CHP400: Community Health Program - lI Research Methodology. Data analysis Hypothesis testing Statistical Inference test t-test and 22 Test of Significance.
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CHP400:Community Health Program - lI
Research Methodology.
Data analysis
Hypothesis testing
Statistical Inference
testt-test and 2
Test of Significance
2testTest of Significance
Content
Definition of Hypothesis Statistical Hypotheses Hypothesis testing Performing the t – test Interpretation of the t-test The contingency table Performing the chi square test The assumptions / Limitations of the Chi square test Causal Associations
A tentative explanation for an observation, phenomenon, or scientific problem that can be tested by further investigation.
Something taken to be true for the purpose of argument or investigation.
A mere assumption or guess
Statistical HypothesisDefinition
Since that is often impractical, researchers typically examine a random sample from the population.
Statistical Hypotheses
The best way to determine whether a statistical hypothesis is true would be to examine the entire population.
If sample data are not consistent with the statistical hypothesis, the hypothesis is rejected.
Hypothesis testing is the use of statistics to determine the probability that a given hypothesis is true.
Hypothesis testing consists of four steps.
Statistical Hypothesis testing
1- Null hypothesis: H0 and Alternative hypothesis: HA
2- Test statistic:3- Compute the “-value”4- Compare the ()-value to an
acceptable significance value()
Statistical Hypothesis testing1- Null hypothesis: H0 and Alternative hypothesis: HA
Formulate the null hypothesis (commonly, that the observations are “not different” or” not associated”)
2- Test statistic:3- Compute the “P-value”4- Compare the ()-value to an acceptable significance value()
and the alternative hypothesis (commonly, that the observations show a “real difference” or “association”)
Statistical Hypothesis testing1- Null hypothesis: H0 and Alternative hypothesis: HA
2- Test statistic:
Identify a “statistical test” that can be used to assess the truth of the null hypothesis(2 test, t-test etc.); and calculate the “test statistic”3- Compute the “P-value”4- Compare the ()-value to an acceptable significance value()
Statistical Hypothesis testing1- Null hypothesis: H0 and Alternative hypothesis: HA
2- Test statistic:
3- Compute the “P-value”Compute the -value, which is the probability that a test statistic equals observed value or even more extreme would be obtained assuming that the null hypothesis were true. The smaller the -value, the stronger the evidence against the null hypothesis. 4- Compare the ()-value to an acceptable significance value()
Statistical Hypothesis testing1- Null hypothesis: H0 and Alternative hypothesis: HA
2- Test statistic:3- Compute the “P-value”
4- Compare the ()-value to an acceptable significance value()
Compare the -value to an acceptable significance value (sometimes called an alpha value). If , that the observed effect is statistically significant, the null hypothesis is ruled out, and the alternative hypothesis is valid
The t-test
SignificanceTest of Significance Test of
1. Test for single mean Whether the sample mean is equal to the
predefined population mean ? 2. Test for difference in means Whether “the mean energy expenditure ” in
obese women is equal to “the mean energy expenditure ” in lean women?
3. Test for paired observation Whether the treatment conferred any
significant benefit ?
The t-testTest of Significance
The t-testExamines the difference between means
2nd groupmean
Is there a difference?
1st groupmean
The t-testTest of Significance
mediumvariability
highvariability
lowvariability
The difference in mean is the same for
all three cases
What does difference mean?
Which one showsthe greatestdifference?
The t-testTest of Significance
Difference between means
Sample size
Variability of data
t-test t t ++
What does difference mean?
The t-testTest of Significance
a statistical difference is a function of the difference between means relative to the variability
=difference between group means
Variability of groups
=
=
X
1 2
X
X
1 2
X( )SE
t-value
=
X X
1 2
112pS n1 n2
What does difference mean?
The t-testTest of Significance
Given below are the 24 hrs total energy expenditure (MJ/day) in groups of lean and obese women. Examine whether the obese women’s mean energy expenditure is significantly higher ?.
Lean
6.1 7.2 7.5 10.9
8.5 5.5 7.6
7.9 9.1 8.1
8.3 8.4 10.2
t-test for difference in means
Obese 8.8 9.2 9.2 9.7 9.7 10.0 11.5 11.8 12.8
The t-testTest of Significance
Null Hypothesis( )
Obese women’s mean energy expenditure is equal to the lean women’s mean energy expenditure.
Data Summary
lean Obese
N 13 9
8.10 10.30
S 1.38 1.25
Xt-test for difference in means
Solution
H0
The t-testTest of Significance
= 0.05df = 13 + 9 - 2 = 20Critical Value(s):
X X
X X
Solution
t-test for difference in means
H0 : 1 - 2 = 0
HA : 1 - 2 ≠ 0
The t-testTest of Significance
α (2 tail) 0.1 0.05 0.02 0.01 0.005 0.002 0.001
df
1 6.3138 12.7065 31.8193 63.6551 127.3447 318.4930 636.0450
2 2.9200 4.3026 6.9646 9.9247 14.0887 22.3276 31.5989
3 2.3534 3.1824 4.5407 5.8408 7.4534 10.2145 12.9242
4 2.1319 2.7764 3.7470 4.6041 5.5976 7.1732 8.6103
17 1.7396 2.1098 2.5669 2.8983 3.2224 3.6458 3.9651
18 1.7341 2.1009 2.5524 2.8784 3.1966 3.6105 3.9216
19 1.7291 2.0930 2.5395 2.8609 3.1737 3.5794 3.8834
20 1.7247 2.0860 2.5280 2.8454 3.1534 3.5518 3.8495
21 1.7207 2.0796 2.5176 2.8314 3.1352 3.5272 3.8193
49 1.6766 2.0096 2.4049 2.6800 2.9397 3.2651 3.5004
50 1.6759 2.0086 2.4033 2.6778 2.9370 3.2614 3.4960
51 1.6753 2.0076 2.4017 2.6757 2.9343 3.2579 3.4917
106 1.6593 1.9826 2.3620 2.6230 2.8670 3.1689 3.3847
107 1.6592 1.9824 2.3617 2.6225 2.8664 3.1681 3.3838
149 1.6551 1.9760 2.3516 2.6092 2.8494 3.1458 3.3570
150 1.6551 1.9759 2.3515 2.6090 2.8491 3.1455 3.3565
151 1.6550 1.9758 2.3513 2.6088 2.8489 3.1451 3.3561
199 1.6525 1.9720 2.3452 2.6008 2.8387 3.1317 3.3401
200 1.6525 1.9719 2.3451 2.6007 2.8385 3.1315 3.3398
t-test for difference in means
The t-testTest of Significance
= 0.05df = 13 + 9 - 2 = 20Critical Value(s): (tab t 9+13-2 = 20 df = t 0.05,20 = 2.086)
X X
X X
Solution
t-test for difference in means
H0 : 1 - 2 = 0
HA : 1 - 2 ≠ 0
(2.086)
The t-testTest of Significance
Calculating the Test Statistic:
=
X X
1 2
11
n1
2pS
t
S2
p=S2
1
(n1-1)+(n2-1)
(n1-1)+ (n2-1)2
S2p
S22
S21
n1
n2
:pooled-variance
: variance of sample 1
: variance of sample 2
:size of sample 1
:size of sample 2
n2
t-test for difference in means
S2
The t-testTest of Significance
S2p=
S21
(n1-1)+(n2-1)
(n1-1)+ (n2-1)S22
First, estimate the common variance as a weighted average of the two sample variances using the degrees of freedom as weights
Calculating the Test Statistic:
= (13-1)+(9-1)= 1.767
t-test for difference in means
1.38 2(13-1)+1.25 2(9-1)
The t-testTest of Significance
Calculating the Test Statistic:
=
X X
1 2
11
n1
2pS
t
=8.1 – 10.3
1.767
n2
1 1+
13 9_ _
3.82=
t-test for difference in means
The t-testTest of Significance
Inference: The calculated t (3.82) is higher than table t (at 0.05, 20. ie 2.086) . This implies that there is an evidence that the mean energy expenditure in obese group is significantly (p<0.05) higher than that of lean group.
t-test for difference in means
2testTest of Significance
A physician wants to know whether the proportion of esophageal cancer diagnosed as multiple primary tumors (MPT) differs between females and males. He selected a random sample of 100 cases of esophageal cancer, 60 were males and 40 were females. Of the 60 male esophageal cancer patients, 40 were diagnosed as MPT. Of the 40 female esophageal cancer patients, 10 were diagnosed as MPT.
Chi square test
2testTest of Significance
Exposure(estrogen)
Outcome (cancer)Total
Yes No
Yes
No
Total
Out of 30 women who had uterine cancer, 20 claimed to have used estrogens. Out of 30 women without uterine cancer 10 claimed to have used estrogens.
Chi square test
2testTest of Significance
Exposure(estrogen)
Outcome (cancer)Total
Yes No
Yes 20 20 10 10 3030
No 10 10 20 20 3030
Total 3030 3030 6060
Out of 30 women who had uterine cancer, 20 claimed to have used estrogens. Out of 30 women without uterine cancer 10 claimed to have used estrogens.
Chi square test
2testTest of Significance
A: Calculating 2 value1. Calculate the expected frequency (E) for each
cell
2. For each cell subtract the expected frequency from the observed frequency (O):
3. For each cell, square the result of (O-E) and
divide by
the expected frequency (E):
StepsChi square test
row total x column total
total number of observation
E =
O – E
(O – E)E
2testTest of Significance
Exposure(estrogen)
Outcome(cancer)Total
Yes No
Yes 20 20 (15) 10 10 (15) 3030
No 10 10 (15) 20 20 (15) 3030
Total 3030 3030 6060
Out of 30 women who had uterine cancer, 20 claimed to have used estrogens. Out of 30 women without uterine cancer 10 claimed to have used estrogens.
StepsChi square test
2testTest of Significance
4. Add the results of step (3) for all the cells Formula: 2 = [ ( O - E )2 / E] = [(20-15)2/15] + [(10-15) 2//15] +
[(10-15) 2/15] + [(20-15) 2/15]
= 1.67 + 1.67 + 1.67 + 1.67 = 6.68B: Deciding p value 1. Calculate degrees of freedom (r-1) (c-1)
df = (2-1) (2-1) = 1 x 1 =1
2. Looking at 2 table
StepsChi square test
2testTest of Significance
df 0.10 0.05 0.025 0.01 0.005
1 2.706 3.841 5.024 6.635 7.879
2 4.605 5.991 7.378 9.210 10.597
3 6.251 7.815 9.348 11.345 12.838
4 7.779 9.488 11.143 13.277 14.860
22 30.813 33.924 36.781 40.289 42.796
23 32.007 35.172 38.076 41.638 44.181
24 33.196 36.415 39.364 42.980 45.559
30 40.256 43.773 46.979 50.892 53.672
40 51.805 55.758 59.342 63.691 66.766
50 63.167 67.505 71.420 76.154 79.490
60 74.397 79.082 83.298 88.379 91.952
100 118.498
124.342 129.561 135.807 140.169
StepsChi square test
2testTest of Significance
StepsChi square test
Inference: The calculated 2 (6.68) is higher than table 2(at 0.05, 1. ie 3.841) . This implies that there is an evidence that there is association between use of estrogens and uterine cancer
2testTest of Significance Chi square test
Data is from a random sample. Data is from a random sample.
A sufficiently large sample size is A sufficiently large sample size is required (at least 20) required (at least 20)
Actual count data (not percentages) Actual count data (not percentages)
All expected counts are All expected counts are ≥1≥ 80%expected counts are expected counts are ≥5
Observations must be independent. Observations must be independent.
Does not prove causality.Does not prove causality.
Assumptions / Limitations
2testTest of Significance
Causal Associations
Relationship between variables
Relationship between variables
Not statistically associated
Not statistically associated
Non-causalNon-causal CausalCausal
Indirectly causal
Indirectly causal
Statistically associated
Statistically associated
Directly causal
Directly causal
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