis yields H 2132 gt 921 With df 2 pvalue lt 0001 Thus reject H0 and conclude there is a signi cant difference in the distributions of deviations for the three suppliers b The box plots and normal probability plot are given here 62 Normal Probablllty Plot Residuals from Fitted Model Residual u ix nu zmxn2n2nxmizh 2 4 a 1 2 Ouanmes 07 standard Normal The box plots and normal probability plots of the residuals indicate that the normality condition may be violated The Levine test yields L 3189 with 01025 lt pvalue lt 0105 l Thus7 there is signi cant evidence that the equal variance condition is also violated c The AOV table is given here Source df SS MS F pvalue Supplier 2 1072318 5361 i9 161 09 01000 Error 24 79819 3313 Total 26 1152217 Reject H0 if F 2 3140 F 161109 gt 31407 reject H0 and conclude there is a signi cant difference in the mean deviations of the three suppliersl 95 C111 on MA 18923 i 210645177 185126193120 95 C111 on MB 15628 i 210645177 152131160125 95 C111 on 0 20394 i 210645177 199197207191 Since the upper bound on the mean for supplier B is more than 20 units less than the lower bound on the mean for suppliers A and B7 there appears to be a practical difference in the three suppliersl However7 because the normality and equal variance assumptions may not be valid7 the Cilils may not be accurate L 86 For Tukey7s w 405612 3177 a W 3177 005815 11522 a Pairs Not Signi cantly Different V27 V3 817 The box plots indicate the distribution of the residuals is slightly right skewed This is con rmed with an examination of the normal probability plot The Hartley test yields 63 88 810 811 Fm 235 lt 711 using an a 005 test Thus the conditions needed to run the ANOVA Ftest appear to be satis ed From the output F 1568 with pvalue lt 00001 lt 005 Thus we reject H0 and conclude there is signi cant evidence of a difference in the average weight loss obtained using the ve different agents a Fisher7s LSD Pairs Not Signi cantly Different 41 23 b Tukey s W Pairs Not Signi cantly Different 41 42 12 23 3S a Based on the box plots and the normal probability plot the condition of normality of the population distributions appears to be satis ed The Hartley test yields Fm w 159 lt 145 value from Table 12 with a 01 There is not signi cant evidence of a difference in the 4 population variances b From the ANOVA table we have pvalue lt 0001 Thus there is signi cant evidence that the mean ratings differ for the four groups c 95 Cl on 133125 i 2048 Wig 7590 95 C1 on 1126437 i2048 wig 5771 95 Cl on WI 40000 i 2048 33 47 95 Cl on W 25000 i 2048V05g753 1832 d The C17s are the same as those given in the output For Tukey7s w q05428 387 a W 58 8 1336 a All pairs are signi cantly different a The Kruskal Wallis test yields H 2662 with df 3 pvalue lt 00001 Thus there is signi cant evidence that the distribution of ratings differ for the four groups b The two procedures yield equivalent conclusions a All middlegrade students from a wide variety of academic abilities and backgrounds b Sixthgrade students with academic ability and background similar to the students in the study c One one class per treatment group d Only sixthgrade students were chosen all students had similar ability and back ground only one sample for each treatment only one teacher e Randomly assign 30 classes to each of the three groups Each group7s classes should vary in grade level The student s in the classes should be randomly selected from a population of classes which vary in ability and background Randomly select 30 teachers with each teacher teaching one class from each group There a number of other factors that should be considered 402003 F 881935 5470 with df 336 pvalue lt 00001 lt 005 There is signi cant evidence of a difference in the average leaf size under the four growing conditions 64 b1 95 CL on 1111231571 2028 202036154 95 CL on 11198158 i 210287V881r139 j36 514111175 95 CL on 110 141931210287Vg l39g36 1117618110 95 CL on 11D 35135 i 210283 32118 38152 The C111 for the mean leaf size for Condition D implies that the mean is much larger for Condition D than for the other three conditions c F 13323 2110 with df 336 5 pvalue 011174 5 There is not signi cant evidence of a difference in the average nicotine content under the four growing conditions d From the given data it is not possible to conclude that the four growing conditions produce different average nicotine contenti e No If the testimony was supported by this experiment then the test conducted in part c would have had the opposite conclusioni 8114 The plots indicate that for both leaf size and nicotine content the normality and equal variance conditions have not been violated 815 ai Because the data appears to be nonnormal in distribution the Levene test will be use 1 L 0184 with df 217 pvalue 014489 There is not signi cant evidence of a difference in population variancesi b No because the distributions appear to be not normally distributed c The data was transformed using the logarithmic transformation The data is still somewhat skewed right for Machine C but the ANOVA will be con ducted anywaysi 9848 2 F 11954 7100 With df 217 5 pvalue 010061 There is signi cant evidence of a difference in the mean diameters of the three ma chinesi d The analysis using the original data yield pvalue 01094 which would not support a difference in the mean diametersi e1 She could have selected an equal number of observations from each machine Equal sample sizes has a moderating in uence on the unequal variances effects on the Ftest1 8116 The KruskalWallis test yields identical results for the transformed and original data be cause the transformation was strictly increasing which maintains the order of the data after the transformation has been performed H 9189 with df 2 pvalue 0100711 Thus our conclusion is the same as was reached using the transformed data 817 a The surnrnarv statistics for the four groups are given here Descrrptrve Stotrstrcs ActlveEx PasslvEx TestOnly Control Varlable N Mean Medlan TrMean StDev SE Mean ActlveEx 6 10125 96 10125 144 059 PasslvEx 6 11375 10750 11375 1896 0774 TestOnly 6 11 708 11 750 11708 1 520 0 621 Control 5 12 350 12 000 12 350 0 962 0 430 h Box p1ots are given here Age months at Whlch Chlld Flrst Walked Age months First Walked L1 1 2 a 4 Group 1AcuveEx 2PassivEx 3TestOnIy 4Control c The contro1 group has the 1argest rnean age The four groups have sirni1ar 1eve1s of varihihtv in ages with the contro1 groups standard deviation somewhat smaller than the other three groups The box p1ots are not very informative because of the very srna11sarnp1e sizes in each of the four groups FL The Frtest from the Minitah AOV has a prvalue 010129 which indicates that there is not sigiihcant evidence in the data that the four groups have diderent rnean ages e The Minitab output for the LSD and Tukey procedure are given here Tukey s pairwise comparisons Family error rate 00500 Individual error rate 00111 Critical value 398 Intervals for column level mean row level mean Act iveEx Control PassivEx Control 4 809 0 359 PassivEx 3714 al609 1 214 3 559 TestUnly 4047 1942 2797 0881 3226 2131 Fisher s pairwise comparisons Family error rate 0191 Individual error rate 00500 Critical value 2093 Intervals for column level mean row level mean ActiveEx Control PassivEx Control 4147 0 303 PassivEx 3082 0947 0 582 2 897 TestUnly 3416 1280 2166 0249 2564 1499 In Tukey s procedures the 6 pairwise comparisons intervals all contain 0 Therefore Tukey s procedure agrees with the conclusion from the Ftest that there is not signif icant evidence of a difference in the four group means Fisher7s pairwise comparisons procedure has the interval for comparing the mean of the Control with the mean of ActiveEX group not containing 0 but with the other 5 pairwise comparisons intervals all containing 0 This would seem to contradict the Ftest and Tukey conclusions However if we use the Protected Fisher LSD procedure then no further comparisons would have been made once the conclusion of the Ftest indicated no signi cant evi dence of a difference Thus all three procedures F test Tukey and Protected Fisher LSD have reached the same conclusion No signi cant evidence at the 05 level of a difference in the means of the four groups 67 f A plot of the average number of walking re exes observed is given here for the three groups Average Number of Walking Reflexes During Weekly 1 Minute Tests Average Number MWalkmg Re exes 5 Week The average number of walking re exes for the active exercise group increases dra matically over the 8 weeks Whereas the values for the Passive and Test Only groups remains relatively constant 8 18 8 57 n E CD Box plots are given here The box plots are all relatively symmetric with about the same level of spread There appears to be some differences in their mediansi This is con rmed by an examination of the summary statistics Median7 StDev7 Q1 and Q3 There is no strong indication from the box plots that the assumptions of equal variance and normality appear to be violate i i The assumption of normality of the residuals does not appear to be violated based on the normal probability plot The data values are mostly close to the straight line with the exception of the largest values for the residuals The Levenels test yields F 04167 df 4 65 pvalue 0797i Thus7 there is not signi cant evidence in the data that the treatment variances are different i The mean rating from the applicant using crutches was signi cantly higher than the mean rating for the applicant who was hard of hearing No other pairs of rating means were found to be signi cantly different Thus there is very little evidence that the various types of physical handicaps produce different mean empathy ratingsi
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