Test of significance and qualitative tests

1. Tests of significance
Qualitative data
Z & Chi square Test
Dr. Shaikh B.M.
JRII
Dept. of Community Medicine

2. outline
Basic terms (recap)……….
Sampling Variation,
Null hypothesis,
Level of significance and confidence,
P value, power of the test, degree of freedom
Tests of significance and type
Selection of the test of significance
Steps in hypothesis testing
Z test (SEDP)
Chi Square test
Limitations of the tests of significance
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3. Sampling Variation
• Research done on samples and not on populations.
• Variability of observations occur among different samples.
• This complicates whether the observed difference is due to
biological or sampling variation from true variation.
• To conclude actual difference, we use tests of significance
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4. Test of significance
• The test which is done for testing the research hypothesis
against the null hypothesis.
Why it is done?
 To assist administrations and clinicians in making decision.
 The difference is real ? or
 Has it happened by chance ?
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5. Null Hypothesis (HO)
• 1st step in testing any hypothesis.
• Set up such that it conveys a meaning that there exists no
difference between the different samples.
• Eg: Null Hypothesis – The mean pulse rate among the two
groups are same (or) there is no significant difference between
their pulse rates.
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6. • By using various tests of significance we either:
–Reject the Null Hypothesis
(or)
–Accept the Null Hypothesis
• Rejecting null hypothesis → difference is significant.
• Accepting null hypothesis → difference is not significant.
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7. Level of significance and confidence
• Significance means the percentage risk to reject a null
hypothesis when it is true and it is denoted by 𝛼.Generally
takenas1%,5%,10%
• (1 − 𝛼) is the confidencelevelin which the null hypothesis will
exist when it is true.
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8. Level of Significance – “P” Value
• p-value is a function of the observed sample results
(a statistic) that is used for testing a statistical hypothesis.
• It is the probability of null hypothesis being true. It can accept
or reject the null hypothesis based on P value.
• Practically, P < 0.05 (5%) is considered significant.
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9. • P = 0.05 implies,
– We may go wrong 5 out of 100 times by rejecting null
hypothesis.
– Or, We can attribute significance with 95% confidence.
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10. 5% Significance level & 95% confidence level
Acceptance and Rejection
regions
𝑅𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛
/𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙
(𝛼 = 0.025 𝑜𝑟 2.5%)
𝑅𝑒𝑗𝑒𝑐𝑡𝑖𝑜𝑛 𝑟𝑒𝑔𝑖𝑜𝑛
/𝑠𝑖𝑔𝑛𝑖𝑓𝑖𝑐𝑎𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙
(𝛼 = 0.025 𝑜𝑟 2.5%)
𝑇𝑜𝑡𝑎𝑙 𝐴𝑐𝑐𝑒𝑝𝑡𝑎𝑛𝑐𝑒 𝑟𝑒𝑔𝑖𝑜𝑛
𝑜𝑟
𝑐𝑜𝑛𝑓𝑖𝑑𝑒𝑛𝑐𝑒 𝑙𝑒𝑣𝑒𝑙
(1 − 𝛼) = 95%
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11. Power of the test &
Degree of freedom
Power of the test :
Type II error is β and 1−β is called power of the test.
Probability of rejecting False H0, i. e. taking correct decision
Degree of freedom:
• Number of independent observations used in statistics (d. f.)
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12. Various tests of significance
1. Parametric – Data is or assumed to be normal distributed.
2. Non parametric – Data is not normal distributed.
 Parametric tests :
For Qualitative data:-
1. Z test
2. 2. Chi-square test or X2
For Quantitative data:-
1. Unpaired ‘t’ test
2. Paired ‘t’ test
3. ANOVA
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13. Selection of the test
• Base
1. Type of data
2. Size of sample
3. Number of samples
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14. Selection of the test
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Qualitative
Data
n > 30
Large
Sample
1, 2 samples
Z test
n </> 30
Small / large
Sample
More than 2
samples
Chi square
test
Quantitative
Data
n </> 30
Small / large
Sample
1 samples
One sample
t-test
2 samples
Unpaired t-
test / Paired t-
test
More than 2
samples
ANOVA

15. Steps in Testing a Hypothesis
• General procedure in testing a hypothesis
1. Set up a null hypothesis (HO).
2. Define alternative hypothesis (HA).
3. Calculate the test statistic (Z, X2, t etc.).
4. Find out the corresponding probability level (P Value) for
the calculated test statistic from relevant tables.
5. Accept or reject the Null hypothesis depending on P
value.
P> 0.05 HO accepted
P < 0.05 HO rejected
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16. Z test
[Standard error of difference between proportions (SEp1-p2)]
• For comparing qualitative data between 2 groups.
• Used for large samples only. (> 30 in each group).
• Standard error of difference between proportions (SEp1-p2) is
calculated by using the formula
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17. • And then the test statistic ‘Z Score’ value is calculated by
using the formula:
• If Z > 1.96 P < 0.05 SIGNIFICANT - H0 rejected
• If Z < 1.96 P > 0.05 NOT SIGNIFICANT- H0 accepted
• Finding p value from Z table
• Go to the row that represents the ones digit and the first digit
after the decimal point (the tenths digit) of your z-value.
• Go to the column that represents the second digit after the
decimal point (the hundredths digit) of your z-value.
• Intersect the row and column from Steps 1 and 2.
1.0000 - 0.9772
=0.0228
< 0.05
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18. Example
• Consider a hypothetical study where cure rate of Typhoid fever
after treatment with Ciprofloxacin and Ceftriaxone were
recorded to be 90% and 80% among 100 patients treated with
each of the drug.
• How can we determine whether cure rate of Ciprofloxacin is
better than Ceftriaxone?
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19. Solution
Step – 1: Set up a null hypothesis – H0:
– “There is no significant difference between cure rates of
Ciprofloxacin and Ceftriaxone.”
Step – 2: Define alternative hypothesis – Ha:
– “Ciprofloxacin is 1.125 times better in curing typhoid fever
than Ceftriaxone.”
Step – 3: Calculate the test statistic – ‘Z Score’
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20. Step – 3: Calculating Z score
– Here, P1 = 90, P2 = 80
– SEP1-P2 will be given by the formula:
So, Z = 10/5 = 2
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21. Step – 4: Find out the corresponding P Value
– Since Z = 2 i.e., > 1.96, hence, P < 0.05
Step – 5: Accept or reject the Null hypothesis
– Since P < 0.05, So we reject the null hypothesis(H0)
There is no significant difference between cure rates of
Ciprofloxacin and Ceftriaxone
– And we accept the alternate hypothesis (Ha)
that, Ciprofloxacin is 1.125 times better in curing typhoid
fever than Ceftriaxone
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22. Chi square (X2) test
• This test is also for testing qualitative data.
• Its advantage over Z test is:
– Can be applied for smaller samples as well as for large
samples.
• Prerequisites for Chi square (X2) test to be applied:
– The sample must be a random sample
– None of the observed values must be zero.
– Adequate cell size
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23. Steps in Calculating (X2) value
1. Make a contingency table mentioning the frequencies in
all cells.
2. Determine the expected value (E) in each cell.
3. Calculate the difference between observed and expected
values in each cell (O-E).
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24. 4. Calculate X2 value for each cell
5. Sum up X2 value of each cell to get X2 value of the table.
6. Find out p value from x2 table
7. If p > 0.05 , difference is not significant, null hypothesis
accepted;
If p < 0.05 , difference is significant , null hypothesis rejected.
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25. Example
• Consider a study done in a hospital where cases of breast
cancer were compared against controls from normal
population against with a family history of Ca Breast.
• 100 in each group were studied for presence of family history.
• 25 of cases and 15 among controls had a positive family
history.
• Comment on the significance of family history in breast
cancer.
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26. Solution
• From the numbers, it suggests that family history is 1.66
(25/15) times more common in Ca breast.
• So is it a risk factor in population?
• We need to test for the significance of this difference.
• We shall apply X2 test.
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27. Solution
Step – 1: Set up a null hypothesis
– H0: “There is no significant difference between incidence of
family history among cases and controls.”
Step – 2: Define alternative hypothesis
– Ha: “Family history is 1.66 times more common in Ca
breast”
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28. Step – 3: Calculating X2
1. Make a contingency table mentioning the frequencies in all
cells
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29. Step – 3: Cont..
2. Determine the expected value (E) for each cell.
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30. O – observed values , E – expected value
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31. Step – 4: Determine degree of freedom.
• DoF is determined by the formula:
DoF = (r-1) x (c-1)
where r and c are the number of rows and columns
respectively
• Here, r = c = 2.
• Hence, DoF = (2-1) x (2-1) = 1
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32. Step – 5: Find out the corresponding P Value
– P values can be derived by using the X2 distribution
tables
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33. Step – 6: Accept or reject the Null hypothesis
• In given scenario,
X2 = 3.125
• This is less than 3.84 (for P = 0.05 at dof =1)
• Hence Null hypothesis is Accepted, i.e.,
“There is no significant difference between incidence of
family history among cases and controls”
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34. Limitations of tests of significance
 Testing of hypothesis is not decision making itself; but it helps
for decision making
 Test does not explain the reasons as why the difference
exists, tests do not tell about the reason causing the difference.
 Tests are based on the probabilities and as such can not be
expressed with full certainty.
 Statistical inferences based on the significance tests can
not be said to be entirely correct evidences concerning
the truth of the hypothesis.
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35. THANK YOU
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