Hypothesis - Biostatistics

Biostatistics

Lecture 8

Lecture 7 Review–

Using confidence intervals and p-values tointerpret the results of statistical analyses

• • 

• 

Null hypothesisP-value

Interpretation of confidence intervals & p- values

Null hypothesis

•  A null hypothesis is one that proposes there isno difference in outcomes

•  We commonly design research to disprove a nullhypothesis

P-value:- comparing two groups

What is theprobability (P-

value) of finding the

observed difference

How likely is itwe would see

adifference this big

IFIF

The null hypothesisis true?

There was NO realdifference betweenthe populations?

Interpretation of p-values1!

Weak evidence againstthe null hypothesis0.1!

Increasing evidence againstthe null hypothesis with

decreasing P-value0.01!

0.001!Strong evidence against

the null hypothesis

0.0001!

P-v

alue

!

Objective To assess the effect of combined hormone replacement therapyon health related quality of life.

Design Randomised placebo controlled double blind trial.

(HRT)

Ta ble 3EuroQoL Visual Analogue Scores (EQ-VAS) by treatment group.Figures are means (SE)

one year (95%

Combined HRT

(n=1043*)

Placebo

(n=1087*)Adjusted

difference at

CI)

P-value

EQ-VAS 77.9 (0.5) 78.5 (0.4) -0.59

(-1.66 to 0.47)

0.28

Five trials of drugs to reduce serum cholesterol

A reduction of 0.5 mmol/L or more correspondsto a clinically important effect of the drug

Trial Drug Cost No. of patients

per group

Observed difference in mean

cholesterol (mmol/L)

s.e. of difference (mmol/L)

95% CI for population

difference in mean

cholesterol

P-value

1 A Cheap 30 -1.00 1.00 -2.96 to 0.96 0.32

2 A Cheap 3000 -1.00 0.10 -1.20 to -0.80 <0.001

3 B Cheap 40 -0.50 0.83 -2.13 to 1.13 0.55

4 B Cheap 4000 -0.05 0.083 -0.21 to 0.11 0.55

5 C Expensive 5000 -0.125 0.05 -0.22 to -0.03 0.012

Lecture 8 – Proportions andintervals

Binary variables (RECAP)

confidence

• 

•  Single proportion–  Standard error, confidence interval

•  Incidence & prevalence

•  Difference in two proportions–  Standard error, confidence interval

Categorical variables - Binary

Binary variable – two categories only

(also termed – dichotomous variable)

Examples:-

  Outcome – Diseased or Healthy; Alive

or Dead…

  Exposure - Male or Female; Smoker or non-smoker;

Treatment or control group….

Inference

Proportion of population diseased – π??

Proportion of sample diseased, p=d/n

Number of subjects who do experience outcome (diseased) = dNumber of subjects who do not experience outcome (healthy) = h

Total number in sample = n = h + d

Inference - example

Proportion of population with vivax malaria - π

Proportion of sample with vivaxp = d/n = 15/100 = 0.15 (15%)

malaria,

Number of sample with vivax malaria = d = 15Number of sample without vivax malaria = h = 85Total number in sample = n = 15 + 85 = 100

Single proportion - Inference

• Obtain a sample estimate, p, of the population proportion, π

• REMEMBER different samples would give different estimatesof π (e.g. sample 1 p1, sample 2 p2,…)

• Derive:

– Standard error

– Confidence interval

Standard error & confidence intervalof a single proportion

• Standard error (SE) for single proportion:-(from the Binomial distribution)

π (1 − π ) p(1 − p)s.e.( p ) = ~

n n

• 95% CI for single proportion:-(approximate method based on the normal distribution)

–

–

Lower limit = p - 1.96×s.e.(p)

Upper limit = p + 1.96×s.e.(p)

Standard error & confidence intervala single proportion – malaria exampleof

•

•

Estimated proportion of vivax

Standard error of p

malaria (p) = 15/100 = 0.15

p(1 − p)

0.15(1 − 0.15)s e ( p ). . = = 0.036=

n 100

• 95% Confidence interval for population proportion (π)

–

–

Lower limit = p - 1.96×s.e.(p) = 0.15 – 1.96×0.036 = 0.079

Upper limit = p + 1.96×s.e.(p) = 0.15 + 1.96×0.036 = 0.221

Interpretation..

“We are 95% confident, the population proportion of people

vivax malaria is between 0.079 and 0.221

(or between 7.9% and 22.1%)”

with

Definition of a confidenceREMEMBER…..

interval

If we were to draw several independent,

random samples (of equal size) from the

sample population and calculate 95%confidence intervals for each of them, 0.

4

0.35

0.3

Population0.2

5

then on average 19 out of every 20 (95%)

such confidence intervals would

contain the true population

proportion (π), and one of every 20

0.2

0.15

0.1

(5%) would not.0.05

0

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20

Sam ple

Sam

ple

pro

po

rtio

n a

nd

95%

CI proportion = 0.16

(16%)

WARNING….Confidence Interval of a

single proportionThe normal approximation method breaks

down1)

2)

if:Sample

Sample

size (n) is small

proportion (p) is close to 0 or 1

Require:

np ≥ 10 or n(1-p) ≥ 10

Stata lets you calculate an ‘exact’ CI

Confidence Interval for a single proportion in Stata

• cii 100 15

•

•••

-- Binomial Exact --

[95% Conf. Interval]Variable | Obs Mean Std. Err.

-------------+---------------------------------------------------------------

| 100 .15 .0357071 .0864544 .2353075

• cii 100 1

•

••

•

-- Binomial Exact --

[95% Conf. Interval]Variable | Obs Mean Std. Err.

-------------+---------------------------------------------------------------

| 100 .01 .0099499 .0002531 .0544594

Interpretation of proportions:

Incidence versus Prevalence

Prevalence

Proportion of people in a defined population that have a given disease at a specified point in time

•

Prevalence = no. of people with the disease at particular point in time

no. of people in the population at a particular point in time

Examples:-

• Prevalenceliving in the

Prevalence

Thailand.

Prevalence

of chronic pain among people aged 25+ years andGrampian region, UK.

of typhoid among villagers living in Tak province,•

• of diagnosed asthma in individuals aged 15 to 50

years, registered with a particular general practice in Carlton.

Incidence risk(Cumulative incidence)

Proportion of new cases in a disease free population in a given time period

•

Incidence risk = no. of new cases of disease in a given time period

no. of people disease-free at beginning of time period

Examples:-

• Incidence risk of death in five years following diagnosis withprostate cancer

Incidence risk of breast cancer over 10 years of follow-up in

women 40-69 years of age and free from breast cancer in

1990

•

Incidence rate(NOT a proportion)

Number of new cases in a disease free population per person per unit time

• that occur

Incidence rate = no. of new cases of disease

total person-years of observation

Examples:-

• Incidence rate of all-cause mortality of men in the Melbourne

Collaborative Cohort Study = 9.0 per 1000 men per year

‘9 out of every 1000 men die each year’

(

Comparing two proportions

Comparing two proportions2×2 table

•••

ProportionProportionProportion

of all subjects experiencing outcome, p = d/nin exposed group, p1 = d1/n1

in unexposed group, p0 = d0/n0

Be alert (not alarmed): watch for transposing the table and swapping columns or rows

With outcome

(diseased)

Without outcome

(disease-free)

Total

Exposed

(group 1)

d1 h1 n1

Unexposed

(group 0)

d0 h0 n0

Total d h n

Comparing two proportionsExample:- TBM trial (Thwaites GE et al 2004)

Adults with tuberculous meningitis randomly allocated intotreatment groups:

2

1.

2.

Dexamethasone

Placebo

Outcome measure: Death during nine months following start of

treatment.

Research question:

Can treatment with dexamethasone reduce the risk of deathadults with tuberculous meningitis?

among

Comparing two proportionsExample – TBM trial

Death during 9 months post start of treatment

Treatment group Yes No Total

Dexamethasone

(group 1)

87 187 274

Placebo

(group 0)

112 159 271

Total 199 346 545

Difference in two population proportions, π1-π0

Estimate of difference in population proportions = p1 – p0

Example:- TBM trial

Dexamethasone

p1 = d1/n1 = 87/274 = 0.318

Placebo

p0 = d0/n0 = 112/271 = 0.413

p1 – p0 = 0.318 – 0.413 = -0.095 (or -9.5%)

Difference in two proportions - Inference

• Obtain a sample estimate, p1-p0, of the difference in population proportions, π1Dπ0

• REMEMBER different samplesof π1Dπ0 (e.g. sample 1 p11-p10,

would give different estimatessample 2 p21-p20,…)

• Derive:

– Standard error of difference in sample proportions

– Confidence interval of difference in population proportions

Standard error & confidence intervalfor difference between two

proportions• Standard error (SE) for difference between sample proportions:-

[s.e.( p )]2 + [s.e.( p )]2s.e.( p ) =− p1 0 1 0

• 95% CI for difference between population

Lower limit = (p1-p0) - 1.96×s.e.(p1-p0)

Upper limit = (p1-p0) + 1.96×s.e.(p1-p0)

proportions:-

Standard error & confidence interval

for difference between two proportions

Example:- TBM trial

Estimate of difference in population proportions

= p1-p0 = -0.095

s.e.(p1-p0) = 0.041

95% CI for difference in population proportions (π1-π0):

-0.095 ± 1.96×0.041

-0.175 up to -0.015 OR -17.5% up to -1.5%

Interpretation:-

“We are 95% confident, that the difference in population proportions is

between -17.5% (dexamethasone reduces the proportion of deaths by a

large amount) and -1.5% (dexamethasone marginally reduces the

proportion of deaths)”.

Comparing proportions usingcsi 87 112 187 159

Stata

| Exposed Unexposed | Total-----------------+------------------------+------------

Cases |

Noncases |87

187112

159|

|199

346-----------------+------------------------+------------

Total |

||||

274 271 |

||||

545

Risk .3175182 .4132841 .3651376

Point estimate [95% Conf. Interval]|------------------------+------------------------

Risk difference

Risk ratio Prev.

frac. ex. Prev.

frac. pop

|

|||

-.0957659

.7682808

.2317192

.1164974

|

|||

-.1762352 -.0152966.6139856

.0386495.9613505

.3860144

+-------------------------------------------------chi2(1) = 5.39 Pr>chi2 = 0.0202

Remember the warning about how the table is presented-Stata requires presentation with outcome by rows and exposure by columns

Results are close to those obtained by hand

Difference between two proportions:-

Risk difference

Example:- TBM trial

Outcome measure: Death during nine months

treatment.

following start of

Dexamethasone

p1 (incidence risk) = d1/n1 = 87/274 = 0.318

Placebo

p0 (incidence risk) = d0/n0 = 112/271 = 0.413

p1 – p0 (risk difference) = 0.318 – 0.413 = -0.095 (or -9.5%)

Lecture 8 – Objectives

• Define binary variables, prevalence and incidence risk

• Calculate and interpret a proportion and 95% confidenceinterval for the population proportion

• Calculate and interpret the difference in sample proportionsand 95% confidence interval for difference in population proportions

Thank You

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