Hypothesis Testing, Power, Sample Size and Confidence ...biostat.mc.vanderbilt.edu/wiki/pub/Main/AnesShortCourse/Hypothesis... · Hypothesis Testing, Power, Sample Size and Con dence

Hypothesis Testing, Power, Sample Size and Confidence Intervals (Part 1)

Hypothesis Testing, Power, Sample Size andConfidence Intervals (Part 1)

B.H. Robbins Scholars Series

June 3, 2010

1 / 44


OutlineIntroduction to hypothesis testing

Scientific and statistical hypothesesClassical and Bayesian paradigmsType 1 and type 2 errors

One sample test for the meanHypothesis testingPower and sample sizeConfidence interval for the meanSpecial case: paired data

One sample methods for a probabilityHypothesis testingPower, confidence intervals, and sample size

Two sample tests for meansHypothesis testsPower, confidence intervals, and sample size

2 / 44


Introduction to hypothesis testing

Introduction

I Goal of hypothesis testing is to rule out chance as anexplanation for an observed effect

I Example: Cholesterol lowering medicationsI 25 people treated with a statin and 25 with a placeboI Average cholesterol after treatment is 180 with statins and 200

with placebo.

I Do we have sufficient evidence to suggest that statins lowercholesterol?

I Can we be sure that statin use as opposed to a chanceoccurrence led to lower cholesterol levels?

3 / 44



Scientific and statistical hypotheses

Hypotheses

I Scientific HypothesesI Often involve estimation of a quantity of interestI After amputation, to what extent does treatment with

clonidine lead to lower rates of phantom limb pain than withstandard therapy? (Difference or ratio in rates)

I What is the average increase in alanine aminotransferase(ALT) one month after doubling the dose of medication X?(Difference in means)

I Statistical HypothesisI A statement to be judged. Usually of the form: population

parameter X is equal to a specified constantI Population mean potassium K, µ = 4.0 mEq/LI Difference in population means, µ1 − µ2 = 0.0 mEq/L

4 / 44



Scientific and statistical hypotheses

Statistical Hypotheses

I Null Hypothesis: H0

I A straw man; something we hope to disproveI It is usually is a statement of no effects.I It can also be of the form H0 : µ =constant, or H0: probability

of heads equal 1/2.

I Alternative Hypothesis: HA

I What you expect to favor over the null

I If H0 : Mean K value = 3.5 mEq/LI One sided alternative hypothesis: HA : Mean K > 3.5 mEq/LI Two-sided alternative hypothesis: HA : Mean K 6= 3.5 mEq/L

(values far away from the null)

5 / 44



Classical and Bayesian paradigms

Classical (Frequentist) Statistics

I Emphasizes hypothesis testing

I Begin by assuming H0 is true

I Examines whether data are consistent with H0

I Proof by contradictionI If, under H0, the data are strange or extreme, then doubts are

cast on the null.

I Evidence is summarized with a single statistic which capturesthe tendency of the data.

I The statistic is compared to the parameter value given by H0

6 / 44





I p-value: Under the assumption that H0 is true, it is theprobability of getting a statistic as or more in favor of HA overH0 than was observed in the data.

I Low p-values indicate that if H0 is true, we have observed animprobable event.

I Mount evidence against the null, and when sufficient, rejectH0.

I NOTE: Failing to reject H0 does not mean we have gatheredevidence in favor of it (i.e., absence of evidence does notimply evidence of absence)

I There are many reasons for not rejecting H0 (e.g., smallsamples, inefficient designs, imprecise measurements, etc.)

7 / 44





I Clinical significance is ignored.

I Parametric statistics: assumes the data arise from a certaindistribution, often a normal or Gaussian.

I Non-parametric statistics: does not assume a distribution andusually looks at ranks rather than raw values.

8 / 44




Bayesian Statistics

I We can compute the probability that a statement, that is ofclinical significance, is true

I Given the data we observed, does medication X lower themean cholesterol by more than 10 units?

I May be more natural than the frequentist approach, but itrequires a lot more work.

I Supported by decision theory:

I Begin with a (prior) belief → learn from your data → Form anew (posterior) belief that combines the prior belief and thenew data

I We can then formally integrate information accrued fromother studies as well as from skeptics.

I Becoming more popular.

9 / 44



Type 1 and type 2 errors

Errors in Hypothesis Testing

I Type 1 error: Reject H0 when it is trueI Significance level (α) or Type 1 error rate: is the probability of

making this type of errorI This value is usually set to 0.05 for random reasons

I Type 2 error: Failing to reject H0 when it is falseI The value β is the probability of a type 2 error or type 2 error

rate.

I Power: 1− β: probability of correctly rejecting H0 when it isfalse

State of H0

Decision H0 is true H0 is false

Do not reject H0 Correct Type 2 error (β)

Reject H0 Type 1 error (α) Correct

10 / 44



Type 1 and type 2 errors

Notes Regarding Hypothesis TestingI Two schools of thought

I Neyman-Pearson: Fix Type 1 error rate (say α = 0.05) andthen make the binary decision, reject/do not reject

I Fisher: Compute the p-value and quote the report in thepublication.

I We favor Fisher, but Neyman-Pearson is used all of the time.I Fisher approach: discussion of p-values does not require

discussion of type 1 and type 2 errorsI Assume the sample was chosen randomly from a population

whose parameter value is captured by H0. The p-value is ameasure of evidence against it.

I Neyman-Pearson approach: having to make a binary call(reject vs do not reject) regarding significance is arbitrary

I There is nothing magical about 0.05I Statistical significance has nothing to do with clinical

significance

11 / 44


One sample test for the mean

Hypothesis testing


I Assumes the sample is drawn from a population where valuesare normally distributed (normality is actually not necessary)

I One sample tests for mean µ = µ0 (constant) don’t happenvery often except when data are paired (to be discussed later)

I The t-test is based on the t-statistic

t =estimated value - hypothesized value

standard deviation of numerator

I Standard deviation of a summary statistic is called thestandard error which is the square root of the variance of thestatistic

12 / 44



Hypothesis testing


I Sample average: x = 1n

∑ni=1 xi

I The estimate of the population mean based on the observedsample

I Sample variance: s2 = 1n−1

∑ni=1(xi − x)2

I Sample standard deviation: s =√s2

I H0 : µ = µ0 vs. HA : µ 6= µ0

I One sample t-statistic

t =x − µ0

SE

I Standard error of the mean, SE = s√n

13 / 44



Hypothesis testing

One sample t-test for the mean

I When data come from a normal distribution and H0 holds, thet ratio follows the t− distribution. What does that mean?

I Draw a sample from the population, conduct the study andcalculate the t-statistic.

I Do it again, and calculate the t-statistic again.

I Do it again and again.

I Now look at the distribution of all of those t-statistics.

I This tells us the relative probabilities of all t-statistics if H0 istrue.

14 / 44



Hypothesis testing

Example: one sample t-test for the mean

I The distribution of potassium concentrations in the targetpopulation are normally distributed with mean 4.3 andvariance .1: N(4.3, .1).

I H0 : µ = 4.3 vs. HA : µ 6= 4.3. Note that H0 is true!I Each time the study is done,

I Sample 100 participantsI Calculate:

t =x − 4.3

SE

I Conduct the study 25 times, 250 times, 1000 times, 5000times

15 / 44



Hypothesis testing

Distribution of 25 t−statistics

t−values

Fre

quen

cy

−6 −4 −2 0 2 4 6

01

23

4


Fre

quen

cy

−6 −4 −2 0 2 4 6

05

1015

2025


t−values

Fre

quen

cy

−6 −4 −2 0 2 4 6

020

4060

8010

0


Fre

quen

cy

−6 −4 −2 0 2 4 6

010

020

030

040

050

0

16 / 44



Hypothesis testing


I With very small samples (n), the t statistic can be unstablebecause the sample standard deviation (s) is not a preciseestimate of the population standard deviation (σ).

I So, the t-statistic has heavy tails for small n

I As n increases, the t-distribution converges to the normaldistribution with mean equal to 0 and with standard deviationequal to one.

I The parameter defining the particular t-distribution we use(function of n) is called the degrees of freedom or d.f.

I d.f. = n - number of means being estimated

I For the one-sample problem, d.f.=n-1

I Symbol is tn−1

17 / 44



Hypothesis testing

−3 −2 −1 0 1 2 3

0.0

0.1

0.2

0.3

0.4

Density for the t−distribution

t−value

Den

sity

t (d.f.=5)t (d.f.=10)t (d.f.=100)N (0,1)

18 / 44



Hypothesis testing


I One sided test: H0 : µ = µ0 versus HA : µ > µ0

I One tailed p-value:I Probability of getting a value from the tn−1 distribution that is

at least as much in favor of HA over H0 than what we hadobserved.

I Two-sided test: H0 : µ = µ0 versus HA : µ 6= µ0

I Two-tailed p-value:I Probability of getting a value from the tn−1 distribution that is

at least as big in absolute value as the one we observed.

19 / 44



Hypothesis testing


I Computer programs can compute the p-value for a given nand t-statistic

I Critical valueI The value in the t (or any other) distribution that, if exceeded,

yields a ’statistically significant’ result for type 1 error rateequal to α

I Critical regionI The set of all values that are considered statistically

significantly different from H0.

20 / 44



Hypothesis testing

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

T−distribution (d.f.=10) and one−sided critical region (0.05)

t−value

dens

ity

1.812

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

T−distribution (d.f.=10) and two−sided critical regions (0.05)

dens

ity

2.228

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

T−distribution (d.f.=100) and one−sided critical region (0.05)

t−value

dens

ity

1.66

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

T−distribution (d.f.=100) and two−sided critical regions (0.05)

dens

ity

1.984

21 / 44



Power and sample size

Power and Sample Size for a one sample test of means

I Power increases whenI Type 1 error rate (α) increases: type 1 (α) versus type 2 (β)

tradeoffI True µ is very far from µ0

I Variance or standard deviation (σ) decreases (decrease noise)I Sample size increases

I T-statistic

t =x − µ0

σ/√n

I Power for a 2-tailed test is a function of the true mean µ, thehypothesized mean µ0, and the standard deviation σ onlythrough |µ− µ0|/σ

22 / 44




Power and Sample Size for a one sample test of means

I Sample size to achieve α = 0.05, power=0.90 is approximately

n = 10.51

(σ

µ− µ0

)2

I Power calculators can be found at statpages.org/#Power

I PS is a very good power calculator (Dupont and Plummer):http://biostat.mc.vanderbilt.edu/PowerSampleSize

23 / 44




Example: Power and Sample Size

I The mean forced expiratory volume in 1 second in apopulation of asthmatics is 2.5 L/sec, and the standarddeviation is assumed to be 1

I How many subjects are needed to reject H0 : µ = 2.5 in favorof H0 : µ 6= 2.5 if the new drug is expected to increase theFEV to 3 L/sec with α = 0.05 and β = 0.1

I µ0 = 2.5, µ = 3.0, σ = 1

n = 10.51

(1

3.0− 2.5

)2

= 42.04

I We need 43 subjects to have 90 percent power to detect a 0.5difference from 2.5.

24 / 44



Confidence interval for the mean

Confidence Intervals

I Two-sided, 100(1− α)% CI for the mean µ is given by

(x − tn−1,1−α/2 · SE , x + tn−1,1−α/2 · SE )

I tn−1,1−α/2 is the critical value from the t-distribution withd.f.=n-1

I For large n, tn−1,1−α/2 is equal to 1.96 for α = 0.05

I 1− α is called the confidence level or confidence coefficient

25 / 44



Confidence interval for the mean

Confidence IntervalsI 100(1− α)% confidence interval (CI)

I If we were able to repeat a study a large number of times, then100 · (1− α) percent of CIs would contain the true value.

I Two-sided 100(1− α)% CII Includes the null hypothesis µ0 if and only if a hypothesis test

H0 : µ = µ0 is not rejected for a 2-sided α significance leveltest.

I If a 95% CI does not contain µ0, we can reject H0 : µ = µ0 atthe α = 0.05 significance level

n x σ p-value 95% CI20 27.31 54.23 0.036 (1.930, 52.690)20 27.31 59.23 0.053 (-0.410, 55.030)20 25.31 54.23 0.051 (-0.070, 50.690)17 27.31 54.23 0.054 (-0.572, 55.192)

I CIs provide more information than p-values

26 / 44



Special case: paired data

Special case: Paired data and one-sample tests

I Assume we want to study whether furosemide (or lasix) hasan impact on potassium concentrations among hospitalizedpatients.

I That is, we would like to test H0 : µon−furo − µoff−furo = 0versus HA : µon−furo − µoff−furo 6= 0

I In theory, we could sample n1 participants not on furosemideand compare them to n2 participants on furosemide

I However, a very robust and efficient design to test thishypothesis is with a paired sample approach

I On n patients, measure K concentrations just prior to and 12hours following furosemide administration.

27 / 44



Special case: paired data

Special case: Paired data and one-sample testsI The effect measure to test H0 versus HA, is the mean, within

person difference between pre and post- administration Kconcentrations.

I Wi = Yon−furo,i − Yoff−furo,i

I Note that W = Y on−furo − Y off−furoI The average of the differences is equal to the difference

between the averagesI H0 : µw = 0 versus HA : µw 6= 0 is equivalent to the above

H0 and HA

I W = −0.075 mEq/L and s = 0.08

t99 =−0.075− 0

0.08/√

100= 9.375

I The p-value is less than 0.0001 → a highly (!!!!) statisticallysignificant reduction

28 / 44


One sample methods for a probability

Hypothesis testing

One Sample Methods for a Probability

I Y is binary (0/1): Its distribution is bernoulli(p) (p is theprobability that Y = 1).

I p is also the mean of Y and p(1− p) is the variance.

I We want to test H0 : p = p0 versus HA : p 6= p0

I Estimate the population probability p with the sampleproportion or sample average p̂

p̂ =1

n

n∑i=1

Yi

29 / 44



Hypothesis testing

One Sample Methods for a Probability

I A z-test is an approximate test that assumes the test statistichas a normal distribution i.e., it is a t-statistic with the d.f.very large

z =p̂ − p0√

p0(1− p0)/n

I The z-statistic has the same form as the t-statistic

z =estimated value - hypothesized value

standard deviation of numerator

where√

p0(1− p0)/n is the standard deviation of thenumerator which is the standard error assuming the H0 is true.

I (see t-statistic distributions)

30 / 44



Hypothesis testing

One Sample test for a probability: Is our coin fair?I Y ∼ bernoulli(p): H0 : p = 0.5 versus HA : p 6= 0.5I Flip the coin 50 times. Heads (Y=1) shows up 30 times

(p̂ = 0.6).

z =0.6− 0.5√

(0.5)(0.5)/50= 1.414

I The p-value associated with Z is 2 × the area under thenormal curve to the right of z=1.414 (e.g. the area to theright of 1.414 plus the area to the left of -1.414)

I The critical value for a 2-sided α = 0.05 significance level testis 1.96

I The p-value associated with this test is approximately 0.16I Note that if p is very small or very large or if n is small, use

exact methods (e.g. Fishers exact test or permutation test)

31 / 44



Hypothesis testing

−4 −2 0 2 4

0.0

0.1

0.2

0.3

0.4

Z−test for a proportion: Z−statistic=1.414

z−value

dens

ity

Critical region

32 / 44



Power, confidence intervals, and sample size

Power and confidence intervalsI Power increases when

I n increasesI p departs from p0

I p0 departs from 0.5

z =p̂ − p0√

p0(1− p0)/n

I Confidence intervalI 95%CI: (p̂ − 1.96 ·

√p̂(1− p̂/n, p̂ − 1.96 ·

√p̂(1− p̂/n)

I For the coin flipping example: p̂ = 0.6 and the 95% CI isgiven by

0.6± 1.96 ·√

0.6× 0.4/50 = (0.464, 0.736)

which is consistent with the 0.16 p-value that we hadobserved for H0 : p = 0.5.

33 / 44


Two sample tests for means

Hypothesis tests

Two sample test for means

I Two groups of patients (not paired)

I These are much more common than 1 sample tests

I We assume data come from a normal distribution (althoughthis is not completely necessary)

I For now, assume the two groups have equal variability inresponse distribution

I Test whether population means are equal

I Example: All patient in population 1 are treated withclonidine after limb amputation and all patients in population2 are treated with standard therapy.

I Scientific question:I What is the difference in the mean pain scale scores at 6

months following the amputation?

34 / 44



Hypothesis tests


I H0 : µ1 = µ2 which can be generalized to H0 : µ1 − µ2 = 0or H0 : µ1 − µ2 = δ

I The quantity of interest (QOI) is µ1 − µ2

I If we want to test H0 : µ1 − µ2 = 0 and if we assume the twopopulations have equal variances, then the t- statistic is givenby:

t =point estimate of the QOI− 0

standard error of the numerator

I The estimate of the QOI: x1 − x2

35 / 44



Hypothesis tests


I For two independent samples variance of the sum or ofdifferences in means is equal to the sum of the variances

I The variance of the QOI is then given by σ2

n1+ σ2

n2

I We need to estimate a single σ2 from the two samples

I We use a weighted average of the two sample variances

s2 =(n1 − 1)s2

1 + (n2 − 1)s22

n1 + n2 − 2

I The true standard error of the difference in sample means:

σ√

1n1

+ 1n2

I Estimate with s√

1n1

+ 1n2

36 / 44



Hypothesis tests


I The t-statistic is given by,

t =x1 − x2

s√

1n1

+ 1n2

I Under H0 t, has a t-distribution with n1 + n2 − 2 degrees offreedom.

I The -2 comes from the fact that we had to estimate thecenter of 2 distributions

37 / 44



Hypothesis tests

Example: two sample test for means

I n1 = 8, n2 = 21, s1 = 15.34, s2 = 18.23, x1 = 132.86,x2 = 127.44

s2 =7(15.34)2 + 20(18.23)2

7 + 20= 307.18

s =√

307.18 = 17.527

se = 17.527

√1

8+

1

21= 7.282

t =5.42

7.282= 0.74

on 27 d.f.

38 / 44



Hypothesis tests

Example: two sample test for means

I The two-sided p-value is 0.466I You many verify with the surfstat.org t-distribution calculator

I The chance of getting a difference in means as large or largerthan 5.42 if the two populations have the same mean in 0.466.

I No evidence to suggest that the population means aredifferent.

39 / 44




Power and sample size: two sample test for meansI Power increases when

I ∆ =| µ1 − µ2 | increasesI n1 or n2 increasesI n1 and n2 are closeI σ decreasesI α increases

I Power depends on n1, n2, µ1, µ2, and σ approximatelythrough

∆

σ√

1n1

+ 1n2

I When using software to calculate power you can put in 0 forµ1 and ∆ for µ2 since all that matters is their difference

I σ is often estimated from pilot data

40 / 44




Power and sample size: two sample test for means

I ExampleI From available data, ascertain a best guess of σ : assume it is

16.847.I Assume ∆=5, n1 = 100, n2 = 100, α = 0.05I The surfstat software computes a power of 0.555

I The required sample size decreases withI k = n2

n1→ 1

I ∆ largeI σ smallI α largeI Lower power requirements

41 / 44




Power and sample size: two sample test for means

I An approximate formula for required sample sizes to achievepower=0.9 with α = 0.05 is

n1 =10.51σ2(1 + 1

k )

∆2

n2 =10.51σ2(1 + k)

∆2

σ ∆ K n1 n2 n

16.847 5 1.0 239 239 47816.847 5 1.5 199 299 49816.847 5 2.0 177 358 53716.847 5 3.0 160 478 638

I Usually, websites are recommended for these calculations.

42 / 44




Confidence interval: two sample test for means

I Confidence interval

[(x1 − x2)− tn1+n2−2,1−α/2 × s ×√

1

n1+

1

n2,

(x1 − x2) + tn1+n2−2,1−α/2 × s ×√

1

n1+

1

n2]

∆ s n1 n2 LCI UCI

5 16.847 100 100 3.01 6.995 16.847 75 125 2.95 7.055 16.847 50 150 2.70 7.30

43 / 44




Summary

I Hypothesis testing, power, sample size, and confidenceintervals

I One sample test for the meanI One sample test for a probabilityI Two sample test for the mean

44 / 44

Documents

Hypothesis Testing, Power, Sample Size and Confidence ...biostat.mc.vanderbilt.edu/wiki/pub/Main/AnesShortCourse/Hypothesis... · Hypothesis Testing, Power, Sample Size and Con dence