Chi Square Test

Inferential Statistics Details of t-test & chisquare test

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INTRODUCTION 

T-TEST - Numerical Data - Comparing the Mean of two groups

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CHI-SQUARE - Categorical Data

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BOTH TESTS ARE USED FOR UNPAIRED OBSERVATION 2

T - TEST Also referred as Student t-test Suppose we have taken a sample of 100 women and found mean Hb as 11.4 gm. Now the question is whether this sample is from a population whose standard deviation is 2 gm and mean Hb is 12 or not?

What should we be looking at?  Is there a difference b/w population mean and sample mean?  In this case difference is there  Again the question is; Is this difference significant ? Or it is due to chance alone 

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 

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Of course the difference should be large enough to be called significant Now the difference can be written as Sample Mean – population mean What other factor can effect the importance of the difference ? If there is great variation in the population (Standard deviation with small sample size) the difference would not be of that much importance If there is small variation (standard deviation with large sample size) a slight degree of difference can become important 4

So we should add this factor as well to decide about the importance of difference  The

mean

formula becomes Sample Mean – Population Standard deviation/√N

And this is given the name of t test

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Hence the t test = T= mean

Sample Mean – population

Standard deviation/√N Putting the values in the formula t = 11.4 -12 = -0.6 2/√100 2/10 Calculated value of t = - 0.6/0.2 = - 3.0 Critical value (cut off) = ± 2.00 What should be the conclusion Reject Ho or not As the calculated value is more than critical we reject chance explanation and conclude that the sample is not from a population whose mean Hb level is 12 gm. 6

Example 2 : In a certain province the proportion of women who are delivered through caesarian section is very high. A study is, therefore, conducted to discover why this is the case. As small height is known to be one of the risk factors related to difficult deliveries, the researcher may want to find out if there is a difference between the mean height of women in this province who had normal deliveries and of those who had Caesarian sections.

1. Calculating the T-Value 2. Using a T-Table 3. Interpreting the Result 7

USING THE T-TABLE To determine if null hypothesis is rejected or not  

Significance level (p-value) - 0.05 Degree of freedom - measure derived form the sample size - student t-test the number of degree of freedom is calculated as the sum of two sample sizes minus 2 d.f. = 60 + 52 - 2 = 110

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t-value belonging to the p-value and the degree of freedom is located in the table - t-value belonging to p -0.05 and d.f. = 120 and we find it is 1.98

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INTERPRETING THE RESULT  

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Decision Rule Calculated t-value is larger than the value derived from the table. p is smaller than the value indicated at the top of the column. We than reject the null hypothesis and conclude a statistical significant difference b/w the two means. Calculated t-value is smaller, p is larger than the value indicated, we accept null hypothesis and conclude observed difference is not statistically significance.

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INTERPRETING THE RESULT t-value calculated in step 1 is 3.6 larger than t-value derived in step 2 is (1.98)  p is smaller than 0.05, we reject the null hypothesis  Observed difference of two cms b/w mean heights of women with normal del. and CS is statistically significant difference. 

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CHI-SQUARE (x ) TEST 2

Suppose you are presented with two vaccines A and B for the prevention of measles. Both the manufacturer claim that their vaccine is better. How would you resolve this issue? If we want to decide the credibility of vaccine objectively. You would ask both the manufacture to provide some vaccine to conduct a clinical trial. 100 randomly selected children were given vaccine A and same number of children were vaccinated with B. Out of those vaccinated by A fifteen (15) developed infection 20 children vaccinated by B developed infection. Which vaccine is better? 11

If Vaccine A Then Producer of vaccine B may claim that this difference is by chance. However the producer of vaccine A will call this difference as significant. The dispute will not be resolved even if the difference is large enough until a cut off value is decided by some neutral person and that neutral umpire is test of significance.

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By putting information from the data following 2x2 table can be filled

Type of va 13

By putting information from the data following 2x2 table can be filled

Type of va 14

By putting information from the data following 2x2 table can be filled

Type of vac 15

By putting information from the data following 2x2 table can be filled

Type of vac 16

 According

to null Hypothesis

If

both vaccine had same effect then we should expect the same number of infection in both the groups. Accordingly the following table shows the shape of the table we should expect if null hypothesis is true

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Type of v A

If both vaccine same then half of infection will be in vaccine A and Half in vaccine B and the table will look like this 18

Type of va Alternatively we can determine the expected value in cell a by the principle of unit. i.e. Out of 200 the number of infection are 35 Out of 1 the number of infection will be 35/200 Out of 100 the number of infection will be

A

35 x 100

= 17.5

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If you look care fully then you will observe that in the first step we have divided the coloumn total of cell a with grand total and then multiplied it with row total of cell a. So this two step procedure for expected value can be written as a single step Expected value(EV) of cell a = Column total x Row total Grand Total

Thus we can use this formula to calculate expected value of any cell Now we fill the table with all expected values 20

Type of va According to null hypothesis we should not find any difference in observed value in cells a, b, c, and d. and expected value. Here we find some difference. This difference may be small (insignificant) or large (significant). Hence the next logical step should be that we should subtract expected value from observed value to find the difference. We would get a table like this

A

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Cell

Observed Expected Value Value O E

a

15

b

85

c

20

d

80

O – E (O – E)2 (O – E)2/ E

Total

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Total

Cell

Observed Expected Value Value O E

a

15

17.5

b

85

82.5

c

20

17.5

d

80

82.5

O – E (O – E)2 (O – E)2/ E

Total

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Total

Cell

Observed Expected Value Value O E

O – E (O – E)2 (O – E)2/ E

a

15

17.5

-2.5

b

85

82.5

2.5

c

20

17.5

2.5

d

80

82.5

-2.5

Total

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Total

Cell

Observed Expected Value Value O E

O – E (O – E)2 (O – E)2/ E

a

15

17.5

-2.5

6.25

b

85

82.5

2.5

6.25

c

20

17.5

2.5

6.25

d

80

82.5

-2.5

6.25

Total

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Total

Cell

Observed Expected Value Value O E

O – E (O – E)2 (O – E)2/ E

a

15

17.5

-2.5

6.25

6.25/ 17.5

b

85

82.5

2.5

6.25

6.25/ 82.5

c

20

17.5

2.5

6.25

6.25/ 17.5

d

80

82.5

-2.5

6.25

6.25/82.5

Total

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Total

Cell

Observed Expected Value Value O E

O – E (O – E)2 (O – E)2/ E

Total

a

15

17.5

-2.5

6.25

6.25/ 17.5

0.36

b

85

82.5

2.5

6.25

6.25/ 82.5

0.07

c

20

17.5

2.5

6.25

6.25/ 17.5

0.36

d

80

82.5

-2.5

6.25

6.25/82.5

0.07

Total = 0.86

Prof. Dr. M. I. Siddiqui – Clinision Workshop

Table 4.12 D.F .0.50 1 0.45 2 1.39 3 2.37 4 3.36 5 4.35 6 5.35 7 6.35 8 7.34 9 8.34 10 9.34

.10 2.74 4.61 6.25 7.78 9.24 10.65 12.02 13.36 14.68 15.99

.050. 020 3.84 5.41 5.99 7.82 7.82 9.84 9.49 11.67 11.07 13.39 12.59 15.03 14.07 10.62 15.51 18.17 16.92 19.68 18.31 21.16

.010 6.04 9.21 11.34 13.28 15.09 16.81 18.48 20.09 21.67 23.21

.0030 7.68 10 12.84 14.36 16.75 18.55 20.28 21.96 23.59 25.19

.001 10.83 13.62 16.27 18.47 20.31 22.46 24.32 26.13 27.88 29.59

( Source. JE Park K. Park, Text Book of Preventive and social Medicine) 28

Calculated value of difference = 0.86 Critical value = 3.84 from the table of Chi square at 5% level of significance Decision ? As the difference is less than the minimum difference to be called significant. We conclude that there is no significant difference between two vaccine and we fail to reject null hypothesis. In simple English both the vaccine are of same efficacy and the difference is by chance and we can decide to buy the one which is cheaper.

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CHI-SQUARE (x ) TEST 2

Example: Suppose that a study of the factors affecting the utilization of antenatal clinics you found that 51 women out of 80 who lived within 10 km of the clinic came for antenatal care, compared to only 35 out of 75 of those who lived more than 10 km away. This suggests that antenatal care (ANC) is used more often by women who live close to the clinics.

Distance from 1) Calculate the x2 value 2) Using a x2 Table 3) Interpreting the Result 30

Distance ANC 31

By putting expected value of cell from our calculation

Distance fr ANC 32

If there are 155 user then the ANC was used by 86 If there was 1 user then ANC was used by 86/155 If there were 80 people then the ANC was used by 80X86/155 (equation 1) 44.4 We can write this expression as 80 = row total (RT) 86 = coloumn Total (CT) 155 = Grand total (GT) So we can write equation one as follow EV= RTXCT/GT Hence we can calculate expected value of any cell with this expression or formula

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CALCULATE X2 VALUE 

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Expected frequency (E) for each cell. E = row total x column total / grand (overall) total Each cell, subtract the expected frequency from the observed frequency (O) O-E For each cell square the result of (O-E) and divide by expected frequency E. Add the result of the above step for all the cells

34 Cont…

CALCULATE X2 VALUE 

Formula for calculating chi-square value: x2 = ∑ ( 0 - E)2 / E

O is the observe frequency (indicated in the table) E is the expected frequency to be calculated ∑ (the sum of) directs you to add together the products of (O-E)2 for all the cell of the table

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For two by two table (which contain 4 cells) the formula is

x2 = [(01 - E1)2 / E1] + [(02 - E2)2 / E2] + [(03 - E3)2 / E3] + [(04 - E4)2 / E4] 35

USING X2 TABLE Decide a p-value example 0.05  Degree of freedom df = (r-1) x (c-1) 

for a 2 by 2 table the no. of d.f. is 1 (i.e. d.f. = (2-1) x (2-1) = 1)

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INTERPRETING THE RESULT null hypothesis is rejected if < 0.05 which is the case if x2 is larger than theoretical x2 in table 

Step 1 (a) expected frequency for each cell E1 = 86 x 80 / 155 = 44.4

Distance fr 37 Cont…

INTERPRETING THE RESULT Step 1(b) to (1d) x2 = (51- 44.4)2 / 44.4 + …… + …… = 0.98 + 1.22 + 1.05 + 1.30 = 4.55  Step 2 

- (d.f.) is 1 - table of chi-square decided p-value = 0.05 - d.f. is 1, we look along row in the column where p=0.05. This gives us value of 3.84. Our value of 4.55 is > 3.84, which means that the p-value is < 0.05 38 Contd..

INTERPRETING THE RESULT  Step

3 We can now conclude that the women living within the distance of 10 km from the clinic used antenatal care significantly more often than women living more than 10 km away. 39

INTERPRETATION OF EXAMPLE 

64% of the women living within a distance of 10 km from the clinic used antenatal care during pregnancy, compared to only 47% of women living 10 km or further away from the nearest clinic. This difference is statistically significant (x2 = 4.55; - < 0.05)

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NOTE •

x2 test applied if the sample is large enough

•

General rule total should be at least 40 and the expected frequency in each of the cells should be at least 5.

•

x2 test can be used to compare more than two groups

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A table with 3 or more rows or columns would be designed rather than a two by two tables

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In above example < 5 km, 5 - 10 km, > 10 km data would be put in 2 x 3 tables.

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d.f. would be (3 -1) x (2 - 1) = 2 41

QUICK FORMULA 

For two-by-two tables there is a quick method for calculating the chi-square value, which can replace step 1 described above. If the various numbers in the costs table are represented by the following letters Condition Total +

-

Exposure

a

b

e

Yes

c

d

f

No

g

h

n

The quick formula for calculating the Chi-square value is

x2 = n (ad - bc)2 / efgh 42

Objectives By the end of session the participant will be able to Define test of significance Name test of significance according to type of data Able to calculate T test Able to compute X2 Interpret the result Interpret statistical results in simple English language 43

Tutorial Exercise for test of significance A surgeon wants to compare the two surgical procedure for hysterectomy, Abdominal Hysterectomy (AH) and Vaginal Hysterectomy (VH) with a research question of infection rate between two surgical procedures. She operates on 400 patient with VH approach and 40 developed infection. Out of 100 AH cases 20 developed infection Answer the following questions

a) What should be the title of research for publication in a journal? b) Frame Null hypothesis in this case? c) What level of significance you will set?. d) What test of significance you will apply and why? e) Calculate test statistics f) Interpret your result in statistical language g) Describe the result in simple English

Home Assignment Home Assignment Do exercise 12.1, 12.2 , 12.3 12.10, 12.13, 12.19

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