Suggested Answers To Exercises. estimates variability in a sampling distribution of random means. About 68% of means fall

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Clementine Carroll
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1 Beyond Significance Teting ( nd Edition), Rex B. Kline Suggeted Anwer To Exercie Chapter. The tatitic meaure variability among core at the cae level. In a normal ditribution, about 68% of the core fall within the interval M ±, and about 95% of the core fall within the interval M ±. The tandard error M i alo a tandard deviation but it etimate variability in a ampling ditribution of random mean. About 68% of mean fall within the interval µ ± M, and about 95% of mean fall within the interval µ ± M. It value i determined by and N. In general, ample ize affect M, but not. (Thi aume that N i not very mall, uch a N =, in which cae i undefined.). Given = and M = 6.00, we can ay that N = 00 and df = 99. The tatitic = ay that the quare root of the mean quared ditance of core from M averaged over the degree of freedom (99) equal The tatitic M = 6.00 ay that the etimated tandard deviation in a ampling ditribution of random mean each baed on 00 cae i The value of both tatitic in a different ample are almot certainly not the ame. 3. Given M = 00.00, = 9.00, N = 5, M =.800, and t -tail,.05 (4) =.064, we calculated the 95% CI for µ a [96.8, 03.7]. For α =.0, t -tail,.0 (4) =.797, the margin of error i.800 (.797) = 5.035, o the 99% CI for µ i ± 5.035, or [94.97, 05.03] which i wider than the 95% interval. 4. The 95% CI for µ D baed on a dependent ample analyi of the data in Table. i [.3, 4.3]. For an independent-ample analyi where n = n = 5: pool = 6.5, M M = [6.5 (/5 + /5) ] / =.58, t -tail,.05 (8) =.306

2 95% CI for µ µ i.00 ± 3.646, or [.65, 5.65] The 95% CI for µ µ i wider becaue the cro-condition correlation r i aumed to be zero in the independent ample analyi. 5. The core for group in Table.4 lited from lowet to highet are The trimmed core for p tr =.0 core are and the mean of the middle 6 core i M tr = The 0% Winorized core are and the Winorized mean and variance of thee 0 core are, repectively, M Win = 6.80 and Win = The Winorized mean M Win mut be calculated in order to derive the Winorized variance Chapter 3 Win, which i part of the equation for the tandard error of M tr (Equation.9).. For n = 5 in Table 3.: = [6.5 (/5 + /5)] / =.93 M M t (8) =.00/.93 =.9, p =.037 t -tail,.05 (8) = % CI for µ µ i.00 ±.93 (.048), or [.3, 3.87]. For n = 30 in Table 3.3, M T =.00 and SS A = 30 ( ) + 30 (.00.00) + 30 (.00.00) = df A = 3 =, MS A = 60.00/ = SS W = 9 (7.50) + 9 (5.00) + 9 (4.00) =

3 3 df W = 90 3 = 87, MS W = /87 = 5.50 F (, 87) = 30.00/5.50 = 5.45, p = For a dependent ample analyi of the data in Table 3.4, M T = 3.00 and SS A = 5 ( ) + 5 ( ) + 5 ( ) = df A = 3 =, MS A = 40.00/ = 0.00 SS W = 4 (7.50) + 4 (5.00) + 4 (4.00) = df W = 5 3 =, MS W = 66.00/ = 5.50 cov = 7.50 / (5.00 / ).7348 = cov 3 = 7.50 / (4.00 / ).7303 = cov 3 = 5.00 / (4.00 / ).8385 = M cov = ( )/3 = MS A S = =.467 df A S = (3 )(5 ) = 8, o SS A S = =.3336, or.33 SS S = = df S = 5 = 4, MS S = 3.67 F (, 8) = 0.00/.467 = 4., p = Becaue there i jut one et of difference core in a dependent ample t tet, there i no aumption of equal population variance of thoe difference core. That i, correlated deign with jut two condition etimate 5. For n = 80 in Table 3.6, all expected frequencie are computed a f e = (80 80)/60 = 40 and the chi-quare tatitic i calculated a follow: σ D and no other parameter. χ () = (48 40) /40 + (3 40) /40 + (3 40) /40 + (48 40) /40 D in a

4 4 = 6.40, p =.0 Chapter 4. Sample repreentativene concern how the cae were elected and ha nothing to do with reult of tatitical tet in a particular ample. Repreentativene alo ha nothing to do with ample ize. That i, a ample can be large but not repreentative. The interpretation of p value from tatitical tet, however, aume random ampling, which i rare in mot tudie. Mot ample are convenience ample, and it i not alway clear whether a particular convenience ample i repreentative.. If reliability here mean repeatability (i.e., replication), then p value generally give imprecie information about what i likely to be found in replication ample (replicability fallacy). But if reliability implie preciion then, ye, lower p value indicate that a reult tand out relatively more againt ampling error. Jut how much that reult tand out (i.e., effect ize), however, cannot be determined by inpecting p value. 3. If only it were true that ignificance teting in one ample told u what we are likely to find in replication, then we ave ourelve ome work and not bother with replication. But thi i jut untrue (replicability fallacy). Outcome of tatitical tet do not indicate the probability that a particular reult occurred by chance (odd-againt-chance fallacy). Statitical ignificance doe not automatically mean that ome reult i meaningful. 4. Significance teting doe not indicate the probability that any hypothei i correct, given the data. Thi particular paage eem to decribe the mirror image of the validity fallacy. That i, if p i incorrectly taken a the probability that the reearch hypothei (H ) i true, then a conitent miinterpretation would be that p i the probability that H i fale. 5. Statitical ignificance doe not indicate that a reult i real or that H 0 i definitely fale.

5 5 Chapter 5 Thi i the filter myth. The lat entence i reaonable: whether the oberved effect ize i large enough to meet a tandard of practical or theoretical importance may be the mot important quetion of all, auming that the effect replicate.. Given M = 3.00, = 7.50, M =.00, = 5.00, 30, the d tatitic in Table 5.3 are calculated a follow: d pool =.00/6.5 / = 3.0 (/30 + /30) / =.80 pool = 6.5, and t (58) = 3.0 for n = c (df W ) d pool = 3.80 =.79 4(58). Given M = 3.00, d =.00/7.50 / =.73 and d =.00/5.00 / =.89 SS A = 30 ( ) + 30 (.00.00) = SS W = 9 (7.50) + 9 (5.00) = SS T = = 4.50 T = 4.50/59 = 7.6 d total =.00/7.6 / =.75 = 7.50, M =.00, = 5.00, and r =.735 for a dependent-ample analyi of the data in Table., we calculate: pool = ( )/ = 6.5 d pool =.00/6.5 / =.80 D = (.735)(7.50 / )(5.00 / ) = For the data in Table.4 where M tr = 3.00, = n = 0, and d Win p =.6, we calculate d diff =.00/3.50 / =.07 = 8.489, M W in tr = 7.00, = 9.067, n W in

6 6 Win p = [9 (8.489) + 9 (9.067)]/8 = d Win = 6.00/8.489 / =.40 and d Win = 6.00/9.067 / =.99 Depending on the tandardizer, etimate of δ rob range from.40 to Given M M =.00, 3.00, SS A = 60.00, SS T = 4.50, and t (58) = 3.0 for n = 30, we calculate r pb in Table 5.3 a S T = 4.50/60 = 7.04 r pb = (.00/7.04 / ) (.50.50) / = 3.0/( ) / =.38 r pb = (60.00/4.50) / = For a dependent ample analyi of the data in Table.: SS A = 5 ( ) + 5 (.00.00) = 0.00 SS W = 4 (7.50) + 4 (5.00) = SS T = = df W = 8, MS W = 50.00/8 = 6.5 cov =.735 (.739) (.36) = 4.50 MS A S = =.75 df A S = 4, o SS A S =.75 4 = 7.00 SS S = = = 0.00/60.00 =.77 and partial = 0/( ) = Given d pool =.80, n = 30, and r =.75, and uing Equation 5., we calculate d pool.80 (.75) = + =.664 (9) 30 Approximate 95% CI for δ i.80 ±.664 (.96), or [.47,.3]

7 7 which i narrower than the correponding confidence interval when treating the mean a independent, or [.7,.33]. 7. Given d pool =.46, n = 5, and n = 56 for the data in Table 5.6, and uing Equation 5.0, we calculate d pool = + =.093 (665) 5(56) Approximate 95% CI for δ i.46 ±.093 (.96), or [.8,.64] The noncentral 95% confidence interval for δ reported in Table 5.6 for thee data i [.8, 65], which i imilar to the approximate interval jut calculated. Chapter 6. For the original table, χ () = 8.00 and ˆϕ =.0. But let u multiply the frequencie in the firt row by, ay, 4, which quadruple the ize of the control group only: Relaped Not relaped Control Treatment For thi modified table, χ () =.99, o ˆϕ =.6. For both the original and modified table, however, p C =.60, p T =.40, RD =.0, RR =.50, and OR =.5 (i.e., thee tatitic are not margin bound).. Given RR =.50 and n C = n T = 00, we calculate the approximate 95% confidence interval uing the equation for ln (RR) in Table 6.3 a follow: ln (.50) =.4055 ln (.50) = + = (.60) 00 (.40)

8 ±.47 (.96), or [.70,.6940] in natural log unit ln (.70) = e.70 =.4 and ln (.6940) = e.6940 =.00 The approximate 95% confidence interval for π C /π T i [.4,.00]. 3. I ued an online contingency table chi-quare calculator to derive the reult χ (3) = 9.83 for the data in Table 5.7; you can alo ue Equation 3.6 to compute the ame tatitic. Becaue the mallet dimenion of the table i and N = 667 (ee Equation 6.5), Cramer V i calculated a V = (9.83/667) / =. 4. Given bae rate =.375, enitivity =.80, and pecificity =.70, we can arbitrarily aume a population ize of 0,000 cae. (Selection of any other number will have no effect on calculation of predictive value.) You hould verify that all the value jut tated imply the fourfold preented next: Clinical Normal Screening tet + 3,000,875 4,875 Screening tet 750 4,375 5,5 3,750 6,50 0,000 Of all 4,875 poitive creening tet reult, a total of 3,000 are correct, o poitive predictive value = 3,000/4,875 =.65, or about 6%. Of all 5,5 negative creening tet reult, a total of 4,375 are correct, o pecificity = 4,375/5,5 =.854, or about 85%. 5. Given enitivity =.80, pecificity =.70, LPR =.667, NLR =.86, and bae rate =.50, we calculate: Pretet odd =.50/(.50) =.00 Pottet odd after a poitive reult = =.667

9 9 Chapter 7 Pottet probability after a poitive reult =.667/( +.667) =.77 Pottet odd after a negative reult = =.86 Pottet probability after a negative reult =.86/( +.86) =. With a family hitory of the diorder, the probability of having that diorder increae from.50 to.77 after a poitive creening tet reult, but the probability of having the diorder decreae from.50 to. following a negative tet reult.. For the analyi in Table 7., ˆψ i defined by (½,, ½) and equal 3.00, MS W = 5.50, d with = 3.00/(5.50) / =.79, t -tail,.05 () =.79, and a per Equation 7.8: ( ) ˆψ = = % CI for ψ i 3.00 ±.845 (.79), or [.00, ] Dividing the endpoint of the exact 95% CI for ψ by 5.50 / =.345 give u the endpoint of the approximate 95% CI for δ ψ, which i [.0857,.477]. At two-decimal accuracy, the reult jut calculated match thoe reported in Table 7. for ˆψ.. For the analyi in Table 7.3, ˆψ = 3.00 and t -tail,.05 (4) =.776. To find the variance of the contrat difference core, we mut firt apply the weight (½,, ½) to the raw core in Table 3.4: Cae Y Y Y 3 (Y + Y 3 )/ Y A B 4.00 C D E

10 0 The variance of the contrat difference core i.650, o uing Equation 7.9: ˆψ =.650 / 5 = % CI for ψ i 3.00 ±.570 (.776), or [.474, 4.586] Dividing the endpoint of the exact 95% CI for ψ by 5.50 / =.345 give u the endpoint of the approximate 95% CI for δ ψ, which i [.6044,.9540]. At two-decimal accuracy, the reult jut calculated match thoe reported in Table 7.3 for ˆψ. 3. For ˆψ in Table 7., SS ˆψ = 30.00, F ˆψ = 30.00/5.50 = , t ˆψ =.3355, SS non-ψ ˆ = 0.00, F = 0.00/5.50 =.88, df ˆ =, and df W =. Uing Equation : non-ψ ˆ non-ψ r ˆψ =.3355 = partial r ˆψ =.3355 = Thu, the correlation between ˆψ and the outcome variable i.53. After removing the effect of ˆψ from total variance, the correlation between ˆψ and the outcome variable i.559. The quare of the value jut lited equal the repective value of for thi contrat reported in Table 7.. ψ and partial 4. For the dependent ample analyi in Table 7.3, SS ˆψ = 0.00, SS re (ψ) ˆ = 7.00, SS ˆψ = 30.00, and SS ˆ re (ψ) = Next we calculate: partial partial ψ = 0.00/( ) =.588 ψ = 30.00/( ) =.874 Thu, ˆψ explain about 58.8% of the reidual variance controlling for ˆψ and the ubject ψ

11 effect, and ˆψ explain about 87.4% of the reidual variance controlling for ˆψ and the ubject effect. 5. For the reult in Table 7., MS A = 0.00, MS error = MS W = 5.50, and uing Equation 7.: ˆσ A = /5 ( ) =.90 ˆσ error = 5.50, ˆσ total = = 8.40 ˆρ I =.90/8.40 =.345 Uing the direct method in Table 7.4 given SS T = 06.00, we calculate ˆρ I 3 ( ) = = The omnibu explain thu about 34.5% of the variance auming random election of the three level of factor A. 6. For the analyi in Table 7.3, MS A = 0.00, MS error = MS re (A) =.33/8 =.46, SS T = 06.00, and MS S = 54.67/4 = Uing the direct method in Table 7.4: ( ) ˆω A = = Given MS ˆψ = 0.00, MS re (ψ) ˆ = 7.00/4 =.750, MS ˆψ = 30.00, and MS re (ψ) ˆ = 4.33/4 =.083, next we ue Equation 7.: ˆσ ψ = /5 ( ) =.550 ˆσ ψ = /5 ( ) = partial ω ˆ ˆψ = = partial ω ˆ ˆψ = =

12 To ummarize, the omnibu effect explain about 3.% of the total variance. The firt contrat explain about 3.9% of the reidual variance controlling for the econd contrat and the ubject effect, and the econd contrat explain about 64.0% of the variance controlling for the firt contrat and the ubject effect. All of thee reult are adjuted for capitalization on chance and aume an additive model. In contrat, the oberved proportion of explained variance for ame effect are partial ψ =.874 (ee Table 7.3). A =.377, partial ψ =.588, and 7. For both tak of attention in Table 7.7, df A = for the omnibu effect. For the tak of Chapter 8 imple attention, F A (, 8) =.07, o we calculate uing Equation A = = o group memberhip (ectay uer, cannabi uer, nonuer) explain about 4.% of the total oberved variance in thi tak. For the tak of elective attention, F A (, 8) = 7.3, o 7.3 A = = which indicate a larger effect ize, pecifically, group memberhip account for about 38.9% of the oberved total variance in the election attention tak.. Aume n = 0, MS W = 5.00, D = drug, and G = gender. For mean in the left ide of Table 8.3, we calculate uing the equation in Table 8.: SS D = 0 ( ) + 0 ( ) =,50.00 SS G = 0 ( ) + 0 ( ) = 9,000.00

13 3 SS DG = 0 ( ) + 0 ( ) + 0 ( ) + 0 ( ) =, SS W = = 4, SS T = 6, SS D at women = 0 ( ) + 0 ( ) = 3,5.00 SS D at men = 0 ( ) + 0 ( ) = 5.00 SS D at G = 3, = SS D + SS DG =, , = 3, The interaction contrat for thi exercie are lited next: B B B 3 (I) B B B 3 (II) A 0 A ½ ½ A 0 A ½ ½ The um of the product of the weight from the correponding cell mean i (½) + 0 ( ) + ( ) (½) + ( ) ( ½) + 0 () + ( ½) = 0 Thu, thee two contrat are orthogonal in a balanced deign. 3. From Table 8.5, SS AB = and df AB =. For each contrat pecified in (I) and (II) and uing Equation 8.8, we calculate from the mean in Table 8.4: ˆψ AB ( I ) = = 8.00 SS ˆψ AB ( I ) = 3 (8.00) ( + )( ) = ˆψ AB ( II ) =.5 (9.00) (9.00).5 (5.00) (3.00) = 6.00 SS ˆψ AB ( I I ) = 3 ( 6.00) ( )( ) = 36.00

14 4 Baed on thee reult: SS AB = SS ˆψ AB ( I ) + SS ˆψ AB ( II ) = = A et of weight that pecify a three-way interaction contrat i a follow: C C B B B B A A Applying thee weight to the cell mean for thi problem give u ˆψ ABC = = 0 which ay that there i no three-way interaction. The difference between all pair of imple two-way interaction are alo zero. For example, ˆψ AB at C = ( ) ( ) =.00 ˆψ AB at C = ( ) ( ) =.00 ˆψ AB at C ˆψ AB at C =.00 (.00) = 0 Becaue there are no difference between any pair of imple interaction, there i no threeway interaction. Thu, Equation 8.9 hold for thee data. 5. For the mean in Table 8.4, where n = 3, we calculate uing Equation 8.5: SS W, B, AB = [ (7.00) + 3 ( ) ] [ (4.00) + 3 ( ) ] [ (7.00) + 3 ( ) ] [ (3.00) + 3 ( ) ] [ (4.00) + 3 ( ) ] [ (7.00) + 3 ( ) ] = df W, B, AB = + + = 6, MS W, B, AB = 96.00/6 =.5

15 5 6. Given MS A = 48.00, MS B = 40.00, MS AB = 6.00, MS W = 4.00, a = 3, b = 6, and n = 5, and uing the equation in Table 8.8, we calculate proportion of explained variance for AB a: ˆσ = A 6 (5) ( ) =.467 and 5 ˆσ B = ( ) =.33 3 (6)(5) ˆσ = AB 5 ( ) =.400 and ˆσ error = ˆσ total = = ˆω = AB =.057 and partial ˆω = AB =

Comparing Means: t-tests for Two Independent Samples

Comparing Means: t-tests for Two Independent Samples Comparing ean: t-tet for Two Independent Sample Independent-eaure Deign t-tet for Two Independent Sample Allow reearcher to evaluate the mean difference between two population uing data from two eparate