Dependent Samples: Hypothesis Test
For Hypothesis tests for dependent samples, we
1.
2.
3.
list the pairs of data in 2 columns (or rows),
take the difference between each pair of data,
analyze the differences
xd
d


n
i
where di is the difference for the ith pair
of data and n is the # of pairs of data.
sd is the standard deviation of the differences.
x  d
Then, t  d
has a t distribution with df = n-1
sd
under the usual assumptions.
n
Dependent Samples: P-value Calculation
For Hypothesis tests, we want to find p-values. This requires
Minitab.
Let the observed test statistic be t*
Now, p-value =
P(tn-1 > t*),
when Ha: d > 0
P(tn-1 < t*),
when Ha: d < 0
or, 2P(tn-1 > |t*|),
when Ha: d  0
where n = number of pairs in the analysis.
Diabetes Knowledge Scores
Scores before and after a lecture. Let d = after-before
Before
75 62 67 70 55 59 60 64 72 59
After
77 65 68 72 62 61 60 67 75 68
Difference 2
3
1
2
7
2
0
(After-Before)
2  3  1  2  7  2  0  3  3  9 32
xd 

10
10
sd = 2.741 (Minitab)
= 3.2
3
3
9
Dependent Samples Hypothesis Test
Step 1: H0: d = 0
H1: d > 0
xd  0
Step 2: t* = s
has a t-distribution with df = 9.
d
10
3 .2  0
Step 3: t* = 2.741
= 3.7 (round to tenths place)
10
Step 4: p-value = P(tdf=9 > 3.7) = .00246 (Minitab)
Dependent Samples Hypothesis Test
Step 5: Since the p-value is less than any
reasonable level of significance, we reject
the null hypothesis and accept the research
hypothesis.
Step 6: With a .05 level of significance, we conclude
there is enough evidence that the
program yields a mean increase in
knowledge of diabetes.
Dependent Samples Confidence Interval
We can also find a formula to estimate the mean
difference. The result is
 s 
xd  t / 2,df  n1  d 
 n
Diabetes Example: Find a 95% CI.
Recall, xd = 3.2 and sd = 2.741.
Also, there were n = 10 pairs of data so df = 9.
t.025,9 = 2.262 from software.
Thus, 3.2  2.262  2.741  = 3.2  1.961 = (1.2, 5.2 )

10 
We are 95% confident the mean increase in knowledge about
diabetes from the course is between 1.2 and 5.2 points.
Dependent Samples Confidence Interval
Travel Times for two routes (Minitab output)
Paired T-Test and CI: Route I, Route II
Paired T for Route I - Route II
N
Mean StDev SE Mean
Route I
10 26.700 2.406 0.761
Route II
10 25.900 2.183 0.690
Difference
10 0.800 1.317 0.416
95% CI for mean difference: (-0.142, 1.742)
T-Test of mean difference = 0 (vs not = 0): T-Value = 1.92 P-Value = 0.087
It is unclear at the 95% level of confidence which route is faster on average (takes
less time). If route I is faster, it is by no more than 0.14 minute, on average. If route
II is faster on average, it is by no more than 1.74 minutes.
Prescription Drug Costs
Are Rx drugs from Canada cheaper than drugs in the
USA?
Because costs of drugs vary considerably, we should
consider using a dependent samples design.
Otherwise, the large variation is costs among the Rx
drugs may hide a difference in typical costs.
A sample of 10 drugs were chosen, then using an
online pharmacy price checking website, the drugs
were priced at a Canadian and a USA pharmacy.
Rx Costs - Independent Samples Analysis
The output from an independent samples analysis is
given on the next slide.
What is the research hypothesis?
Ha: The mean cost of drugs in Canada is lower than
the mean cost of drugs in the USA.
What would we conclude from THIS analysis?
We will ask whether this is appropriate or not later.
Rx Costs - Independent Samples Analysis
Two-Sample T-Test and CI: Canada, USA
N Mean StDev SE Mean
Canada
10 3.16 3.28
1.0
USA
10 4.56 3.58
1.1
Difference = mu (Canada) - mu (USA)
Estimate for difference: -1.41
T-Test of difference = 0 (vs <):
T-Value = -0.91 P-Value = 0.186 DF = 18
Both use Pooled StDev = 3.4348
Rx Costs - Independent Samples Analysis
From this output, we would conclude that there is not
enough evidence to convince us that the mean cost
of drugs from Canada is lower than the mean cost of
drugs in the USA.
Do you believe this? Why or why not?
Rx Costs - Dependent Samples Analysis
Because we priced the same 10 drugs, we should be
using a dependent samples (paired) analysis.
This also controls for variability among drug costs.
The output from a dependent samples analysis is
given on the next slide.
What is the research hypothesis?
Ha: The mean cost of drugs in Canada is lower than
in the USA.
Rx Costs - Dependent Samples Analysis
Paired T-Test and CI: Canada, USA
Paired T for Canada - USA
N Mean StDev SE Mean
Canada
10 3.16 3.28 1.04
USA
10 4.56 3.58 1.13
Difference 10 -1.405 0.763 0.241
95% upper bound for mean difference: -0.962
T-Test of mean difference = 0 (vs < 0):
T-Value = -5.82 P-Value = 0.000
Rx Costs - Dependent Samples Analysis
Notice that there is now overwhelming evidence to
support the research hypothesis. The p-value is
listed as 0.000 (< 0.001 is how it would be reported).
This implies there is almost no chance of seeing this
pattern in a sample of 10 pairs of drug costs if the
mean costs are not different between Canada and the
USA. (ie: there is no chance of seeing the sample
data if the null hypothesis is true).
So we would claim the research hypothesis has been
proven beyond a reasonable doubt.
Rx Costs - Dependent Samples Analysis
What about assumptions? In dependent samples
analysis, we need to have
• a random sample of paired observations
• differences must be normally distributed, or we need
a large sample (30 or more differences as a rule of
thumb)
The sample was selected when I looked at names of
drugs listed on the website, so it is not truly a random
sample. This is one criticism.
How would we check the normality assumption?
Is Data Normal or not?
Stat  Basic Statistics  Normality Test gives a way
to check. (Enter the variable)
Your hypotheses are:
Ho: The data comes from a normal population of data.
Ha: The data comes from a population that is not a
normal distribution.
Caveat: Statisticians NEVER accept the null
hypothesis, yet that is exactly what this test does!
Is Data Normal or not?
If the p-value is small, we must use another technique
(called nonparametrics).
If the p-value is large, then researchers assume the ttest is ok to use.
The t-test is called robust to this assumption, so only
when the original population is very different from a
normal distribution does it make a big difference (the
p-value is not affect much when the population is
slightly non-normal).
Is Data Normal or not? Minitab
Probability Plot of differences
Normal
99
Mean
StDev
N
AD
P-Value
95
90
Percent
80
70
60
50
40
30
20
10
5
1
-3.5
-3.0
-2.5
-2.0
-1.5
-1.0
differences
-0.5
0.0
0.5
-1.405
0.7634
10
0.265
0.609