One Sample T-Test With Howell Data, IQ of Students in Vermont
data howell; infile 'C:\Users\Vati\Documents\StatData\howell.dat';
input addsc sex repeat iq engl engg gpa socprob dropout;
IQ_diff = iq - 100;
run;
*Want to test null hypothesis that mean IQ is 100;
proc means mean stddev n skewness kurtosis t prt CLM; var iq IQ_diff; run;
The MEANS Procedure
Variable
Mean
Std Dev
N
Skewness
Kurtosis
t Value
Pr > |t|
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
iq
100.2613636
12.9849553
88
0.3943203
-0.1634709
72.43
<.0001
IQ_diff
0.2613636
12.9849553
88
0.3943203
-0.1634709
0.19
0.8507
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Lower 95%
Upper 95%
Variable
CL for Mean
CL for Mean
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
iq
97.5101145
103.0126128
IQ_diff
-2.4898855
3.0126128
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
--------------------------------------------------------------------------------------------------
Data CI1;
t = 0.19 ;
n = 88 ;
df = n-1;
d = t/sqrt(n);
ncp_lower = TNONCT(t,df,.975);
ncp_upper = TNONCT(t,df,.025);
d_lower = ncp_lower/sqrt(n);
d_upper = ncp_upper/sqrt(n);
output; run; proc print; var d d_lower d_upper; run;
One Sample T-Test With Howell Data, IQ of Students in Vermont
Obs
d
d_lower
d_upper
1
0.020254
-0.18876
0.22915
2
Vermont students’ mean IQ (M = 100.26, SD = 12.98) did not differ significantly
from 100, t(87) = 0.19, p = .85, d = .02. A 95% CI for the difference in means runs
from 97.51 in 103.01 in IQ units and from -.19 to .23 in standard deviation units.
--------------------------------------------------------------------------------------------------
Experiment 2 of Karl's Dissertation
Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups
data Mus; infile 'C:\Users\Vati\Documents\StatData\tunnel2.dat';
input nurs $ 1-2 L1 3-5 L2 6-8 t1 9-11 t2 12-14 v_mus 15-16 v_rat 17-18;
*t_mus=sqrt(1.575 * t1 + .5); *t_rat=sqrt(1.575 * t2 + .5);
*L_mus=LOG10(1.575 * L1 + 1); *L_rat=LOG10(1.575 * L2 + 1);
v_diff=v_mus - v_rat; *t_diff=t_mus - t_rat; *L_diff=L_mus - L_rat;
proc sort; by nurs;
proc means mean stddev n skewness kurtosis t prt;
var V_mus V_rat V_diff; by nurs;
run;
-------------------------------------------- nurs=MM --------------------------------------------The MEANS Procedure
Variable
Mean
Std Dev
N
Skewness
Kurtosis
t Value
Pr > |t|
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
v_mus
22.4375000
12.8164933
16
-0.3163856
-0.8764539
7.00
<.0001
v_rat
7.5625000
5.8874867
16
0.1115399
-1.5610184
5.14
0.0001
v_diff
14.8750000
10.5506714
16
0.0587866
-1.4495685
5.64
<.0001
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
-------------------------------------------- nurs=MR --------------------------------------------Variable
Mean
Std Dev
N
Skewness
Kurtosis
t Value
Pr > |t|
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
v_mus
22.3125000
10.2092687
16
-0.4074942
-0.0747930
8.74
<.0001
v_rat
7.5625000
5.7383941
16
0.9244088
1.1906350
5.27
<.0001
v_diff
14.7500000
7.5674743
16
-0.0213611
-1.1933542
7.80
<.0001
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
-------------------------------------------- nurs=RR --------------------------------------------Variable
Mean
Std Dev
N
Skewness
Kurtosis
t Value
Pr > |t|
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
v_mus
23.4375000
8.5085741
16
0.3691328
-0.1333649
11.02
<.0001
v_rat
24.7500000
8.0952661
16
0.6327068
0.8319591
12.23
<.0001
v_diff
-1.3125000
8.4041161
16
0.1199781
0.3553554
-0.62
0.5416
ƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒƒ
Notice that house mice reared with house mice or with deer mice avoided the rat-scented
tunnels, but those reared with rats did not. Avoiding the scent of rat makes a lot of sense when you
consider that rats eat mice.
proc corr nosimple; var V_Mus; with V_rat; by nurs; run;
We need these correlations to run Algina’s code for computing d with confidence intervals.
-------------------------------------------------------------------------------------------------Experiment 2 of Karl's Dissertation
Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups
4
-------------------------------------------- nurs=MM --------------------------------------------The CORR Procedure
1 With Variables:
1
Variables:
v_rat
v_mus
Pearson Correlation Coefficients, N = 16
Prob > |r| under H0: Rho=0
v_mus
v_rat
0.58052
0.0184
-------------------------------------------------------------------------------------------------Experiment 2 of Karl's Dissertation
Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups
5
-------------------------------------------- nurs=MR --------------------------------------------The CORR Procedure
1 With Variables:
1
Variables:
v_rat
v_mus
Pearson Correlation Coefficients, N = 16
Prob > |r| under H0: Rho=0
v_mus
v_rat
0.68185
0.0036
-------------------------------------------------------------------------------------------------Experiment 2 of Karl's Dissertation
Correlated t-tests, Visits to Mus Tunnel vs Rat Tunnel, Three Nursing Groups
6
-------------------------------------------- nurs=RR --------------------------------------------The CORR Procedure
1 With Variables:
1
Variables:
v_rat
v_mus
Pearson Correlation Coefficients, N = 16
Prob > |r| under H0: Rho=0
v_mus
v_rat
0.48854
0.0548
--------------------------------------------------------------------------------------------------
Data CI2;
m1=22.4375 ;
m2= 7.5625 ;
s1=12.8164933 ;
s2= 5.8874867 ;
r= 0.58052 ;
n=16
;
prob=.95 ;
v1=s1**2;
v2=s2**2;
s12=s1*s2*r;
se=sqrt((v1+v2-2*s12)/n);
pvar=(v1+v2)/2;
nchat=(m1-m2)/se;
es=(m1-m2)/(sqrt(pvar));
df=n-1;
ncu=TNONCT(nchat,df,(1-prob)/2);
ncl=TNONCT(nchat,df,1-(1-prob)/2);
ul=se*ncu/(sqrt(pvar));
ll=se*ncl/(sqrt(pvar));
output;
proc print;
title1 'll is the lower limit and ul is the upper limit';
title2 'of a confidence interval for the effect size';
title3 'MM group only' ;
var es ll ul ;
run;
ll is the lower limit and ul is the upper limit
of a confidence interval for the effect size
MM group only
Obs
es
ll
ul
1
1.49151
0.73888
2.21981
7
House mice who had been reared with house mice visited the tunnels scented with house
mouse significantly more often (M = 22.44, SD = 12.81) than tunnels scented with rat (M = 7.56, SD =
5.89), t(15) = 5.64, p < .001, d = 1.49, 95% CI [.74, 2.22].
--------------------------------------------------------------------------------------------------
Independent Samples T-Tests on Mouse-Rat Tunnel Difference Scores
Foster Mom is a Mouse or is a Rat
data Mus2; set Mus;
if nurs NE 'RR' then Mom = 'Mouse';
else if nurs = 'RR' then Mom = 'Rat';
proc ttest; class Mom; var v_diff; run;
*proc mixed; *class Mom; *model v_diff = Mom / ddfm = satterth;
*repeated / group=Mom; *lsmeans Mom / pdiff CL; *run;
title2 'Foster Mom is a Mouse, Pups Exposed to Rat Scent (MR) or Not (MM)'; run;
The TTEST Procedure
Variable:
Mom
Mouse
Rat
Diff (1-2)
Mom
Mouse
Rat
Diff (1-2)
Diff (1-2)
v_diff
N
Mean
Std Dev
Std Err
Minimum
Maximum
32
16
14.8125
-1.3125
16.1250
9.0320
8.4041
8.8321
1.5966
2.1010
2.7043
0
-17.0000
31.0000
17.0000
Std Dev
95% CL Std Dev
Method
Mean
95% CL Mean
14.8125
-1.3125
16.1250
16.1250
Pooled
Satterthwaite
11.5561
-5.7907
10.6816
10.7507
Method
Variances
Pooled
Satterthwaite
Equal
Unequal
18.0689
3.1657
21.5684
21.4993
9.0320
8.4041
8.8321
DF
t Value
Pr > |t|
46
32.141
5.96
6.11
<.0001
<.0001
7.2410
6.2082
7.3393
12.0078
13.0070
11.0930
Equality of Variances
Method
Folded F
Num DF
Den DF
F Value
Pr > F
31
15
1.15
0.7906
--------------------------------------------------------------------------------------------------
data CI3;
t= 5.96 ;
df = 46 ;
n1 = 32 ;
n2 = 16 ;
***********************************************************************************;
d = t/sqrt(n1*n2/(n1+n2));
ncp_lower = TNONCT(t,df,.975);
ncp_upper = TNONCT(t,df,.025);
d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2));
d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2));
output; run; proc print; var d d_lower d_upper; run;
Obs
d
d_lower
d_upper
1
1.82487
1.11164
2.52360
Mice that had been fostered onto a mouse mother had a significantly stronger preference for
visiting mouse-scented rather than rat-scented tunnels (M = 14.81, SD = 9.03) than did mice who had
been fostered onto a rat mother (M = -1.31, SD = 8.40), t(32.1) = 6.11, p < .001, d = 1.82, 95% CI
[1.11, 2.52].
The graphics provided by SAS in the htm output can be very helpful with respect to evaluating
the normality assumption. For each group we get a histogram overlaid with 1.) a smoothed curve
representing the distribution of the observed scores, and 2.) a normal curve with the same mean and
standard deviation of the observed scores.
Independent Samples T-Tests on Mouse-Rat Tunnel Difference Scores
Foster Mom is a Mouse, Pups Exposed to Rat Scent (MR) or Not (MM)
data Mus3; set Mus; if nurs NE 'RR';
proc ttest; class nurs; var v_diff; run;
The TTEST Procedure
Variable:
nurs
MM
MR
Diff (1-2)
nurs
N
Mean
Std Dev
Std Err
Minimum
Maximum
16
16
14.8750
14.7500
0.1250
10.5507
7.5675
9.1810
2.6377
1.8919
3.2460
0
3.0000
31.0000
26.0000
Method
MM
MR
Diff (1-2)
Diff (1-2)
v_diff
Mean
Pooled
Satterthwaite
95% CL Mean
14.8750
14.7500
0.1250
0.1250
9.2529
10.7176
-6.5042
-6.5329
Method
Variances
Pooled
Satterthwaite
Equal
Unequal
20.4971
18.7824
6.7542
6.7829
Std Dev
95% CL Std Dev
10.5507
7.5675
9.1810
7.7938
5.5901
7.3367
DF
t Value
Pr > |t|
30
27.204
0.04
0.04
0.9695
0.9696
16.3292
11.7121
12.2721
Equality of Variances
Method
Num DF
Den DF
F Value
Pr > F
15
15
1.94
0.2096
Folded F
The sample sizes are equal and the ratio of the larger variance to the smaller variance less
than 4, so I employ the pooled variances t test.
--------------------------------------------------------------------------------------------------
Independent t on WTLOSS Data
data wtloss;
input program $ loss @@ ; cards;
A 25 A 21 A 18 A 20 A 22 A 30
B 15 B 17 B 9 B 12 B 11 B 19 B 14 B 18 B 16 B 10 B 5 B 13
proc ttest; class program; var loss; run;
The TTEST Procedure
Variable:
program
A
B
Diff (1-2)
program
A
B
Diff (1-2)
Diff (1-2)
loss
N
Mean
Std Dev
Std Err
Minimum
Maximum
6
12
22.6667
13.2500
9.4167
4.2740
4.0927
4.1502
1.7448
1.1815
2.0751
18.0000
5.0000
30.0000
19.0000
Method
Mean
Pooled
Satterthwaite
95% CL Mean
22.6667
13.2500
9.4167
9.4167
Method
Variances
Pooled
Satterthwaite
Equal
Unequal
18.1814
10.6496
5.0177
4.7023
Std Dev
27.1519
15.8504
13.8157
14.1310
4.2740
4.0927
4.1502
DF
t Value
Pr > |t|
16
9.7083
4.54
4.47
0.0003
0.0013
Equality of Variances
Method
Folded F
Num DF
Den DF
F Value
Pr > F
5
11
1.09
0.8357
95% CL Std Dev
2.6678
2.8992
3.0909
10.4824
6.9489
6.3163
--------------------------------------------------------------------------------------------------
data CI4;
t= 4.54 ;
df = 16 ;
n1 = 6 ;
n2 = 12 ;
***********************************************************************************;
d = t/sqrt(n1*n2/(n1+n2));
ncp_lower = TNONCT(t,df,.975);
ncp_upper = TNONCT(t,df,.025);
d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2));
d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2));
output; run; proc print; var d d_lower d_upper; run;
Independent t on WTLOSS Data
Obs
d
d_lower
d_upper
1
2.27
0.99588
3.50073
12
Those who completed weight-loss Program A lost significantly more weight (M = 22.67 lb., SD
= 4.27, n = 6) than did those who completed weight-loss Program B (M = 13.25 lb., SDs = 4.09, n =
12), t(9.7) = 4.47, p = .001.
When considering these results, one must question how the selective attrition might have
affected the means. Perhaps weight loss was greater in Program A simply because Program A was
more effective in chasing off those who were not losing weight.
options pageno=min nodate formdlim='-';
proc format; value bk 0='Kerry' 1='Bush'; run;
title 'Income, IQ, and Presidential Voting'; run;
data Voting; infile 'C:\D\StatData\Bush-Kerry2004.txt';
input IQ state $ income vote candidate $; format vote bk. ;
proc corr; var vote iq income; run;
proc ttest; class vote; var iq income; run;
Data CI;
t= 2.61 ;
df = 49 ;
n1 = 20 ;
n2 =31 ;
d = t/sqrt(n1*n2/(n1+n2));
ncp_lower = TNONCT(t,df,.975);
ncp_upper = TNONCT(t,df,.025);
d_lower = ncp_lower*sqrt((n1+n2)/(n1*n2));
d_upper = ncp_upper*sqrt((n1+n2)/(n1*n2));
output; run; proc print; var d d_lower d_upper; run;
-------------------------------------------------------------------------------------------------Obs
d
d_lower
d_upper
1
2.27
0.99588
3.50073
Income, IQ, and Presidential Voting
The CORR Procedure
3
Variables:
vote
IQ
Simple Statistics
income
Variable
vote
IQ
income
N
Mean
Std Dev
Sum
Minimum
Maximum
51
51
51
0.60784
99.92157
30.81918
0.49309
2.39034
4.92449
31.00000
5096
1572
0
94.00000
23.44800
1.00000
104.00000
48.34200
Pearson Correlation Coefficients, N = 51
Prob > |r| under H0: Rho=0
vote
IQ
income
1.00000
-0.34902
0.0121
-0.63470
<.0001
IQ
-0.34902
0.0121
1.00000
0.26385
0.0614
income
-0.63470
<.0001
0.26385
0.0614
1.00000
vote
-------------------------------------------------------------------------------------------------Income, IQ, and Presidential Voting
2
The TTEST Procedure
Variable:
vote
Kerry
Bush
Diff (1-2)
vote
Kerry
Bush
Diff (1-2)
Diff (1-2)
IQ
N
Mean
Std Dev
Std Err
Minimum
Maximum
20
31
101.0
99.2581
1.6919
1.9861
2.4217
2.2628
0.4441
0.4349
0.6490
95.0000
94.0000
104.0
102.0
Method
Pooled
Satterthwaite
Mean
101.0
99.2581
1.6919
1.6919
Method
Variances
Pooled
Satterthwaite
Equal
Unequal
95% CL Mean
100.0
98.3698
0.3878
0.4407
Std Dev
101.9
100.1
2.9961
2.9431
1.9861
2.4217
2.2628
DF
t Value
Pr > |t|
49
46.079
2.61
2.72
0.0121
0.0091
95% CL Std Dev
1.5104
1.9352
1.8902
2.9009
3.2370
2.8197
Equality of Variances
Method
Num DF
Den DF
F Value
Pr > F
30
19
1.49
0.3688
Folded F
Obs
g
d_lower
d_upper
1
0.74857
0.16375
1.32623
The estimated mean IQ of residents of states which voted for Kerry (M = 101.0, s = 1.99, n =
20) was significantly higher than that of states which voted for Bush (M = 99.3, s = 2.42, n = 32),
t(46.1) = 2.71, p = .009, d = .75, 95% CI [.16, 1.33].
One could use the point biserial r as the effect size estimator here, or its square (which can be
interpreted as a proportion of variance in IQ explained by the model). The value of that r here is
-0.35.
Variable:
vote
Kerry
Bush
Diff (1-2)
vote
Kerry
Bush
Diff (1-2)
Diff (1-2)
income
N
Mean
Std Dev
Std Err
Minimum
Maximum
20
31
34.6722
28.3334
6.3388
5.0018
2.8794
3.8441
1.1184
0.5172
1.1025
28.8310
23.4480
48.3420
34.2380
Method
Mean
34.6722
28.3334
6.3388
6.3388
Pooled
Satterthwaite
Method
Variances
Pooled
Satterthwaite
Equal
Unequal
95% CL Mean
32.3312
27.2772
4.1232
3.8114
Std Dev
37.0131
29.3896
8.5543
8.8662
5.0018
2.8794
3.8441
DF
t Value
Pr > |t|
49
27.205
5.75
5.14
<.0001
<.0001
95% CL Std Dev
3.8038
2.3009
3.2111
7.3055
3.8488
4.7902
-------------------------------------------------------------------------------------------------Income, IQ, and Presidential Voting
3
The TTEST Procedure
Variable:
income
Equality of Variances
Method
Folded F
Num DF
Den DF
F Value
Pr > F
19
30
3.02
0.0067