Introduction to Statistics

Dr Linda Morgan

Clinical Chemistry Division

School of Clinical Laboratory Sciences

Outline

• Types of data• Descriptive statistics• Estimates and confidence intervals• Hypothesis testing• Comparing groups• Relation between variables• Statistical aspects of study design• Pitfalls

Types of data

• Categorical data– Ordered categorical data

• Numerical data– Discrete– Continuous

Descriptive statisticsCategorical variables

• Graphical representation – bar diagram

• Numbers and proportions in each category

smoking habit

heavy smokerlight smokerex-smokernon-smoker

Pe

rce

nt

50

40

30

20

10

0

Descriptive statisticsContinuous variables

• Distributions– Gaussian– Lognormal– Non-parametric

• Central tendency– Mean– Median

• Scatter– Standard deviation– Range– Interquartile range

Maternal age

42.5

40.0

37.5

35.0

32.5

30.0

27.5

25.0

22.5

20.0

17.5

15.0

60

50

40

30

20

10

0

Std. Dev = 4.84

Mean = 28.0

N = 223.00

Gaussian (normal) distribution

• Central tendency

Mean = x

n

• Scatter

Variance = (x-mean)2

n –1

Standard deviation = variance

Gaussian (normal) distribution

Plasma renin concentration

36.0

34.0

32.0

30.0

28.0

26.0

24.0

22.0

20.0

18.0

16.0

14.0

12.0

10.0

8.0

6.0

4.0

2.0

20

10

0

Std. Dev = 5.36

Mean = 9.3

N = 73.00

Lognormal distribution

Log plasma renin concentration

1.501.381.251.131.00.88.75.63.50.38

30

20

10

0

Std. Dev = .21

Mean = .91

N = 73.00

Lognormal distribution

Lognormal distribution

• Mean = log x n

• Geometric mean = antilog of mean (10mean)

• Median– Rank data in order– Median = (n+1) / 2th observation

Variability

• Variance = (x-mean)2

n –1

• Standard deviation = variance

• Range

• Interquartile range

Variability of Sample Mean

• The sample mean is an estimate of the population mean

• The standard error of the mean describes the distribution of the sample mean

• Estimated SEM = SD/ n• The distribution of the sample mean is

Normal providing n is large

Standard error of the difference between two means• SEM = SD/ n• Variance of the mean = SD2/n• Variance of the difference between two

sample means

= sum of the variances of the two means= (SD2/n)1 + (SD2/n)2

• SE of difference between means

= [(SD2/n)1 + (SD2/n)2 ]

Variability of a sample proportion

• Assume Normal distribution when np and n(1-p) are > 5

• SE of a Binomial proportion =

(pq/n) where q = 1-p

Standard error of the difference between two

proportions• SE (p1 – p2 )

= [variance (p1) + variance (p2) ]

= [ (p1 q1 /n1) + (p2 q2 /n2) ]

Confidence intervals of means

• 95% ci for the mean =

Sample mean 1.96 SEM

• 95% ci for difference between 2 means =

(mean1 – mean2 ) 1.96 SE of difference

Confidence intervals of proportions

• 95% ci for proportion

= p 1.96 (pq/n)

• 95% ci for difference between two proportions

= (p1 – p2 ) 1.96 x SE (p1 – p2 )

Hypothesis testing

• The null hypothesis

• The alternative hypothesis

• What is a P value?

Comparing 2 groups of continuous data

• Normal distribution:

paired or unpaired t test

• Non-Normal distribution:

transform data

OR

Mann-Whitney-Wilcoxon test

Paired t test

We wish to compare the fasting blood cholesterol levels in 10 subjects before and after treatment with a new drug.

What is the null hypothesis?

Paired t testSubject Fasting cholesterol DNumber Predrug Postdrug01 6.7 4.4 2.302 7.8 7.0 0.803 8.1 6.0 2.104 5.5 5.8 -0.305 8.6 9.0 -0.406 6.7 6.1 0.607 7.1 7.3 -0.208 9.9 9.9 009 8.2 6.3 1.910 6.5 7.1 -0.6

Paired t test

• Calculate the mean and SEM of D

• The null hypothesis is that D = 0

• The test statistic t =

mean(d) – 0

SEM (d)

Paired t test

• Mean = 0.62• SEM = 0.351• t = 1.766• Degrees of freedom = n - 1 = 9• From tables of t,

2-tailed probability (P) is between 0.1 and 0.2• How would you interpret this?

Comparing 2 groups of categorical data

• In a study of the effect of smoking on the risk of developing ischaemic heart disease, 250 men with IHD and 250 age-matched healthy controls were asked about their current smoking habits.

• What is the null hypothesis?

Results

• 70 of the 250 patients were smokers

• 30 of the healthy controls were smokers

Smoker Non-smoker

Total

IHD 70 180 250

Control 30 220 250

Total 100 400 500

Smoker Non-smoker Total

IHD 70

50

180

200

250

Control 30

50

220

200

250

Total 100 400 500

Calculate expected values, E, for each cell

Calculate (observed – expected) value, D

Smoker Non-smoker Total

IHD 70 – 50 =

20

180–200=

-20

Control 30-50=

-20

220-200=

20

Total

Calculate D2/E

Smoker Non-smoker Total

IHD 400/50=

8

400/200=

2

Control 400/50=

8

400/200=

2

Total

Calculate the sum of D2/E

8 + 8 + 2 + 2 = 20

This is the test statistic, chi squaredCompare with tables of chi squared with (r-1)(c-1) degrees of freedom In this case, chi squared with 1 df has a P value of < 0.001

How do you interpret this?

Statistical analysis using computer software

SPSS as an example

Planning

• Experimental design

• Suitable controls

• Database design

Statistical power

• The power of a study to detect an effect depends on:– The size of the effect– The sample size

• The probability of failing to detect an effect where one exists is called

• The power of a study is 100(1-)%• Wide confidence intervals indicate low

statistical power

Statistical power

• The necessary sample size to detect the effect of interest should be calculated in advance

• Pilot data are usually required for these calculations

Statistical power - example

• 30% of the population are carriers of a genetic variant. You wish to test whether this variant increases the risk of Alzheimers Disease.

• For P < 0.05, and 80% power, number of controls and cases required:

Control carriers Case carriers Sample size 30% 50% 10030% 40% 35030% 35% 1400

Multiple testingNumber of Probability of Tests false positive

1 0.052 0.103 0.144 0.195 0.2310 0.4020 0.64

Bonferroni correction: Divide 0.05 by the number of tests to provide the required P value for hypothesis testing at the conventional level of statistical significance

Data trawling

• Decide in advance which statistical tests are to be performed

• Post hoc testing of subgroups should be viewed with caution

• Multiple correlations should be avoided

HELP!

• “In house” support

• Cripps Computing Centre

• Trent Institute for Health Service Research

• Practical Statistics for Medical Research

Douglas G Altman