Basic Concepts of Medical Statistics 10-6-2016 · Basic Concepts of Medical Statistics Section for Clinical Epidemiology Faculty of Medicine Ramathibodi Hospital Mahidol University

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Basic Concepts of Medical Statistics

Section for Clinical EpidemiologyFaculty of Medicine Ramathibodi Hospital

Mahidol University

Outline

• Types of data• Descriptive statistics• Inferential statistics & Hypothesis testing• Statistical analysis for categorical data• Statistical analysis for continuous data• Statistical software

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Types of Data

Sample of Case Report Form from a clinical study

Perspect Clin Res. 2014 Oct‐Dec; 5(4): 159–166.

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Sample of Data obtained from clinical study

Types of Data

Types of Data

Categorical (Qualitative)

Nominal Ordinal

Numerical (Quantitative)

Discrete Continuous

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Categorical data

• Categories1. Nominal

• Without ‘order’• Blood group – A / B / AB / O• Sex – Male / Female• Cancer status – Have cancer / Do not have cancer

• Special case of 2 choices: binary/dichotomous data

2. Ordinal• With ‘order’ – we can see which is bigger/smaller

• Cancer staging – I / II / III / IV• Degree of disease – Mild / Moderate / Severe

Numerical data

1. Discrete• Stepping, usually without fractions

• Number of positive lymph nodes after surgery• Number of heart beats per minute

2. Continuous• Age (years)• Cholesterol level (mg/dL)

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Importance of types of data

• Types of data are important in several aspects of choosing the right statistical analysis

• Summarizing• Estimation• Hypothesis testing

Post test for types of data

Data Types of data

Weight (kg)

History of smoking (Y/N)

Serum potassium level

Length of stay in hospital (day)

Patient status (dead/alive)

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Post test for types of data

Data Types of data

Weight (kg)

History of smoking (Y/N)

Serum potassium level

Length of stay in hospital (day)

Patient status (dead/alive)

Numerical

Categorical (dichotomous)

Numerical

Numerical

Categorical (dichotomous)

Descriptive Statistics

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Descriptive statistics

• “Summaries of data”• Methods for describing a set of data by using graphs or summary measures

• Provide an ‘overview’ of general features of a set of data

Descriptive statistics

Summarizing Data

Categorical Data

‐ Frequency (n)

‐ Percentage (%)

Numerical Data

‐Measures of central tendency

‐Measures of dispersion

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Summarizing Categorical Data

• Ex1. Summarizing of sex among 70 patients

Sex Frequency Percentage

Male 56 80%

Female 14 20%

Total 70 100%

Summarizing Categorical Data

• Ex2. Summarizing of staging of cancers

Stages of cancers Frequency Percentage

I 120 15%

II 320 40%

III 160 20%

IV 200 25%

Total 800 100.0%

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Summarizing Numerical Data

Summarizing Numerical

Data

Measures of Central Tendency

Mean Median

Measures of Dispersion

Standard Deviation

Range

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Measures of Central Tendency

• Describe where the data are.

1. Mean2. Median

Measures of Dispersion

• Describe the spreading of data

1. Standard Deviation (SD)

2. Range, Interquartile Range (IQR)

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Distribution of Data

• Normal• Central tendency: use mean• Dispersion: use standard deviation

• Non-normal distribution• Central tendency: use median• Dispersion: use range

Checking for normal distribution

There are many ways of checking for distribution of the data, such as:

• Histogram visualization• Comparison of mean and median• Comparison of mean and standard deviation

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Histogram

Comparison between mean and median

• Mean 62.4• Median 30.5

• Distribution of CD4 data is skewed to the right (mean > median)

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Compare mean and standard deviation

Mean SD Mean-2SD, Mean+2SD

Height (cm) 164.7 7.8 149.1, 180.3

CD4 count 62.4 74.4 -86.4, 211.2

Summarizing Numerical Data

Mean (SD)

Age (year) 49.6 (14.3)

Weight (kg) 95.6 (21.7)

Height (cm) 171.5 (9.2)

BMI 32.5 (7.1)

CD4 62.4 (74.4)

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Inferential Statistics

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• Process of drawing conclusions about a population on the basis of the information contained in a sample taken from that population.

Population Parameters

Sample Statistics

Usefulness of Inferential Statistics

• Estimation of Population Parameter from Sample Statistics• Test hypothesis of Population from Sample

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Parameter Estimation

Point estimate

Range estimate

Hypothesis Testing

Single pop.

Two pop.

More than two pop.

Parameter estimation

• Estimation of population parameter based from sample statistics• For example:

• Estimation for continuous data• Estimation of the mean age of all patients who have breast cancer in Thailand.

• Estimation for categorical data• Estimation of the prevalence of patients with diabetes mellitus (DM) in the Thai

population.

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Example: Estimation

• Researcher would like to estimate the prevalence of hypertension in the Thai population.

• A random sample of 350 was taken from the general population. Forty-two out of 350 people in the sample had hypertension.

• Therefore, the point estimate of prevalence = 42/350 = 0.12 = 12%

Confidence interval

• The main purpose of confidence intervals is to indicate the precision of the point estimates.

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Confidence interval

• Prevalence of HT in 350 patients is 12% with 95% CI of 9% to 15%. • This means that the true prevalence of HT of population lies somewhere in range 47 to 62 years.

Confidence Interval

Sample Size Mean age 95% CI

30 54 15 .. 80

500 55 50 .. 58

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Categorical data

Population proportion

Sample proportion

Continuous data

Population mean

Sample mean

Example: Prevalence of Disease

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• Researcher would like to estimate the prevalence of hypertension in the Thai population.

• A random sample of 350 was taken from the general population. Forty-two out of 350 people in the sample had hypertension.

• Point estimate from sample proportion = 42/350 = 12%

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Solution:

a) The point estimate of the proportion is 42/350=0.12 or 12%

b) The Z value for a 95% CI is 1.96.

c) The SE of sample proportion is

SE = /

SE = . . / .

. . . to . . .

That is from 0.09 to 0.15 or 9% to 15%.

95% CI for the proportion is

Point estimate

Conclusion:

• Therefore, we can be 95% confident from this sample that the true prevalence of hypertension in Thailand lies somewhere in the range from 9% to 15%.

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Categorical data

Population proportion

Sample proportion

Continuous data

Population mean

Sample mean

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-Researcher would like to estimate the mean age of patients with DM in Bangkok.-A random sample of 750 patients with DM was taken and the mean age of the patients in the sample was estimated.-The mean age of the patients in the sample was 54.42 with a standard deviation of 12.93.

Example:

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Solution:

a) The point estimate of the mean age is 54.42. b) For a 95% CI, the value of t with 749 df is 1.647.

c) The SE of sample mean is

SE = / = . / = 0.47

Therefore, a 95% CI for the mean age is

54.42-(1.647x0.47) to 54.42+(1.647x0.47),

that is from 53.65 to 55.19.

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Therefore, we can be 95% confident from this sample that the true mean age of patients with DM in Bangkok lies somewhere in the range 53.65 to 55.19 years.

Conclusion:

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Recommendation

If the primary objective is to estimate the population characteristics, the point estimates should be reported with their confidence intervals to indicate their precision.

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Parameter Estimation

Point estimate

Range estimate

Hypothesis Testing

Single pop.

Two pop.

More than two pop.

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According to a survey study in Thailand in 2000, the proportion of people with chronic kidney disease (CKD) is 0.091 or 9.1%. Researchers want to test if the proportion of people with CKD in this year is the same as the proportion of people with CKD in 2000 which is 0.091.

Hypothesis for single population

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It is known that calcium supplement is associated with increased bone mineral density (BMD) in the adult population. Researchers wanted to test if the means of BMD between postmenopausal women who received and who did not receive calcium supplement are different.

Hypothesis for two populations

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Hypothesis testing

The purpose of hypothesis testing is to make a decision about a population by examining a sample from that population.

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Hypothesis testing

Types of hypotheses Types of error Steps for hypothesis testing

Figure 9 Flow chart for hypothesis testing

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Types of hypotheses

- Null hypothesis (H0) :It is a statement that a population parameter is assumed to be true or there is no difference between groups.

- Alternative hypothesis (Ha):It is a statement that is opposed to the null hypothesis.

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Null hypothesisHo: μBMD(calcium+) = μBMD(calcium-)

Alternative hypothesisHa: μBMD(calcium+) ≠ μBMD(calcium-)

For example:

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Hypothesis testing



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Types of Errors

There are two possible types of errors that may occur in hypothesis testing:

- Type I error- Type II error

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Type I error () is the probability of deciding that the means of BMD between women who receive and who did not receive calcium supplement are different, when in fact in the population, the means of BMD between these two groups are not different.



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Type II error (β) is the probability of deciding that the means of BMD between women who receive and who did not receive calcium supplement are not different, when in fact in the population, the means of BMD between these two groups are different.



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Ho: μBMD(calcium+) = μBMD(calcium-)

Ha: μBMD(calcium+) ≠ μBMD(calcium-)

Statistical Decision In PopulationBased on Sample Ho True Ho False-----------------------------------------------------------Reject H0 1-

(Type I error) (Power of test)

Do not reject H0 1- (Confidence) (Type II error)

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-----------------------------------------------------------Actual condition of population

Test result Without disease With disease-----------------------------------------------------------Positive 1-

(False positive) (True positive/sensitivity)

Negative 1- (True negative/ (False negative)

specificity)-----------------------------------------------------------

Diagnostic test

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Hypothesis testing



Steps for Hypothesis testing

Step 1 Generate null and alternative hypotheses



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Step 2 Determine the significance level

The commonly used value of the significance level is 0.05. Usually this value does not exceed 0.10.

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Step 3 Select appropriate statistics

Depends on : • Objective of analysis• Type of data• Number of groups of samples• Dependent or independent groups of samples

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Step 4 Calculate the test statistic

The general form of the test statistic can be expressed as:

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Step 5 Convert the test statistic to p value

The p value is the probability of obtaining our observed data (or more extreme data) when the null hypothesis is true.

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Step 6 Draw a conclusion

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If the p value is less than or equal to the significance level, we reject the null hypothesis.

If the p value is greater than the significance level, we do not reject the null hypothesis.

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For example:



Significance level P value Decision

0.05 0.30

0.05 0.01

Fail to reject

Reject

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Hypothesis testing for two pop.

Paired samples

Independent samples

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Paired Samples

Paired samples arise in a number of ways:When the same subjects are observed two times, usually at different times or in different circumstances.For example, we wanted to compare the SBP before and after use of oral contraceptive (OC) in pre-menopausal women.

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Independent SamplesIndependent samples arise when each set of observations is made on different groups of subjects. Examples for two independent samples:

- We want to compare the systolic blood pressure between men and women.

- We want to compare the cholesterol levels between the groups of patients with and without chronic kidney disease (CKD).

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Statistical analysis for numerical data

Normal distribution Non-normal distribution

Parametric methods Nonparametric methods

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Distribution Parameter Statistical test

Normal

Non-normal

Statistical test for hypothesis testing of paired samples

Mean

Median

Paired t-test

Wilcoxon matched signed-rank test

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Researchers wanted to test if the mean weights of HIV patients before and after 12 weeks of receiving an antiretroviral therapy regimen are different.

Class example I:

From a study of Monosuthi and et al.

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Selection an appropriate statistical test

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Question Answer

No. of samples 2 samples

Characteristic of samples Paired samples

Distribution of data Normal distribution

Parameter of interest Mean

Statistical test Paired t-test

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Draw a conclusion

Conclusion, the mean weights of HIV patients before and after receiving antiretroviral therapy are not equal.

The p value is less than 0.001.

H0: μbefore=μafter

HA: μbefore≠μafter

So we reject the null hypothesis.

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Distribution Parameter Statistical test

Normal

Non-normal

Statistical test for hypothesis testing of two independent groups

Mean

Median

Student t-test

Mann-Whiney test

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Figure 5 Flow chart of hypothesis testing for independent means by using student t-test

Student t-test

Equal variances

t-test with equal variances

Unequal variances

t-test with unequal variances

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Equal variances

Unequal variances

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A preliminary test for difference of variances should be done, before hypothesis testing for two independent population means, is carried out. The hypothesis testing for difference of variances between two population is based upon the F test or variance ratio test.

Recommendation

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Class example II:

Researchers wanted to test if the means/medians of weights of HIV patients who received NVP, and HIV patients who received EFV, are different.

From a study of Monosuthi and et al.

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Question Answer


Characteristic of samples Independent samples


Parameter of interest Mean

Statistical test Student t-test

Selection an appropriate statistical test

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Two types of t-test

• Equal variance• Non-equal variance

• To check for variance between groups, we usually use variance ratio test

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Draw a conclusion

Conclusion, the variances of weights of patients between the NVP and the EFV groups are equal.

The p value for this example is 0.095.

Therefore, we cannot reject the null hypothesis.

H0: σ2NVP=σ2

EFV

HA: σ2NVP≠σ2

EFV

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Select an appropriate statistical test

Question Answer


Characteristic of samples Independent samples


Statistical test Student t-test

Variances between 2 groups Equal

Statistical test Student t-test with equal variance

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Draw a conclusion

Conclusion, the mean weight of the NVP group is equal to the mean weight of the EFV group.

The p value in this example is 0.42.

So, we cannot reject the null hypothesis.

H0: μNVP=μEFV

HA: μNVP≠μEFV

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Distribution Parameter Condition Statistical test

Normal

Non-normal

Summary of hypothesis testing for two groups

Mean

Mean

Median

Median

Paired

Paired

Independent

Independent

Paired t-test

Student t-test

Wilcoxon matched signed-rank test

Mann-Whitney test

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Summary of selection the appropriate statistical test

Types of data: categorical or numerical data

For numerical data: Distribution of data: normal or non-normal

Number groups of samples: one, two, …

For two or more groups:Characteristics: dependent or independent

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Hypothesis testing

Continuous data

Single pop.

Two pop.

Categorical data

Single pop.

Two pop.


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Figure 2 Flow chart of statistical analysis for categorical data in two populations

Statistical analysis for categorical data in two pop.

Hypothesis testing

Independent samples

Paired samples

Estimation of strength of association

Independent samples

Paired samples

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Statistical analysis

For independent samples, the Chi-square test is conducted to examine the association between two categorical variables.H0: Two categorical variables are independent.

orH0: The proportions of the interested event between two independent groups are not different.

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Independent Samples- A case-control study was conducted to look at the effects of traditional medicine used on the risk of hip fracture. - Researchers wanted to assess the association between traditional medicine used and the risk of osteoporotic hip fracture.

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Pearson's chi-squared test (χ2)

• ∑

• Oi = Observed frequency• Ei = Expected (theoretical) frequency• n = Number of cells in the table

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Table 1. The observed frequencies of the 2x2 table for assessing the association between traditional medicine used and the risk of hip fracture.

Traditional medicine users

Hip fracture Total

Yes No

Yes 20 8 28

No 208 216 424

Total 228 224 452

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Traditional medicine users

Hip fracture Total

Yes No

Yes 28

No 424

Total 228 224 452

228452

28452

• Expected Probability of being in

this cell =

• Expected Frequency of this cell =

452

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Steps for Hypothesis Testing

1. Generate the null hypothesis as follows:

H0: Traditional medicine used is not associated with the risk of hip fracture.

or H0: Proportions of traditional medicine used between two groups of patients, with and without hip fracture are not different.

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2. Generate the alternative hypothesis as follows:

Ha: Traditional medicine used is associated with the risk of hip fracture.

or Ha: Proportions of traditional medicine used between two groups of patients, with and without hip fracture are different.

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3. Set the level of significance

4. Select an appropriate statistical test

We follow the standard convention and set the level of significance to 0.05.

We are testing a hypothesis about the association between two categorical variables, and the samples are independent, so a statistical test for this case is the Chi-square test.

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. tabulate tredmed hip, chi2 column exact expected

+--------------------+| Key ||--------------------|| frequency || expected frequency || column percentage |+--------------------+

traditiona | hip fracturel medicine | no yes | Total-----------+----------------------+---------- yes | 8 20 | 28

| 13.9 14.1 | 28.0 | 3.57 8.77 | 6.19

-----------+----------------------+---------- no | 216 208 | 424

| 210.1 213.9 | 424.0 | 96.43 91.23 | 93.81

-----------+----------------------+----------Total | 224 228 | 452

| 224.0 228.0 | 452.0 | 100.00 100.00 | 100.00

Pearson chi2(1) = 5.2588 Pr = 0.022

Result of the chi-square test:

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Steps for Hypothesis Testing5. Draw a conclusion

The p value is equal to 0.02, thus, we reject the null hypothesis and conclude that traditional medicine used is significantly associated with the risk of hip fracture. In other words, the proportions of traditional medicine used between two groups of patients, with and without hip fracture are different.

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Test for independence – small samples

• The Chi-square test is not an appropriate method of test for independence if the expected value is less than 5 for more than 20% of the total cells.• The Fisher’s exact test is an alternative method of test for independence when the requirements of the Chi-square test are not met.

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The expected frequency of ever used OC and have cervical cancer (n11) can be calculated as:

OC used Cervical cancer Total

Yes No

Ever used 68 150 218

Never used 254 875 1129

Total 322 1025 1347

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The expected frequency of ever used OC and did not have cervical cancer (n12) can be calculated as:

OC used Cervical cancer Total

Yes No

Ever used 68 150 218

Never used 254 875 1129

Total 322 1025 1347

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HRT Hip fracture Total

Yes No

Yes 1 2 3

No 213 214 427

Total 214 216 430

Table 2. The observed frequencies of the 2x2 table for assessing the association between HRT and the risk of hip fracture.

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HRT Hip fracture Total

Yes No

Yes 1.5 1.5 3

No 212.5 214.5 427

Total 214 216 430

Table 3. The expected frequencies of the 2x2 table for assessing the association between HRT and the risk of hip fracture.

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%Expected freq. less than 5 Statistical test

Greater than 20% Fisher’s exact test

Less than or equal to 20% Chi-square test

Statistical tests for hypothesis testing about the association between two categorical variables

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Figure 3 Flow chart of statistical analysis for categorical data in two populations

Statistical analysis for categorical data in two pop.

Hypothesis testing

Independent samples

Paired samples

Estimation of strength of association

Independent samples

Paired samples

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Estimation of the strength of association

Risk ratio (RR) Odds ratio (OR)

Figure 4 Types of the strength of association

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Risk ratio (RR)

- It is a ratio of the proportions of disease in exposed versus non-exposed groups.- It is often used in a cohort study and randomized controlled trial.

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From the 2x2 contingency table, the risk ratio can be calculated as:

Exp Non-Exp Total

Case a b a+b

Control c d c+d

Total a+c b+d a+b+c+d

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VL<400 copies Treatment Total

EFV NVP

Yes 7 1 8

No 17 29 46

Total 24 30 54

The relative risk can be calculated as:

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The 95%CI can be calculated as:

95%CI=1.15, 66.32

VL<400 copies Treatment Total

EFV NVP

Yes 7 1 8

No 17 29 46

Total 24 30 54

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The RR is equal to 8.75 with 95% CI 1.15 to 66.32. It means that patients who received EFV had 8.75 times the virologicalsuccess as compared with patients who received NVP with a 95% confidence that it would lie between 1.15 to 66.32 times.

Draw a conclusion

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Odds ratio (OR)

It is often used in a case-control study and cross-sectionalstudy.

odds ratio is a ratio of the odds of having disease in exposed group versus the odds of having disease in non-exposed groups.

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Exp Non-Exp Total

Case a b a+b

Control c d c+d

Total a+c b+d a+b+c+d

Therefore, the odds ratio can be calculated as:

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Table 4. 2x2 contingency table for estimation of the odds ratio of the hip fracture and traditional medicine used

Hip fracture Traditional medicine Total

Yes No

Yes 20 208 228

No 8 216 224

Total 28 424 452

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The odds ratio can be calculated as:

The 95% CI can be calculated as:

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The OR is equal to 2.60 with 95% CI 1.06 to 6.96. It means that hip fracture is 2.60 times as likely to occur among patients who used traditional medicine than patients who did not use traditional medicine with a 95% confidence that it would lie between 1.06 to 6.96 times.

Draw a conclusion

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Statistical Software

Name Website Price Features Ease of use

Note

SPSS http://www.ibm.com/software/analytics/spss/

$$$$$ ++++ ++++ Need to purchase separate modules for complicated analyses (such as Survival Analysis)Available from MU (http://softwaredownload.mahidol/)

Stata http://www.stata.com/

$$$$ ++++ +++ Ramathibodi access (CEB server)

R http://www.r‐project.org/

(Free) +++ + R‐commander is nice add on

SAS http://www.sas.com/ $$$$$ ++++ 0 Need programming skill

Excel ‐ $$ + ++++ Not really a statistical software, but can do many statistical calculations

Stata

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SPSS

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Statistical Model/Regression, Logistic Regression

• Statistical Model =

Documents

Basic Concepts of Medical Statistics 10-6-2016 · Basic Concepts of Medical Statistics Section for Clinical Epidemiology Faculty of Medicine Ramathibodi Hospital Mahidol University