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Unformatted text preview: Statistics 101A
Professor Esfandiari Review of twosample test and paired sample test of the mean
A Case Study I Introduction to the study: In a study that was conducted in more than six states
the objective was to ﬁnd out if exposing the students to a combination of law related
education (LRE) and social studies (the regular social studies book), enhances the middle
school students’ knowledge of the United States Constitution. The experimental group studied the regular social studies book plus LRE. The control
group only studied the regular social studies book. This study was conducted on 931
experimental and 5 31 control students. The students were in seventh and eighth grade.
They were exposed to the program for one year. The experimental and the control groups
were pretested (tested before the program started) and posttested (tested after the
program was ﬁnished) on their knowledge of the United States Constitution. The students
could not be randomly assigned to the groups because in school settings you cannot
randomly assign students. So, the treatments were randomly assigned to the classrooms.
So, the study did not follow a true experimental design because the individual students
could not be randomly assigned to the experimental and the control group. Rather it
followed a quasiexperimental design because the classrooms were ﬁrst numbered, then
around two thirds were randomly selected to be in the experimental group and the rest
were assigned to the control group. 11 Questions we could ask that represent the twosample test of the mean?
Question 1: Did the experimental and control group have a similar baseline on the
knowledge of the US Constitution. We want to know if both groups have the same
baseline prior to starting the intervention. The null and alternative hypothesis would be: H0: pt (pretest for control)  u (pretest for experimental) = 0 Question 2: Did the experimental and control group show similar gain on the knowledge
of the US Constitution? Gain is deﬁned as the posttest score minus the pretest score. H0: u (gain for control) — u (gain for experimental) = 0 How can we answer these questions? Based on CLT for large and independent random samples of size 30, 40 or more the
distribution of X1 — X2 follows the normal distribution with mean (111 ~ v.2) equal to zero and the standard error equal to (1/ SlA2/n1 + SZAZ/n2). If the sample sizes are smaller than 30, we use the t distribution with df = n1 + n 2 — 2.
As the sample size increases the t and Z value become similar to each other. t= X1 — X2 /\/(Sl"2/n1 + SZAZ/nZ)
Assumptions of the twosample test of the mean
° Normality of the data, checked through the pplot
° Equality of variance (can be examined through sideby—side boxplot). We will
discuss a more formal test later.
' Independence established through random assignment of subjects to groups.
The null hypothesis could also be tested through the 95% conﬁdence interval. 95% CI: )‘(1 — >22 +/1.96* x/(SIAZ/n1+ szAz/nz) Now let us examine the computer printouts that relate to question 1 Group Statistics Deviation Mean . grate? dscorefs 0" COWO' 503 27.4592 9.16221 .40 852
now e ge O _ constitution expe “mental 951 31. 7336 10.35 756 .33 411 Indepql den! samus Test Levene's Test for Equalrty
0' Variances I45! '01 Eualit of Means
95% Cnnﬂdenm Interval of
Sig. Mean sm. Error "‘9 W‘mm
r Si. (1 railed) Difference Drrmna: m
10. 152 .00]. ‘7.795 1462 .000 4.2744 54331 "5.34993 AIIQKEO
000 «1.2744 .52 775 S.30985 4.23m prelesr stores an Equal variances knuwle d9: of assumed constitution Equal variz ms
no! assum Ed 1133.078 The P value of 0.000 as well as the conﬁdence interval that does not include zero
indicates that we reject the null and conclude that the two groups did not have the same
baseline knowledge on the US Constitution. We are 95% conﬁdent that the experimental
group scored between 3.2% to 5.3% higher than the control prior to starting the program. This is an important message indicating that if we want to see whether the program was
effective we should not be looking at the posttest scores rather, we should compare the
gain scores. However, if there had been no difference between the pretest scores, we
could have compared the posttest scores. Now let us examine the printout related to comparing the gain scores Group Statistics
Std. Std. Error
rou  Deviation Mea n
gain in the knowledge control 503 9.2644 13.484 82 .60 126
°f the Us c°"5“t”t'°“ experimental 961 15. 7190 17.50453 .56 466 Indeuzndznl Simﬂs Ten Levunz's Te 5! for Equalrry
D' Vzriantes I'IES‘ for E Hill! of M23 75
95% Conhdznm Inlerval 0'
5.9. sm. Error ”“ 0mm“!
5 . (245m) mrrmnce 3:341:25 "‘5 22.013 4.223 1452 Don 4.4546 .39357 —x.zo744 4.70132
Equal VIIIIIKES "0‘ assumm 7.82 5 1253.359 .000 “5.4545 .81484 8.07253 V4.33643 The null is rejected, P =0.000 and we are 95% conﬁdent that the experimental group
gained between 4.7% to 88.2% higher on the knowledge of the US Constitution than the control group. Tests of the assumption of normality and equality of variance for the gain scores:
The points are pretty close to the line implying that the expected and actual cumulative
percentiles are close and so the data are pretty close to normal. The two boxplots are
similar in terms of scatter. This is also conﬁrmed by the close standard deviations (13.5
and 17.5). By a rule of thumb, as long as the ratio of the two standard deviations is less
than two, we do not need to be concerned about violating the assumption of equality of variance. Expected Cum Prob 1.00 .75 .50 .25 Normal P—P Plot of gain in the knowled 1.00 .25 .50 .75 Observed Cu m Prob 100 80 60 4O 20 gain in the knowledge ofthe US constitution
O N = 1 503 961
Missing control experimental group
II Questions we could ask that represent paired sample test of the mean Did the attitude of the control and the experimental students change toward law and
authority from pretest to posttest. The null hypothesis can be stated as:
For the control group H0: u post  u pre = 0 H0: f) = 0 Once you subtract the pretest from the posttest scores, the data become similar to the one
sample test of the mean so that In one sample test of the mean we had t=( X—u)/S( X)withS( X)=S/\/N In the paired same t—test we have T =( d— O)/ S( d ) with S( d ) = S(d)/\/ N ( d is the average of the deviation scores in
the sample). Let us now examine the outputs for the control and the experimental groups Paired Sam pies Statisticsa —
Deviation Mea n Pair 1 pretest on authority 60.9746 439 11.49722 .54873
posstest on authority 58.72 11 439 12.18115 .5813 7 a. group = control Paired Samples Test“ Paler Diff eeeee as
95% Confidence Interval of
5m 5m Em" me Difference Sig‘
Deviation Mean df (Z—lailed) Pair 1 pretest an authority— a. group = canrrol Paired Samples Correlationsat — Correlation “
Pair 1 pretest on authority & a. group = control Conclusions for the control grou p: Null is rejected and students became less
positive about law and authority after the program. However, close examination of the
data indicates that the mean decreased only 2% and the reason that we have statistical
signiﬁcance is because of the large sample. But, this difference is not practically
signiﬁcant. THIS IS A VERY IMPORTANT STATISTICAL MESSAGE THAT
STATISTICAL SIGNIFICANCE DOES NOT IMPLY PRACTICAL SIGNIFICANCE. Another important message is that the pretest and posttest data are correlated to each
other (correlation is equal to 0.468) and this is the major factor that differentiates paired
sample data from two—sample test data sometimes also referred to as independent sample
data. Pretty much the same conclusions can be drawn about the experimental group. A
better method of data analysis would have been to compare the gain in attitude toward
law and authority for the control and experimental group. The objective was to provide
you with an example of paired sample test of the mean. Paired Samp les Statist icsa Std. Std. Error
Deviation Mean Pair 1 pr8t$t 0n aUthOI’itY 61.72 95 890 11.02421 .3695 3
POSStCSY 0” aUthOI’ilY 60.98 31 890 11.21350 .37588 a group = experimental Paired Samples Tm“ Paired Differences
95% Confidence interval of
5m 5m Eng, the Difference Sig‘
Deviation Mean df (Z—lailed) Pair 1 pretest on aumomy
.7464 11.19152 .37 514 .0101 114 827 335 ,047 a group . experimental Paired Samples Correlationsal _
Pair 1 pretest on authority& 890
posstest on authority a. group = experimental III The conditions under which you want to use a paired sample or a dependent
t—test 1) Repeated measures, measuring the same person, object, etc. twice. You make a similar
measurement at two different times. examples are measuring some attribute before and
after an experiment or intervention. 2) Matched pairs, examples are giving the same test to subjects who come in pairs like
wifehusband, twins, etc. You can also make these matched pairs. For example, suppose
that you want to test the effectiveness of two different methods for teaching twosample
test of the mean, But, you think that students' prior knowledge on hypothesis testing may
be a confounding factor. So, to control this factor, you give them a pretest on hypothesis
testing and you rank them based on how they did. Then you pair them on the basis of
their ranks and pair one would be the people with ranks l and 2, pair 2 would be people
with ranks 3, and 4, and so on. Then you randomly assign the subjects in pair one into
methods 1 and 2, then you randomly assign pair 2 into methods one and two and so on.
This way you make sure that the subjects that are assigned to the two methods for
teaching the twosample t—test have the same knowledge base on hypothesis testing, 3) Administering two different treatments to the same subjects on a random basis. An
example would be asking the same people to taste two different drinks, but, for each
person you randomly change the order in which they receive the drink. d=(2+2+ ...... ..+ 6)/10=40/10=4 s" 2(a) = (d — H)A2/ n— 1 = 180/9 = 20 S(d) = V20.888 = 4.47 s( a)= S(d)/\/n=4.47/\/10=4.47/3.16=1.41
t= CT/s(c1)=4/1.41=2.83 We need to look up the t value corresponding to nine or (n —1 ) degrees of freedom to see whether
we need to reject or fail to reject the null hypothesis. Since 2.83 is larger than 2.228 , we reject
the null and conclude that there was a statistically signiﬁcant difference from time 1 (x1) to time
2 (x2). We could make the same difference based on the 95% conﬁdence interval. 95% conﬁdence interval = d+/— 2.228 * S( d) = 4 +/ 2.228 * 1.41 = (+0.85, +7.14)
The 95% CI does not capture zero, we reject the null and conclude and we are 95% conﬁdent that
from time one to time two there was an average increase of 0.85 to 7.14 in the data. T—Distribution Table
Df or = 0.1 0.05 0.025 0.01 0.005 0.001 0.0005
1.282 1.645 1.960 2.326 2.576 3.091 3.291 00 1 3.078 6.314 12.706 31.821 63.656 318289636578
2 1.886 2.920 4.303 6.965 9.925 22.328 31.600
3 1.638 2.353 3.182 4.541 5.841 10.214 12.924
4 1.533 2.132 2.776 3.747 4.604 7.173 8.610
5 1.476 2.015 2.571 3.365 4.032 5.894 6.869
6 1.440 1.943 2.447 3.143 3.707 5.208 5.959
7 1.415 1.895 2.365 2.998 3.499 4.785 5.408
8 1.397 1.860 2.306 2.896 3.355 4.501 5.041
9 1.383 1.833 2.262 2.821 3.250 4.297 4.781
10 1.372 1.812 2.228 2.764 3.169 4.144 4.587
11 1.363 1.796 2.201 2.718 3.106 4.025 4.437
12 1.356 1.782 2.179 2.681 3.055 3.930 4.318
13 1.350 1.771 2.160 2.650 3.012 3.852 4.221
14 1.345 1.761 2.145 2.624 2.977 3.787 4.140
15 1.341 1.753 2.131 2.602 2.947 3.733 4.073
16 1.337 1.746 2.120 2.583 2.921 3.686 4.015
17 1.333 1.740 2.110 2.567 2.898 3.646 3.965
18 1.330 1.734 2.101 2.552 2.878 3.610 3.922
19 1.328 1.729 2.093 2.539 2.861 3.579 3.883
20 1.325 1.725 2.086 2.528 2.845 3.552 3.850
21 1.323 1.721 2.080 2.518 2.831 3.527 3.819
22 1.321 1.717 2.074 2.508 2.819 3.505 3.792
23 1.319 1.714 2.069 2.500 2.807 3.485 3.768
24 1.318 1.711 2.064 2.492 2.797 3.467 3.745
25 1.316 1.708 2.060 2.485 2.787 3.450 3.725
26 1.315 1.706 2.056 2.479 2.779 3.435 3.707
27 1.314 1.703 2.052 2.473 2.771 3.421 3.689
28 1.313 1.701 2.048 2.467 2.763 3.408 3.674
29 1.311 1.699 2.045 2.462 2.756 3.396 3.660
30 1.310 1.697 2.042 2.457 2.750 3.385 3.646
60 1.296 1.671 2.000 2.390 2.660 3.232 3.460 120 1.289 1.658 1.980 2.358 2.617 3.160 3.373
00 1.282 1.645 1.960 2.326 2.576 3.091 3.291 ...
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This note was uploaded on 05/02/2011 for the course STAT 101A taught by Professor Mahtashesfandiari during the Winter '11 term at UCLA.
 Winter '11
 MahtashEsfandiari

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