Quiz STAT
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lecture__9_t_tests_sp15.pptxStatistical Tests
Types of Statistical Tests
More powerful
Requires certain assumptions:
Normality
Homoscedasticity
(Equal variances)
If assumptions not fulfilled:
Try to transform
If still not fulfilled, use non-parametric tests
2
Statistical
test
Non-Parametric
Parametric
Testing for mean
Only one sample is taken
Involves one quantitative variable
The null hypothesis tests that the mean of a population parameter for a given variable is equal to one numeric value
Assumptions: We will assume the numeric variable is approximately normal.
One Sample T-test
Note: If assumption is not fulfilled, use median test
Example: It is believed that the mean age of smokers in San Bernardino is 47. Researchers from LLU believe that the average age is different than 47. In order to test this hypothesis the researchers took a sample of 20 random smokers and found that average age is 51 with a standard deviation of 10. What should they conclude?
H0: µ = 47
HA: µ ≠ 47
1.Confidence Interval Method
95% confidence interval is (46.32, 55.68)
Decision: Since the null value, 47, falls within the confidence interval, we fail to reject the null hypothesis.
Conclusion: We are 95% confident that the mean age of San Bernardino smokers was not significantly different from 47.
One Sample T-test
H0: µ = 47 years
α = 0.05
df = 19
X = 51
s = 10
n = 20
H0
CV
CV
Fail to Reject
2.093
-2.093
2. Test Statistic Method
Decision: Since the test statistic, 1.79 is in the fail to reject region, we fail to reject the null hypothesis
FTR
-2.093 2.093
ts= 1.79
α /2 = 0.025
p-value = Area beyond test statistic
3. P-Value Method
Decision: Since the p-value is > .05, we fail to reject the null hypothesis
Two Independent Sample T-test
Used when comparing the averages of two samples
Involves one quantitative and one qualitative variables where the qualitative variable has two categories
Assumptions:
the samples are randomly selected from normally distributed populations
the samples are selected in an independent manner
- variances are equal
Note: If assumption is not fulfilled, use Mann-Whitney Test
9
Example: The data below shows the average working hours per week for nurses and physicians at local hospital.
Are these averages significantly different at alpha of 0.05?
Solution:
OR
Degree of freedom independent sample t-test (df)
10
1a. Confidence Interval
Decision: Since the 95% includes the H0 value of zero, we will fail to reject the null hypothesis
Conclusion: The average hours per week of the nurses and physicians were not significantly different
11
1b. Confidence Interval
Nurses
Physicians
12
1b. Confidence Interval
Decision: Since the 95% overlap, we fail to reject the null hypothesis
Conclusion: The average hours per week of the nurses and physicians were not significantly different
Physicians
Nurses
13
2. Test Statistic
Decision: Since the test statistic is in the fail to reject region, we will fail to reject the null hypothesis
Conclusion: The average hours per week of the nurses and physicians are not significantly different
Note: CV = 2.074
FTR
ts CV
14
3. P-value
Decision: Since the p-value is greater than 0.05, we fail to reject the null hypothesis
Conclusion: The average hours per week of the nurses and physicians were not significantly different
Alpha = 0.05
15
Paired T-test
Also known as two dependent samples t-test
Used when comparing the averages of two samples
The two samples have to be dependent
Usually used in a before and after studies
Involves two quantitative variables that are dependent on each other
16
Paired T-test
Assumptions:
- The frequency distribution of the population of differences is approximately normal
Note:
- Typically, when both a pretest and a posttest are used, the same subjects are used in the study. Thus, this kind of sampling plan usually leads to dependent samples.
- If assumption is not fulfilled, use Wilcoxon Sign Test
17
Example: Salt-free diets are often prescribed for people with high blood pressure. The following data was obtained from an experiment designed to estimate the reduction in diastolic blood pressure as a result of following a salt-free diet for two weeks. Assume diastolic readings to be normally distributed. Is there a significant reduction in diastolic blood pressure.
Solution:
OR
18
Confidence Interval
Decision: Since the 95% CI of the mean difference includes the null value (Zero), we fail to reject the null hypothesis
Conclusion: There was no a significant reduction in diastolic blood pressure after following a salt-free diet for two weeks
Descriptive statistics for the difference
19
2. Test Statistic
Decision: Since the test statistic is in the fail to reject region, we will fail to reject the null hypothesis
FTR
CV
ts
Note: CV = 2.365
Conclusion: There was no a significant reduction in diastolic blood pressure after following a salt-free diet for two weeks
20
3. P-value
Decision: Since the p-value is greater than 0.05, we fail to reject the null hypothesis
Alpha = 0.05
Conclusion: There was no significant reduction in diastolic blood pressure after following a salt-free diet for two weeks
21
Source: Morton, D.P., et al., The effectiveness of the Complete Health Improvement Program (CHIP) in Australasia for reducing selected chronic disease risk factors: a feasibility study. N Z Med J, 2013. 126(1370): p. 43-54.
22
Source: den Otter, J.J., et al., How to avoid underdiagnosed asthma/chronic obstructive pulmonary disease? J Asthma, 1998. 35(4): p. 381-7.
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20
10
51
=
=
=
n
s
x
10
512.093514.68
20
m
=±=±
/2
S
xt
n
a
m
=±
n
s
x
t
s
m
-
=
79
.
1
20
10
47
51
=
-
=
/20.025
a
=
Employees
n
Average Hours/WeekStandard Deviation
Nurses1248.26.7
Physicians1244.12.3
0
:
nursesphysicians
H
mm
=
0
:0
nursesphysicians
H
mm
-=
12
(1)(1)
nn
=-+-
Sheet1
| Employees | n | Average Hours/Week | Standard Deviation |
| Nurses | 12 | 48.2 | 6.7 |
| Physicians | 12 | 44.1 | 2.3 |
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22
12
12
/2
12
22
()
6.72.3
(48.2-44.1)2.07394.14.24
1212
(-0.14, 8.34)
ss
xxt
nn
a
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±+=±
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12
(1)(1)(121)(121)22
dfnn
=-+-=-+-=
Sheet1
| Employees | n | Average Hours/Week | Standard Deviation |
| Nurses | 12 | 48.2 | 6.7 |
| Physicians | 12 | 44.1 | 2.3 |
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Employees
n
Average Hours/WeekStandard Deviation95% CI
Nurses1248.26.7
43.94,52.46
Physicians1244.12.3
42.64,45.56
6.7
48.22.201
12
48.24.26
(43.94,52.46)
m
m
m
=±
=±
=
2.3
44.12.201
12
44.11.46
(42.64,45.56)
m
m
m
=±
=±
=
Sheet1
| Employees | n | Average Hours/Week | Standard Deviation | 95% CI |
| Nurses | 12 | 48.2 | 6.7 | 43.94,52.46 |
| Physicians | 12 | 44.1 | 2.3 | 42.64,45.56 |
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Sheet1
| Employees | n | Average Hours/Week | Standard Deviation | 95% CI |
| Nurses | 12 | 48.2 | 6.7 | 43.94,52.46 |
| Physicians | 12 | 44.1 | 2.3 | 42.64,45.56 |
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12
12
22
22
12
12
()()
(48.244.1)0
4.1/2.042
6.72.3
1212
s
xx
t
ss
nn
mm
---
--
====
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+
+
ç÷ç÷ç÷ç÷
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Sheet1
| Employees | n | Average Hours/Week | Standard Deviation |
| Nurses | 12 | 48.2 | 6.7 |
| Physicians | 12 | 44.1 | 2.3 |
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Nurses1248.26.7
Physicians1244.12.3
P-value
0.12
Sheet1
| Employees | n | Average Hours/Week | Standard Deviation |
| Nurses | 12 | 48.2 | 6.7 |
| Physicians | 12 | 44.1 | 2.3 |
| P-value | 0.12 |
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Before9310687921029588110
After9210289921019688105
0
:
beforeafter
H
mm
=
0
:0
afterbefore
H
mm
-=
Sheet1
| Before | 93 | 106 | 87 | 92 | 102 | 95 | 88 | 110 |
| After | 92 | 102 | 89 | 92 | 101 | 96 | 88 | 105 |
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n = 8, x = −1.0, and s = 2.39
n=8, x=-1.0, and s=2.39
Before 93 106 87 92 102 95 88 110
After 92 102 89 92 101 96 88 105
Difference -1 -4 2 0 -1 1 0 -5
Before 9310687921029588110
After 9210289921019688105
Difference -1-420-110-5
2.39
1.02.3651.02
8
(1.0,3.0)
s
xt
n
±=-±=-±
-
Sheet1
| Before | 93 | 106 | 87 | 92 | 102 | 95 | 88 | 110 |
| After | 92 | 102 | 89 | 92 | 101 | 96 | 88 | 105 |
| Difference | -1 | -4 | 2 | 0 | -1 | 1 | 0 | -5 |
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1.00
1.18
2.39/8
s
x
t
sn
m
---
===-
Before9310687921029588110
After9210289921019688105
P-value difference0.09
Sheet1
| Before | 93 | 106 | 87 | 92 | 102 | 95 | 88 | 110 |
| After | 92 | 102 | 89 | 92 | 101 | 96 | 88 | 105 |
| P-value difference | 0.09 |
