Download Matched Pair File - Galena Park ISD Moodle

Matched-Pair Design & One-Sample t Test
What is a Matched-Pair Design?
In an experiment, a subject receives two different treatments and the repeated measurements compared
OR
two “matched” subjects each receive a different treatment and their paired measurements are compared.
These types of samples are considered to be DEPENDENT. To test or construct a confidence interval on
the difference between the two measurements (means), one new sample is created by calculating the
differences between each paired measurements (data).
Notation for Paired Differences
d  x1  x2 represents the difference between one pair of data
 d  mean of difference s for the population  1   2
d  sample mean of the difference s  x1  x2
s d  standard deviation of the sample of difference s
s.e.d  
sd
n
 standard error of the sample of difference s
For a One-Sample (Matched-Pair) t Test or Confidence Interval, always remember to subtract one
sample list from the other noting which order on your paper.
CONFIDENCE INTERVAL Example: The 25 students in a liberal arts course in statistical literacy
were given a survey that included questions on how many hours per week they watched television and
how many hours a week they used a computer. Let’s construct a 95% confidence interval for µd the
average difference in hours spent using the computer versus watching television (computer use – TV) for
the population of students represents by this sample. (Minds on Statistics, p.362)
Student Computer
TV Difference
Student Computer
TV Difference
1
30
2.0
28.0
14
5
6.0
-1.0
2
20
1.5
18.5
15
8
20.0
-12.0
3
10
14.0
-4.0
16
30
20.0
10.0
4
10
2.0
8.0
17
40
35.0
5.0
5
10
6.0
4.0
18
15
15.0
0.0
6
0
20.0
-20.0
19
40
5.0
35.0
7
35
14.0
21.0
20
3
13.5
-10.5
8
20
1.0
19.0
21
21
35.0
-14.0
9
2
14.0
-12.0
22
2
1.0
1.0
10
5
10.0
-5.0
23
9
4.0
5.0
11
10
15.0
-5.0
24
14
0.0
14.0
12
4
2.0
2.0
25
21
14.0
7.0
13
50
10.0
40.0
First you must check the assumptions for a One-Sample Matched Pair t Confidence Interval:
1. Subjects (units) were randomly selected and the paired data is dependent.
2. Population of Differences is known to be normally distributed OR n is sufficiently large (n  30) so
that the Sampling Distribution of Differences is normally distributed.
3. Sample size is less than 10% of population size.
Check (verify) Assumptions:
1. We assume students were randomly selected (or we could not carry out this CI) AND the paired
data is dependent b/c the paired responses came from the same subject.
2. The graphical display supports the assumption that the sampling distribution of differences
between computer use and TV is approximately normally distributed.
3. 25 students is less than 10% of all college students.
Histogram
Collection 2
Box Plot
Collection 2
Dot Plot
Collection 2
4
Count
3
2
1
-30 -20 -10
0 10 20
differences
30
40
50
-20
-10
0
10
20
differences
30
40
50
-20
-10
0
10
20
differences
30
40
50
Formula and substitutions for CI
 s diff
x diff  t *
 n



 15.2428 
5.36  2.06

25 

5.36  2.063.04857 
5.36  6.29194
 .9319,11.6519
Using Fathom
estimate of Collection 2
Estimate Mean
Attribute (continuous): differences
Interval estimate for population mean of differences
Sample count: 25
Sample mean: 5.36
Standard deviation: 15.2428
Standard error: 3.04857
Based on the sample, the 95 % confidence interval for the population mean of differences is 5.36 plus or minus 6.29194, ranging from 0.931935 to 11.6519.
If the sampling process w ere performed repeatedly, the confidence intervals generated w ould capture the population mean 95 % of the time.
Interpretation: Based on this sample, we are 95% confident that the average difference between
computer usage and TV viewing for students is COVERED by the interval from -.93 and 11.65 hours per
week.
ASSIGNMENT:
1) Suppose you were given a 95% confidence interval for the difference in two population means. What
could you conclude about the population means if
a) The confidence interval did NOT cover zero?
b) The confidence interval did cover zero?
2) For a sample of n = 20 women aged 18 to 29, responses to the question, “How tall would you like to
be?” are recorded along with actual heights. In the sample, the mean desired height is 66.7 inches; the
mean actual height is 64.9 inches. The standard deviation of the differences is sd = 2.1 inches.
a) What is the mean difference between desired and actual height?
b) What are the necessary conditions for the validity of the confidence interval to be computed?
VERIFY them.
c) Compute a 95% confidence interval for the mean difference between desired and actual
height, formula with substitutions, and write a sentence interpreting this interval. Be specific
about what population is described by the interval.
d) Explain why the following statement is NOT correct: “Based on the confidence interval
computed, we can conclude that all women aged 18 to 29 would like to be taller.”
3) An experiment was done by 15 students in a statistics class at the University of California at Davis to
see if manual dexterity was better for the dominant hand compared to the nondominant hand (left or
right). Each student measured the number of beans they could place in a cup in 15 seconds, once with the
dominant hand and once with the nondominant hand. The order in which the two hands were measured
was randomized for each student.
Student
1
2
3
4
5
6
7
8
9 10 11 12 13 14 15
Dominant hand
22 19 18 17 15 16 16 20 17 15 17 17 14 20 26
Nondominant hand 18 15 13 16 17 16 14 16 20 15 17 17 16 18 25
Difference
a) Explain why the order of the two hands was randomized rather than, for instance, having each
student tests the dominant hand first.
b) Before constructing a confidence interval, what assumptions must be satisfied? VERIFY
them.
c) Compute a 90% confidence interval for the mean difference in the number of beans that can
be placed into a cup in 15 seconds by the dominant and nondominant hands.
d) Write a sentence or two using the interval to address the question of whether manual dexterity
is better, on average, for the dominant hand.
HYPOTHESIS TESTING: Matched-Pair t Test have the same assumptions as the Matched-Pair
t Confidence Interval. The null hypothesis will assume the population mean difference is zero denoted
as Ho: µd = 0. The alternative hypothesis will take on one of the following forms: Ha: µd [>, <, ≠] 0.
Example: Ten pilots performed tasks at a simulated altitude of 25,000 feet. Each pilot performed the
tasks in a completely sober condition and, three days later, after drinking alcohol. The response variable is
the time (in seconds) of useful performance of the tasks for each condition. The longer a pilot spends on
useful performance, the better. The research hypothesis is that useful performance time decreases with
alcohol use, so the data of interest is the decrease (or increase) in performance with alcohol compared to
when sober. (MOS, p.393-4)
Pilot
1
2
3
4
5
6
7
8
9
10
No Alcohol
261
565
900
630
280
365
400
735
430
900
Alcohol
185
375
310
240
215
420
405
205
255
900
Difference = No Alcohol - Alcohol
76
190
590
390
65
-55
-5
530
175
0
Collection 1
Box Plot
State the null and alternative hypotheses:
Ho: µd = 0
where µd = µno alcohol - µalcohol
Ha: µd > 0
State and check assumptions needed to perform hypothesis test.
1. We assume the ten pilots were randomly selected and the paired
-100 0
100 200 300 400 500 600
differences
measurements are dependent.
2. The approx symmetric boxplot supports the assumption the sampling distribution of differences
between no alcohol and alcohol useful performance time is approximately normally distributed.
3. 10 pilots is less than 10% of all pilots.
Perform One-Sample t Test.
The following is a computer print out of the statistical results from a t Test on differences.
Test of mu = 0.0 vs. mu > 0.0
Variable
N
Mean
Diff
10
195.6
StDev
230.5
SE Mean
72.9
T
2.68
p
0.013



195.6  0 
Pxd  195.6  P t 
  Pt  2.68  0.013
230
.
5


10 

Interpretation in the context of the problem using significance level of 0.05.
Since p-value is less than alpha, we will reject the null hypothesis and conclude that alcohol has a
statistical significant effect and decreases useful performance time.
ASSIGNMENT:
4. Paired t for height – momheight
Height
Momheight
Difference
N
93
93
93
Mean
64.342
63.057
1.285
StDev
2.862
2.945
3.136
SE Mean
0.297
0.305
0.325
95% CI for mean difference: (0.639, 1.931)
t-Test of mean difference = 0 (vs > 0); t-Value = 3.95; p-Value = 0.000
a) It has been hypothesized that college students are taller than they were a generation ago and therefore
that college women should be significantly taller than their mothers. State the null and alternative
hypotheses to test this claim. Be sure to define any parameters you use.
b) Using the information in the Minitab output, the test statistic is t = ________. Identify the numbers that
were used to compute the t-statistic. What formula was used to calculate the t-statistic?
c) What are the degrees of freedom for the test statistic?
d) Write the probability statement of the hypothesis test.
e) Draw a sketch that illustrates the connection between the t-statistic and the p-value in this problem.
5. Find the p-value and draw a sketch showing the p-value area for each of the following situations in
which the value of t is the test statistic for the hypotheses given.
a) Ho: µd = 0, Ha: µd > 0, n = 20, t = 2.00
b) Ho: µd = 0, Ha: µd < 0, n = 20, t = -2.00
c) Ho: µd = 0, Ha: µd ≠ 0, n = 20, t = 2.00
d) Ho: µd = 0, Ha: µd ≠ 0, n = 20, t = -2.00
6. In a study of memory recall, eight students from a large psychology class were selected at random and
given 10 minutes to memorize a list of 20 nonsense words. Each was asked to list as many of the words as
he or she could remember both 1 hour and 24 hours later. Is there evidence to suggest that the mean
number of words recalled after 1 hour exceeds the mean recall after 24 hours by more than 3 words? Use
.01 significance level.
1
2
3
4
5
6
7
8
Student
14 12 18 7
11 9
16 15
1 hr later
10 4
14 6
9
6
12 12
24 hours later
Difference
a) State null and alternative hypotheses.
b) State and verify assumptions needed to carry out t test.
c) Complete the probability statement.







x diff   diff 
_______

3
  P t 
  Pt  _______   _______
Px diff  _______   P t 
s diff






8


n 

Interpret the results in the context of the situation.
7. Do girls think they don’t need to take as many science classes as boys? The article “Intentions of
Young Students to Enroll in Science Courses in the Future: An Examination of Gender Differences”
(Science Education (1999):55-76) gives information from a survey of children in grades 4, 5, and 6. The
224 girls participating in the survey each indicated the number of science courses they intended to take in
the future, and they also indicated the number of science courses they thought boys their age should take
in the future. For each girl, the authors calculated the difference between the number of science classes
she intends to take and the number she thinks boys should take.
a) Explain why this data is paired.
b) The mean of the differences was -.83 (indicating girls intended, on average, to take fewer classes than
they thought boys should take), and the standard deviation was 1.51. Construct and interpret a 95%
confidence interval for the mean difference.
8. As part of a study to determine the effects of allowing the use of credit cards for alcohol purchases in
Canada (see “Changes in Alcohol Consumption Patterns Following the Introduction of Credit Cards in
Ontario Liquor Stores,” Journal of Studies of Alcohol (1999):378-382), randomly selected individuals
were given a questionnaire asking them (among other things) how many drinks they had consumed during
the previous week. A year later (after liquor stores started accepting credit cards for purchases), these
same individuals were again asked how many drinks they had consumed in the previous week. The data
shown is consistent with summary statistics presented in the article.
n
1994 Mean
1995 Mean
sd
d
Credit Card Shoppers
96
6.72
6.34
.38
5.52
Non-Credit Card Shopper
850
4.09
3.97
.12
4.58
a) Is this a matched pair data or two independent samples? Explain why.
b) The standard deviation of the difference was quite large. Explain how this could be the case.
c) Calculate a 95% confidence interval for the mean difference in drink consumption for credit card
shoppers between 1994 and 1995. Did the mean number of drinks increase?
d) Test the hypothesis that there was no change in the mean number of drinks between 1994 and 1995
for the non-credit card shoppers. Be sure to calculate and interpret the p-value for this test.