Basic Statistics: MBI-540 · Basic Statistics: MBI-540 Sally W. Thurston, PhD Associate Professor...

Basic Statistics: MBI-540

Sally W. Thurston, PhD

Associate ProfessorDepartment of Biostatistics and Computational Biology

University of Rochester

January 2011

Sally W. Thurston, PhD Basic Statistics: MBI-540

Why statistics?

Allows us to say something (“make inference”) about apopulation based on data from a sample.Example: measure effect of drug on weight loss in 20BALB/c mice

Sample: the 20 micePopulation: all BALB/c mice (that could be subject to thesame condition)Interested in weight loss in BALB/c mice, not just in the 20mice.Statistics allows us to estimate weight loss in the populationusing data from a sample.

Many journal articles ask for statistics

Some comments

Design of experiment is keyNo statistical method can fix bad data!Essential to think about design before data is collected

Be clear about research question(s) before collecting data.Includes: what data to collect to answer question(s).

Know your dataHow the measurements relate to research questionPlots to visualize your data: helpful!

Most statistical models have assumptionsIf violated, statistical “answer” not correct.

How can a statistician help?

Design of experimentTo help, statistician must understand aims, background

AnalysisStatistician can figure out appropriate model, checkassumptions, fit model to data

Research?Talk with statistician early: help with design to prevent biasIf small project, consultation possibleIf detailed project, statistician should be collaborator, notconsultant

Outline

1 Design of experiments: examples

2 Mean, standard deviation, standard error

3 M&M data

4 t-test

5 2-sample t-test

6 Importance of plotting the data

7 1-way ANOVA

8 Summary remarks

Dr. Tuba’s question

Dr. Tuba is interested in the effect of drugs on weight loss inmice after infection.

Drugs are given right after infection6 mice were given the standard drug6 mice were given the new drugDr. Tuba reports p < 0.05 for a t-test of the difference inmean weight loss across the two groups after day 5.

Should Dr. Tuba publish?

Your thoughts?What additional information might you need beforedeciding?

Some things to consider

How did Dr. Tuba choose which mice received each drug?Are the mice independent?Plot the data - why?Did he choose to present results only from the time periodthat showed the most significant effect?

How did Dr. Tuba choose the mice?

Scenario AThe 12 mice were running around in a pen.Dr. Tuba captured the mice one at a time.The first 6 mice captured were given the standard drug.

Problems?Ease of capture might be associated with propensity tolose weight.

Fastest mice might be the most healthyFastest mice might be the lightest

So - mouse selection could be the cause of an apparentdrug effect.

Important to randomize treatment assignment.

Problems?

Ease of capture might be associated with propensity tolose weight.

Scenario BThe mice came from 2 dams, treatment was randomized.Pups from the first dam were given the standard drug.Pups from the second dam were given the new drug.

Problems?The effect of drug is totally confounded with litterTranslation: If we see a difference in outcome between thetwo groups, we can’t tell if it was

caused by the drug - ORdue to a litter effect

Mice in one litter might have a different propensity to loseweight than mice in another litter.

Problems?

The effect of drug is totally confounded with litterTranslation: If we see a difference in outcome between thetwo groups, we can’t tell if it was

Problems?The effect of drug is totally confounded with litterTranslation: If we see a difference in outcome between thetwo groups, we can’t tell if it was

Scenario CThe mice came from 6 dams.The first pup in each dam was given the standard drug.The second pup in each dam was given the new drug.

Problems?Birth order may be associated with the propensity to loseweight.Birth order is confounded with drug.

Better idea: within a litter, randomize first/second born pups todrug.

Problems?

Birth order may be associated with the propensity to loseweight.Birth order is confounded with drug.

Scenario DMice were all first-born pups from different dams.Mice were randomized to treatmentAll mice which received the standard drug were housed inone cage.All mice which received the new drug were housed in one(different) cage.

Problems?There might be a cage effect - viral infection, hotterlocation, less water, · · · .

Better idea: divide animals from a drug group into > 1 cage - orif necessary, house all in same cage.

Problems?

There might be a cage effect - viral infection, hotterlocation, less water, · · · .

Problems?There might be a cage effect - viral infection, hotterlocation, less water, · · · .

Was day 5 one of many days he considered?

1 2 3 4 5

●●

● group 1group 2

Comments?

Simulated data: the mean isthe same for both groups, atall timesLooking for the biggestdifference = multiplecomparisons

1 2 3 4 5

●●

● group 1group 2

Comments?Simulated data: the mean isthe same for both groups, atall timesLooking for the biggestdifference = multiplecomparisons

1 2 3 4 5

●●

● group 1group 2

Comments?

Simulated data: the meandeclines with time for group2, not group 1If expect this, could chooseday 5 a prioriCould use other statisticalmethods to compare slopeover time in the 2 groups.

1 2 3 4 5

●●

● group 1group 2

Comments?Simulated data: the meandeclines with time for group2, not group 1If expect this, could chooseday 5 a prioriCould use other statisticalmethods to compare slopeover time in the 2 groups.

Are the mice independent?

Mice from the same litter are more like each other.Mice housed in the same cage might respond moresimilarlyWe are not interested in the effect of litter (or cage)

If appropriate experimental design, can control for correlatedobservations statistically.Need to record and save data on the litter (or cage)

Are the mice independent?

Mice from the same litter are more like each other.Mice housed in the same cage might respond moresimilarlyWe are not interested in the effect of litter (or cage)

If appropriate experimental design, can control for correlatedobservations statistically.Need to record and save data on the litter (or cage)

Other scenarios (1)

An investigator is comparing response to different doses of 3drugs on 96-well plates.

Each drug is tested at 5 doses.Plate 1 has multiple wells of drug A at each dose level.Plate 2 has multiple wells of drug B at each dose level.Plate 3 has multiple wells of drug C at each dose level.

Is this a good experimental design? Why / why not?

If one drug appears to be better, it might be due to a differencein the plates.

Other scenarios (1)

Other scenarios (2)

Dr. Busy scores mouse activity in a 1-minute period on a scaleof 1-10.

All mice are measured 6 hours after treatment10 mice receive a standard treatment, 10 a new treatment.Treatment assignment is randomizedThe new treatment is expected to be better.The investigator knows the treatment assignment of eachmouse.

Expectation that new treatment is better could unconsciouslybias the investigator’s measurements

Other scenarios (2)

Comments on experimental designs

In the scenarios just described, there is a possibility thatthe intended effect cannot be distinguished fromsomething else that is not of interest.Estimate of intended effect may be biased.

Unless the unintended effect can be estimated(impossible?)

No statistical analysis will fix this problem.Often, the problem can be avoided by good experimentaldesign.

In the scenarios just described, there is a possibility thatthe intended effect cannot be distinguished fromsomething else that is not of interest.Estimate of intended effect may be biased.Unless the unintended effect can be estimated(impossible?)

No statistical analysis will fix this problem.

Often, the problem can be avoided by good experimentaldesign.

In the scenarios just described, there is a possibility thatthe intended effect cannot be distinguished fromsomething else that is not of interest.Estimate of intended effect may be biased.Unless the unintended effect can be estimated(impossible?)

No statistical analysis will fix this problem.Often, the problem can be avoided by good experimentaldesign.

Summary: experimental design

Data should be a random sample from populationIf treatment is assigned, assignment should be randomlydetermined.

Can use table of random numbersBlinding

Person measuring endpoints should not know treatmentassignmentPerson receiving treatment should not know his/hertreatmentParticularly an issue if measurements can be subjective

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

Dr. Forte’s question

A well-known journal article reports that after infection,BALB/c mice given ’superdrug’ lost 4 ounces on averageafter one week.Dr. Forte wants to compare this to ’superdrug’ effects oninfected C57BL/6 mice.Dr. Forte infects 4 C57BL/6 mice, treats with ’superdrug’,and measures weight loss after one week.

His measurements of weight loss: 5, 8, 6, 3 ounces.

The mean

Data (weight loss) represented by x1, x2, · · · , xnn is sample size (number observations)Dr. Forte’s data: x1 = 5, x2 = 8, x3 = 6, x4 = 3.

Population mean (µ): the average xi over all observationsin population

Cannot measure data from all observations in population.µ is unknown (a population parameter)

Estimate µ by µ̂. Here µ̂ = x̄ = sample mean.

For Dr. Forte’s experimentWhat is n?What is the sample?What is the population?

The mean

Data (weight loss) represented by x1, x2, · · · , xnn is sample size (number observations)Dr. Forte’s data: x1 = 5, x2 = 8, x3 = 6, x4 = 3.

Population mean (µ): the average xi over all observationsin population

Cannot measure data from all observations in population.µ is unknown (a population parameter)

Estimate µ by µ̂. Here µ̂ = x̄ = sample mean.

For Dr. Forte’s experimentWhat is n?What is the sample?What is the population?

The mean (cont.)

Dr. Forte’s data: 3, 5, 6, 8.Sample mean, x̄ = 1

i=1 xi

What is Dr. Forte’s sample mean?

x̄ = 14(3 + 5 + 6 + 8) = 5.5

The mean (cont.)

Dr. Forte’s data: 3, 5, 6, 8.Sample mean, x̄ = 1

i=1 xi

What is Dr. Forte’s sample mean?

x̄ = 14(3 + 5 + 6 + 8) = 5.5

Variability

Consider measurements from 2 researchers:Dr. Forte’s measurements: 3, 5, 6, 8.Dr. Pianissimo’s measurements: 5.1, 5.3, 5.7, 5.9Both have x̄ = 5.5.

Whose measurements give you more confidence about yourknowledge of the population mean?

Dr. Pianissimo: data closer together, plausible values for µ insmaller interval.

Variability

Consider measurements from 2 researchers:Dr. Forte’s measurements: 3, 5, 6, 8.Dr. Pianissimo’s measurements: 5.1, 5.3, 5.7, 5.9Both have x̄ = 5.5.

Whose measurements give you more confidence about yourknowledge of the population mean?

Dr. Pianissimo: data closer together, plausible values for µ insmaller interval.

Variance (σ2 = population variance)

Population variance: σ2 = E[(xi − µ)2]

E means “expectation”: average over all observations inpopulation.Translation: σ2 is expected squared difference betweenindividual values and population meanσ = population SD (of xi )

Sample variance, σ̂2

Estimate σ2 by σ̂2.If µ known: σ̂2 = 1

∑ni=1(xi − µ)2

If µ unknown: σ̂2 = s2 = 1n−1

∑ni=1(xi − x̄)2

Dr. Forte’s data: 3, 5, 6, 8, x̄ = 5.5. What is s2?s2 = (3−5.5)2+(5−5.5)2+(6−5.5)2+(8−5.5)2

3 = 4.333

Dr. Pianissimo’s data: 5.1, 5.3, 5.7, 5.9, s2 = 0.133.

∑ni=1(xi − µ)2

∑ni=1(xi − x̄)2

Dr. Forte’s data: 3, 5, 6, 8, x̄ = 5.5. What is s2?

s2 = (3−5.5)2+(5−5.5)2+(6−5.5)2+(8−5.5)2

3 = 4.333

∑ni=1(xi − µ)2

∑ni=1(xi − x̄)2

3 = 4.333

∑ni=1(xi − µ)2

∑ni=1(xi − x̄)2

3 = 4.333

SD, SE

Standard deviation (of x), SD(x) is s =√

Can also write s as sx

Standard error (of x̄), SE(x̄) is√

n = s√n

In other contexts, can speak of standard error of otherquantities, e.g. SE(β̂)

Dr. Forte’s data: 3, 5, 6, 8,x̄ = 5.5, s2 = 4.333.

What are SD(x) and SE(x̄)?SD(x) =

√4.333 = 2.08

SE(x̄) = 2.08√4

= 1.04

Dr. Pianissimo’s data: 5.1, 5.3, 5.7, 5.9x̄ = 5.5, s2 = 0.133.

√0.133 = 0.365

SE(x̄) = 0.365√4

= 0.183

SD, SE

n = s√n

Dr. Forte’s data: 3, 5, 6, 8,x̄ = 5.5, s2 = 4.333.

What are SD(x) and SE(x̄)?

SD(x) =√

4.333 = 2.08SE(x̄) = 2.08√

4= 1.04

√0.133 = 0.365

SE(x̄) = 0.365√4

= 0.183

SD, SE

n = s√n

Dr. Forte’s data: 3, 5, 6, 8,x̄ = 5.5, s2 = 4.333.

√4.333 = 2.08

SE(x̄) = 2.08√4

= 1.04

What are SD(x) and SE(x̄)?

SD(x) =√

0.133 = 0.365SE(x̄) = 0.365√

4= 0.183

SD, SE

n = s√n

Dr. Forte’s data: 3, 5, 6, 8,x̄ = 5.5, s2 = 4.333.

√4.333 = 2.08

SE(x̄) = 2.08√4

= 1.04

√0.133 = 0.365

SE(x̄) = 0.365√4

= 0.183

SD, SE (cont.)

Recall SD(x) is sx =√

1n−1

∑ni=1(xi − x̄)2, SE(x̄) = sx√

If Dr. Forte had collected data from 20 mice instead of 4, howwould SD(x) and SE(x̄) be affected?

SD(x) would be about the same (estimated more precisely)SE(x̄) would decrease (we have more confidence about x̄).

What do SD(x) and SE(x̄) mean?SD(x) is our estimate of σ, the SD of x in the population.SE(x̄) is a measure of our uncertainty about µ.

Can form confidence interval (CI) for µ: x̄ ± tcritSE(x̄)tcrit discussed later (usually ≈ 2)Approximately 95% of such CIs include µ.

SD, SE (cont.)

1n−1

∑ni=1(xi − x̄)2, SE(x̄) = sx√

SD, SE (cont.)

1n−1

∑ni=1(xi − x̄)2, SE(x̄) = sx√

Why divide by n − 1 in s2?

If we knew µ, σ̂2 =Pn

i=1(xi−µ)2

We don’t know µ, so σ̂2 = s2 =Pn

i=1(xi−x̄)2

For a particular sample, x̄ minimizes∑

(xi − µ̂)2 for any µ̂

Using x̄ to estimate µ:Numerate of s2 is (on average) smaller than

∑(xi − µ)2

Need to divide by smaller number to compensate; n − 1 isthe right number

Examine by simulationTake n = 5 observations fromnormal, µ = 0, σ2 = 1.

Calculate s2 =P

(xi−x̄)2

4 andP(xi−x̄)2

Repeat multiple timesNote that true value of σ is 1.

Dividing by n − 1 does wellDividing by n underestimates σ

SD(x): divide by (n−1)

0.5 1.0 1.5 2.0

Not SD(x): divide by n

0.0 0.5 1.0 1.5

Examine by simulationTake n = 5 observations fromnormal, µ = 0, σ2 = 1.

Calculate s2 =P

(xi−x̄)2

4 andP(xi−x̄)2

Repeat multiple timesNote that true value of σ is 1.Dividing by n − 1 does wellDividing by n underestimates σ

SD(x): divide by (n−1)

0.5 1.0 1.5 2.0

Not SD(x): divide by n

0.0 0.5 1.0 1.5

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

M&M data

We collected data on the number of M&M of each color inpackages of mini M&M’s

# brown

5 10 15 20

# orange

4 6 8 10 12 14 16

# yellow

2 4 6 8 10 12 14

# green

6 8 10 12

# blue

4 6 8 10 12 14 16 18

4 6 8 10 12 14 16

32 people totalHistograms show number ofeach color

M&M data

Aside: did everyone get same # M&M’s? NO

Number MM: 50 51 55 56 57 58 59 60 61 62 66 73People: 1 1 3 2 9 2 4 5 1 2 1 1

%(blue + red): All

0.30 0.35 0.40 0.45

67 Focus on % (red + blue)

x1 is % (red + blue) for firstperson, x2 same for secondperson, etc.What is the sample?What is the population?

We could approximate xi with anormal distribution, mean µ,variance σ2

M&M data

Number MM: 50 51 55 56 57 58 59 60 61 62 66 73People: 1 1 3 2 9 2 4 5 1 2 1 1

%(blue + red): All

0.30 0.35 0.40 0.45

M&M data

Number MM: 50 51 55 56 57 58 59 60 61 62 66 73People: 1 1 3 2 9 2 4 5 1 2 1 1

%(blue + red): All

0.30 0.35 0.40 0.45

M&M data

sample 1

0.30 0.35 0.40 0.45

sample 2

0.30 0.35 0.40 0.45

sample 3

0.30 0.32 0.34 0.36 0.38 0.40

sample 4

0.30 0.32 0.34 0.36 0.38 0.40

%(blue + red): samples

Interest: % (red + blue)Divide data into 4 groups of 8observationsCalculate mean, SD, SE(x̄) ineach group

M&M data

Each group has n = 8 observations (32 observations total)Group 1: % (red + blue): Data (x1, · · · , x8) is

0.45, 0.38, 0.39, 0.38, 0.35, 0.33, 0.41, 0.29

Calculate x̄

x̄ = (0.45+0.38+0.39+0.38+0.35+0.33+0.41+0.29)8 = 0.374

Calculate sx (standard deviation)

√(0.45−0.374)2+(0.38−0.374)2+···+(0.29−0.374)2

7 = 0.048

Calculate SE(x̄) = sx√n = 0.048√

8= 0.017

M&M data

0.45, 0.38, 0.39, 0.38, 0.35, 0.33, 0.41, 0.29

Calculate x̄x̄ = (0.45+0.38+0.39+0.38+0.35+0.33+0.41+0.29)

8 = 0.374Calculate sx (standard deviation)

√(0.45−0.374)2+(0.38−0.374)2+···+(0.29−0.374)2

7 = 0.048

8= 0.017

M&M data

0.45, 0.38, 0.39, 0.38, 0.35, 0.33, 0.41, 0.29

Calculate x̄x̄ = (0.45+0.38+0.39+0.38+0.35+0.33+0.41+0.29)

8 = 0.374Calculate sx (standard deviation)

√(0.45−0.374)2+(0.38−0.374)2+···+(0.29−0.374)2

7 = 0.048

8= 0.017

M&M data

Each group has n = 8 observations (32 observations total)Group 1: x̄= 0.374, sx= 0.0479 SE(x̄)= 0.0169Group 2: x̄= 0.3564, sx= 0.0574 SE(x̄)= 0.0203Group 3: x̄= 0.3549, sx= 0.0348 SE(x̄)= 0.0123Group 4: x̄= 0.3465, sx= 0.0342 SE(x̄)= 0.0121

Everyone: x̄= 0.358 sx= 0.0436 SE(x̄)= 0.0077

We might want to know if % red + blue in the population = 13

(later: t-test)

M&M data

Everyone: x̄= 0.358 sx= 0.0436 SE(x̄)= 0.0077

(later: t-test)

M&M data

Everyone: x̄= 0.358 sx= 0.0436 SE(x̄)= 0.0077

(later: t-test)

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

t-test

Recall Dr. Forte’s question: is weight loss in C57BL/6 mice(his data) different than in BALB/c mice (published data,weight loss = 4 ounces)?

H0 is the null hypothesis (“no effect”, “same as before”)HA is the alternative hypothesis (“some effect”, “not H0”)

Example: H0 : µ = µ0 versus HA : µ 6= µ0.µ is mean weight loss for C57Bl/6 mice in populationµ is unknown, but we have an estimate of it (x̄)µ0 is some specific number (here: 4).

How to evaluate H0, HA?Compare x̄ to µ0If x̄ is far from µ0, suggests H0 not true.Need SE(x̄) to evaluate how different x̄ and µ0 are.

t-test

H0 is the null hypothesis (“no effect”, “same as before”)HA is the alternative hypothesis (“some effect”, “not H0”)Example: H0 : µ = µ0 versus HA : µ 6= µ0.

µ is mean weight loss for C57Bl/6 mice in populationµ is unknown, but we have an estimate of it (x̄)µ0 is some specific number (here: 4).

t-test

H0 is the null hypothesis (“no effect”, “same as before”)HA is the alternative hypothesis (“some effect”, “not H0”)Example: H0 : µ = µ0 versus HA : µ 6= µ0.

µ is mean weight loss for C57Bl/6 mice in populationµ is unknown, but we have an estimate of it (x̄)µ0 is some specific number (here: 4).

t-test assumptions

T-test is based on tcalc = x̄−µ0SE(x̄)

Assumptions: xi ∼ independent N(µ, σ2)

Interpretation: ∼ means “distributed as”Says data are independent and have a normal distributionµ = population mean (of the xi ’s)σ2 = population variance (of the xi ’s) (σ = SD)

“How can I tell if my data have a normal distribution?”Assumption is about distribution of data from populationt-test robust to moderate departure from normalityImportant to plot data to check for gross violation fromnormality

t-test assumptions

Assumptions: xi ∼ independent N(µ, σ2)

Interpretation: ∼ means “distributed as”Says data are independent and have a normal distributionµ = population mean (of the xi ’s)σ2 = population variance (of the xi ’s) (σ = SD)

“How can I tell if my data have a normal distribution?”Assumption is about distribution of data from populationt-test robust to moderate departure from normalityImportant to plot data to check for gross violation fromnormality

Samples of n=20 from a normal distribution

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

Normal

normsamp[ctr]

−3 −2 −1 0 1 2 3

One mistake

c(normsamp[1:(nper.plot − 1)], 10.25)

−4 −2 0 2 4 6 8 10

Samples of n=20 from a normal distribution

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

normsamp[ctr]

−3 −1 0 1 2 3

Normal

normsamp[ctr]

−3 −2 −1 0 1 2 3

One mistake

c(normsamp[1:(nper.plot − 1)], 10.25)

−4 −2 0 2 4 6 8 10

Outliers

“My data has one very unusual value. Should I delete it?”

Step 1: check to see if the value is a mistake! If so, fix.Step 2: is there something unusual about thatobservation?

animal was sicka different person took the measurementexperimental conditions different from others

If clear reason why unusual value differs from otherobservations

Value represents different condition: OK to delete (explainin paper)

If cannot determine any reason why value is unusualDeleting observation underestimates population variability:don’t deleteAlternative: report results both with and without the value(explain in paper)

Outliers

“My data has one very unusual value. Should I delete it?”Step 1: check to see if the value is a mistake! If so, fix.Step 2: is there something unusual about thatobservation?

Outliers

“My data has one very unusual value. Should I delete it?”Step 1: check to see if the value is a mistake! If so, fix.Step 2: is there something unusual about thatobservation?

Parametric or non-parametric?

“When should I use a non-parametric version of a t-test?”Some statisticians (nearly) always use non-parametrictests (such as Mann-Whitney)Some statisticians (nearly) always use parametric tests(such as t-test)Knowledge of type of data can help determine if normalityis reasonable

Sometimes data approximately normal after transformation

If no severe departure from normality, either parametric ornonparametric tests probably fine.

t-test example

Dr. Forte: H0 : µ = 4, HA : µ 6= 4.We calculated x̄ = 5.5, SE(x̄) = 1.04.What is tcalc?

Here tcalc = (5.5− 4)/1.04 = 1.44

Under H0, tcalc ∼ tn−1

Here: tcalc ∼ t3If H0 true, tcalc centered around 0

t-test example

Dr. Forte: H0 : µ = 4, HA : µ 6= 4.We calculated x̄ = 5.5, SE(x̄) = 1.04.What is tcalc?

Here tcalc = (5.5− 4)/1.04 = 1.44

Under H0, tcalc ∼ tn−1

Here: tcalc ∼ t3If H0 true, tcalc centered around 0

t-distribution

Recall tcalc = x̄−µ0SE(x̄)

“How can tcalc have a distribution? It’s just a number! ”Your thoughts?

Any single experiment gives a single tcalc .Could take another sample of 4 C57BL/6 mice, measureweight loss, get tcalc for that sample.This could theoretically be repeated multiple times withdifferent miceEach experiment would give a tcalc

A histogram of the tcalc values would look like the t3distribution (assuming enough experiments)

t-distribution

Recall tcalc = x̄−µ0SE(x̄)

“How can tcalc have a distribution? It’s just a number! ”Your thoughts?

Any single experiment gives a single tcalc .Could take another sample of 4 C57BL/6 mice, measureweight loss, get tcalc for that sample.This could theoretically be repeated multiple times withdifferent miceEach experiment would give a tcalc

A histogram of the tcalc values would look like the t3distribution (assuming enough experiments)

Samples from t-distributions with 3 df

−1.0 −0.5 0.0 0.5 1.0 1.5 2.0

n= 100

−6 −4 −2 0 2 4

n= 1000

−4 −2 0 2 4 6

n= 10000

−6 −4 −2 0 2 4 6

n=10: First fewobservations: -0.69,1.73, 0.55, 0.96,-0.79, · · ·

n=100: Starting tolook roughly “normal”n=1000: (12 obsoutside plot. Rangewas: -12.8 to 8.3)n=10,000: (85 obsoutside plot. Rangewas: -24.3 to 35.7)

−1.0 −0.5 0.0 0.5 1.0 1.5 2.0

n= 100

−6 −4 −2 0 2 4

n= 1000

−4 −2 0 2 4 6

n= 10000

−6 −4 −2 0 2 4 6

n=10: First fewobservations: -0.69,1.73, 0.55, 0.96,-0.79, · · ·n=100: Starting tolook roughly “normal”

n=1000: (12 obsoutside plot. Rangewas: -12.8 to 8.3)n=10,000: (85 obsoutside plot. Rangewas: -24.3 to 35.7)

−1.0 −0.5 0.0 0.5 1.0 1.5 2.0

n= 100

−6 −4 −2 0 2 4

n= 1000

−4 −2 0 2 4 6

n= 10000

−6 −4 −2 0 2 4 6

n=10: First fewobservations: -0.69,1.73, 0.55, 0.96,-0.79, · · ·n=100: Starting tolook roughly “normal”n=1000: (12 obsoutside plot. Rangewas: -12.8 to 8.3)

n=10,000: (85 obsoutside plot. Rangewas: -24.3 to 35.7)

−1.0 −0.5 0.0 0.5 1.0 1.5 2.0

n= 100

−6 −4 −2 0 2 4

n= 1000

−4 −2 0 2 4 6

n= 10000

−6 −4 −2 0 2 4 6

n=10: First fewobservations: -0.69,1.73, 0.55, 0.96,-0.79, · · ·n=100: Starting tolook roughly “normal”n=1000: (12 obsoutside plot. Rangewas: -12.8 to 8.3)n=10,000: (85 obsoutside plot. Rangewas: -24.3 to 35.7)

Normal, t distributions: differences?

Normal distributionn= 10000

−6 −4 −2 0 2 4 6

Normal: range was: -3.6 to 3.9

t3 distributionn= 10000

−6 −4 −2 0 2 4 6

t3: range was: -24.3 to 35.7

Different t-distributions

−6 −4 −2 0 2 4 6

df=2df=8normal

With more data, larger ndegrees of freedom (df) =n − 1 increasessmaller area in tails of thedistributionapproaches normaldistribution (which is t with∞ df)

Normal, t distributions

“Why doesn’t tcalc = x̄−µ0SE(x̄) have a normal distribution?”

Recall we assumed xi ∼ N(µ, σ2)

x̄ has a normal distributionIf we knew σ2

SE(x̄) would be σ2/n, a fixed number“tcalc” would have a normal distribution

When σ2 not known, use s2 to estimate itOur estimate of σ2 isn’t perfectResult: tcalc more variable

Normal, t distributions

“Why doesn’t tcalc = x̄−µ0SE(x̄) have a normal distribution?”

Recall we assumed xi ∼ N(µ, σ2)

x̄ has a normal distributionIf we knew σ2

SE(x̄) would be σ2/n, a fixed number“tcalc” would have a normal distribution

When σ2 not known, use s2 to estimate itOur estimate of σ2 isn’t perfectResult: tcalc more variable

t-distribution (3 df)

−6 −4 −2 0 2 4 6

t dist, 3 df2.5% cutpt = 3.18

If H0 trueExpect values nearcenterRed: unusual

If H0 true, control Pr(reject H0) to α(α often chosen as 0.05)Reject H0 if tcalc in shaded area

Each shaded area has 2.5% of thedistributionShaded area defined by tcrit

tcrit is number for which 2.5% ofdistribution has larger values(−tcrit : 2.5% are smaller)tcrit depends on df and α

With 3 df, α = 0.05, tcrit = 3.18Reject H0 if |tcalc | > tcrit

−6 −4 −2 0 2 4 6

Sample from t-distribution: how good is tcrit?

n= 10000

−6 −4 −2 0 2 4 6

How good is tcrit?Do we really get 2.5% in each “tail”?For t3, α = 0.05, tcrit = 3.18.

From sample of 10,000 observation from t3247 observations were < −3.18 (2.47%)258 observations were > 3.18 (2.58%)

If H0 true and α = 0.05, approximately 5% of the time wewould reject H0 (“type I error”)

Sample from t-distribution: how good is tcrit?

n= 10000

−6 −4 −2 0 2 4 6

How good is tcrit?Do we really get 2.5% in each “tail”?For t3, α = 0.05, tcrit = 3.18.

From sample of 10,000 observation from t3247 observations were < −3.18 (2.47%)258 observations were > 3.18 (2.58%)

If H0 true and α = 0.05, approximately 5% of the time wewould reject H0 (“type I error”)

t-test for Dr. Forte

Recall Dr. Forte’s question: is weight loss in C57BL/6mice (his data) different than in BALB/c mice(published data, weight loss = 4 ounces)?His hypotheses: H0 : µ = 4, HA : µ 6= 4.µ: weight loss in population of C57BL/6 miceDr. Forte: tcalc = x̄−µ0

SE(x̄) = (5.5− 4)/1.04 = 1.44

t-test rule: Reject H0 if |tcalc | > tcrit

With 3 df, α = 0.05, tcrit = 3.18What should Dr. Forte conclude?

Fail to reject H0Dr. Forte’s data are consistent with H0 (p = 0.25)No evidence that weight loss in C57BL/6 mice isdifferent from 4 ounces.

t-test for Dr. Forte

Recall Dr. Forte’s question: is weight loss in C57BL/6mice (his data) different than in BALB/c mice(published data, weight loss = 4 ounces)?His hypotheses: H0 : µ = 4, HA : µ 6= 4.µ: weight loss in population of C57BL/6 miceDr. Forte: tcalc = x̄−µ0

SE(x̄) = (5.5− 4)/1.04 = 1.44

t-test rule: Reject H0 if |tcalc | > tcrit

With 3 df, α = 0.05, tcrit = 3.18What should Dr. Forte conclude?Fail to reject H0

Dr. Forte’s data are consistent with H0 (p = 0.25)No evidence that weight loss in C57BL/6 mice isdifferent from 4 ounces.

t-distribution (3df)

−6 −4 −2 0 2 4 6

Dr. Forte’s value: blue line(1.44)Compare Dr. Forte’s value toa t distribution with 3 dfReject H0 if tcalc in shadedareaFail to reject H0

Confidence interval

95% confidence interval (CI): x̄ ± tcrit × SE(x̄)

(Assumes tcrit calculated using α = 0.05)

This example: tcrit = 3.18We calculated x̄ = 5.5, SE(x̄) = 1.04What is Dr. Forte’s 95% CI?

95% CI is 5.5± 3.18× 1.04 = (2.19, 8.81)

This is a 95% CI for µFail to reject H0 because interval includes µ0 = 4.Confidence interval more informative than “fail to reject H0”

Confidence interval

95% confidence interval (CI): x̄ ± tcrit × SE(x̄)

(Assumes tcrit calculated using α = 0.05)

This example: tcrit = 3.18We calculated x̄ = 5.5, SE(x̄) = 1.04What is Dr. Forte’s 95% CI?95% CI is 5.5± 3.18× 1.04 = (2.19, 8.81)

This is a 95% CI for µFail to reject H0 because interval includes µ0 = 4.Confidence interval more informative than “fail to reject H0”

Classical statistics

µ considered fixed.Observed data (xi ) are considered random

Could collect new data: different xiImplies calculated confidence interval is random.

If H0 is true, on average 95% of the CIs will include µ.Can never answer “Is H0 true?”

Can only say something about how unusual our teststatistic is if H0 is true

p-value

−6 −4 −2 0 2 4 6

t dist, 3 df12.5% cutpt = 1.44 p-value = probability of observing

a statistic as extreme or moreextreme as we observed, if H0 istrue

Curve: total area=1Dr. Forte’s tcalc = 1.44.p-value = shaded area = 0.25

p-value

We calculated p = 0.25 for Dr. Forte’s data.

Is this the same as the probability that H0 is true?

Nop-value is calculated assuming H0 is true.

Recall: p-value is probability of observing a statistic as or moreextreme as we observed, if H0 is true

p-value

We calculated p = 0.25 for Dr. Forte’s data.

Is this the same as the probability that H0 is true?

Nop-value is calculated assuming H0 is true.

Recall: p-value is probability of observing a statistic as or moreextreme as we observed, if H0 is true

p-value questions

What do you think of these statements?“The p-value tells you whether or not the observed effect isreal.”

Not true“If the result is not statistically significant, that proves thereis no difference.” Not true

Better:Use knowledge of the magnitude of the estimated effect(and the CI) to help interpret the results.Suppose comparing treatment to control and p = 0.07

If estimated effect large, “Our study suggests an importantbenefit of treatment, but this did not reach statisticalsignificance.”

A p-value is not an end in itself.

p-value questions

What do you think of these statements?“The p-value tells you whether or not the observed effect isreal.” Not true

“If the result is not statistically significant, that proves thereis no difference.” Not true

p-value questions

What do you think of these statements?“The p-value tells you whether or not the observed effect isreal.” Not true“If the result is not statistically significant, that proves thereis no difference.”

Not trueBetter:

Use knowledge of the magnitude of the estimated effect(and the CI) to help interpret the results.Suppose comparing treatment to control and p = 0.07

p-value questions

What do you think of these statements?“The p-value tells you whether or not the observed effect isreal.” Not true“If the result is not statistically significant, that proves thereis no difference.” Not true

p-value questions

What do you think of these statements?“The p-value tells you whether or not the observed effect isreal.” Not true“If the result is not statistically significant, that proves thereis no difference.” Not true

t-test for M&M’s

Consider M&M data from 1st sample of 8 peopleGroup 1: x̄= 0.374, sx= 0.0479 SE(x̄)= 0.0169

Recall xi is % (red + blue) M&M’sTest H0 : µ = 1/3, versus HA : µ 6= 1/3.What is tcalc?

0.374− 0.3330.0169

= 2.41

Compare to tcrit = 2.36 (based on (n-1) df, α = 0.05)Conclusion? Reject H0 (also p = 0.047).

For first 8 people, mean % (red + blue) M&M’s significantlydifferent than 1/3.

t-test for M&M’s

0.374− 0.3330.0169

= 2.41

Compare to tcrit = 2.36 (based on (n-1) df, α = 0.05)Conclusion?

Reject H0 (also p = 0.047).For first 8 people, mean % (red + blue) M&M’s significantlydifferent than 1/3.

t-test for M&M’s

0.374− 0.3330.0169

= 2.41

Compare to tcrit = 2.36 (based on (n-1) df, α = 0.05)Conclusion? Reject H0 (also p = 0.047).

For first 8 people, mean % (red + blue) M&M’s significantlydifferent than 1/3.

t-test for M&M’s

For groups of 8 people, tcrit = 2.36Group 1: x̄= 0.374, sx= 0.0479 SE(x̄)= 0.0169

tcalc = 2.41, p = 0.047Group 2: x̄= 0.3564, sx= 0.0574 SE(x̄)= 0.0203

tcalc = 1.084, p = 0.314

For entire sample (n=32), tcrit = 2.04Everyone: x̄= 0.358 sx= 0.0436 SE(x̄)= 0.0077

tcalc = 3.19, p = 0.003

Larger n: smaller SE(x̄), smaller tcrit , easier to reject H0 if H0not true.

t-test for M&M’s

tcalc = 1.084, p = 0.314

tcalc = 3.19, p = 0.003

t-test for M&M’s

tcalc = 1.084, p = 0.314

tcalc = 3.19, p = 0.003

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

2-sample t-test

Dr. Forte now wants to compare the effect of ’superdrug’on weight loss in knockout mice (new) versus C57BL/6mice (just measured)

NOTE: Better to collect data on both species in the sametime period.

Let x̄1 and s1 be mean and SD of weight loss in 1st sample(C57BL/6), n1 = 4 observations.Let x̄2 and s2 be mean and SD of weight loss in 2ndsample (knockout), n2 observationsµ1 is population mean weight loss in C57BL/6 mice, µ2 inknockout mice.Null hypothesis of “no effect” or “no difference” isH0 : µ1 = µ2, so HA : µ1 6= µ2.

2-sample t-test (cont.)

H0 : µ1 = µ2, HA : µ1 6= µ2

Same as H0 : µ1 − µ2 = 0 and HA : µ1 − µ2 6= 0.Estimate µ1 − µ2 by x̄1 − x̄2.tcalc = x̄1−x̄2

SE(x̄1−x̄2)

Assumptions of the 2-sample test: data from 2 populationsAre independentAre normally distributed (within population)Have the same variance (or: use modified estimate ofstandard error)

Reject H0 if |tcalc | > tcrit (tcrit depends on n1 + n2 and α)

Scenario 1

Pairs of mice are housed in the same cage with same foodsourceInterest is in effect of restricted food intakeMice fed 2x/day, only 1 mouse can eat at once

Is this a reasonable design?

Mice are not independent: if one mouse in a pair eatsaggressively, the other mouse cannot eat as much

Scenario 1

Scenario 2

12 mice receive standard food, 12 a new food.Interest: gain in head circumference, measured at end of 2weeksBob measures mice receiving standard food, Harrymeasures the othersBob and Harry do not know which mice received whichfood source

Is this a reasonable design? No

Food source is totally confounded with person who takesmeasurementsBob might tend to get smaller measurements on sameanimal (bias)Harry might tend to measure less accurately (largervariance)

Scenario 2

Is this a reasonable design? No

Dr. Forte’s 2-sample data

Do you think assumptions arereasonable?

(OK)H0 : µ1 − µ2 = 0, HA : µ1 − µ2 6= 0.tcalc = x̄1−x̄2

SE(x̄1−x̄2)

x̄1 = 5.5, x̄2 = 2.1SE(x̄1 − x̄2) = 1.19 (assumes samevariance in 2 groups)What is tcalc? 5.5−2.1

1.19 = 2.86tcrit = 2.45, from 6 df, α = 0.05What is your conclusion? (Reject H0)

95% CI for µ1 − µ2 is x̄1 − x̄2 ± tcrit × SE(x̄1 − x̄2) = (0.50, 6.30)

Do you think assumptions arereasonable? (OK)H0 : µ1 − µ2 = 0, HA : µ1 − µ2 6= 0.tcalc = x̄1−x̄2

SE(x̄1−x̄2)

x̄1 = 5.5, x̄2 = 2.1SE(x̄1 − x̄2) = 1.19 (assumes samevariance in 2 groups)What is tcalc?

5.5−2.11.19 = 2.86

tcrit = 2.45, from 6 df, α = 0.05What is your conclusion? (Reject H0)

SE(x̄1−x̄2)

1.19 = 2.86tcrit = 2.45, from 6 df, α = 0.05What is your conclusion?

(Reject H0)

SE(x̄1−x̄2)

1.19 = 2.86tcrit = 2.45, from 6 df, α = 0.05What is your conclusion? (Reject H0)

Dr. Forte’s conclusions

Recall: Dr. Forte wants to compare the effect of’superdrug’ on weight loss in knockout vs C57BL/6mice.C57BL/6 mice: (x̄1 = 5.5), knockout mice (x̄2 = 2.1)We rejected H0, p = 0.03

What does “reject H0” mean here?

Data suggests that theaverage weight loss in C57BL/6 mice is significantlygreater than in knockout mice.

What does the p-value mean?If the 2 species had the same mean weight loss, we wouldhave observed a difference as great or greater 3% of thetime.

What does “reject H0” mean here? Data suggests that theaverage weight loss in C57BL/6 mice is significantlygreater than in knockout mice.

What does the p-value mean?

If the 2 species had the same mean weight loss, we wouldhave observed a difference as great or greater 3% of thetime.

What does “reject H0” mean here? Data suggests that theaverage weight loss in C57BL/6 mice is significantlygreater than in knockout mice.

What does the p-value mean?If the 2 species had the same mean weight loss, we wouldhave observed a difference as great or greater 3% of thetime.

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

Plot the data: why?

Comments?

The data distribution looksreasonable

Plot the data: why?

Comments?The data distribution looksreasonable

Plot the data: why?

●●●●●●

Comments?

One large outlying observationA mistake?Non-normal distribution?

Plot the data: why?

●●●●●●

Comments?One large outlying observation

A mistake?Non-normal distribution?

Plot the data: why?

Comments?

Variability within groups notsimilarSame data as first plot -EXCEPT data isexponentiatedSuggest: use logtransformationKnowledge of what theoutcomes are can guidedecision about whether totransform

Plot the data: why?

Comments?Variability within groups notsimilarSame data as first plot -EXCEPT data isexponentiatedSuggest: use logtransformationKnowledge of what theoutcomes are can guidedecision about whether totransform

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

Dr. Crescendo

Dr. Crescendo wants to test the effects of drugs A, B, andC on cell counts in 96-well platesSome things to consider

Is each drug tested on multiple plates? (plate effect)Are the wells at the edge of the plate different from others?(edge effect)Are the wells for drug A always read first? (order effect)

Set up for 1-way ANOVA

Dr. Crescendo took 5 measurements from each of the 3treatments (A,B,C)µ1, µ2, µ3 are population means of treatment effects A, B,and C.Hypotheses:

H0 : µ1 = µ2 = µ3HA : at least 2 means are not equal.

How to test H0, HA?

Pictures for 1-way ANOVA

8group

How do the 2 pictures differ?

Right: 3 means are more different from each otherRight: Variability between means is also more differentANOVA compares variability between treatments tovariability within treatments

8group

How do the 2 pictures differ?Right: 3 means are more different from each other

Right: Variability between means is also more differentANOVA compares variability between treatments tovariability within treatments

8group

How do the 2 pictures differ?Right: 3 means are more different from each otherRight: Variability between means is also more different

ANOVA compares variability between treatments tovariability within treatments

8group

How do the 2 pictures differ?Right: 3 means are more different from each otherRight: Variability between means is also more differentANOVA compares variability between treatments tovariability within treatments

Dr Crescendo’s data

●●

Dr. Crescendo's data

Sample sizes:n1 = n2 = n3 = 5.Sample means: x̄1 =5.49, x̄2 = 6.07, x̄3 = 6.38.Sample variances: s2

1 =0.29, s2

2 = 0.48, s23 = 0.66

1-way ANOVA: basics

ANOVA assumes Var(x1) = Var(x2) = Var(x3)

Call this σ2

Variability within treatments is weighted average ofs2

1, s22, s2

3.degrees of freedom (df) = n1 + n2 + n3 − 3 (here: df=12)Dr. Crescendo: within treatment variance is3.016/12 = 0.25

Variability between treatments is weighted average of howfar apart group means are from overall mean

df = number groups - 1 (here: df=2)Dr. Crescendo: between treatment variance is2.063/2 = 1.03

1-way ANOVA: basics

Call this σ2

1, s22, s2

1-way ANOVA: basics

Call this σ2

1, s22, s2

Testing in 1-way ANOVA (3 groups)

Recall: 1-way ANOVA comparesVariability within treatment: MS(Within) = MS(Error)Variability between treatment: MS(Between) = MS(Group)(Note: MS = mean square)

Dr. Crescendo:MS(Error) = 3.016

12 = 0.25MS(Between) = 2.063

2 = 1.03

ANOVA tableDf Sum Sq Mean Sq

Between 2 2.06337 1.03169Error 12 3.01612 0.25134

Recall: 1-way ANOVA comparesVariability within treatment: MS(Within) = MS(Error)Variability between treatment: MS(Between) = MS(Group)(Note: MS = mean square)

Dr. Crescendo:MS(Error) = 3.016

12 = 0.25MS(Between) = 2.063

2 = 1.03

ANOVA tableDf Sum Sq Mean Sq

Between 2 2.06337 1.03169Error 12 3.01612 0.25134

1-way ANOVA: basics

If H0 trueHow do MS(Within) and MS(Between) compare?How could we estimate σ2?

MS(Between) ≈ MS(Within), and both estimate σ2

MS(Between)MS(Within) close to 1.

If H0 not true, same questionsOnly MS(Within) estimates σ2.

MS(Between) > MS(Within) so MS(Between)MS(Within) > 1

1-way ANOVA: basics

If H0 trueHow do MS(Within) and MS(Between) compare?How could we estimate σ2?MS(Between) ≈ MS(Within), and both estimate σ2

1-way ANOVA: basics

If H0 not true, same questions

Only MS(Within) estimates σ2.

1-way ANOVA: basics

H0 : µ1 = µ2 = µ3, HA : at least 2 means are not equal.

ANOVA tableDf Sum Sq Mean Sq F value Pr(>F)

Between 2 2.06337 1.03169 4.1047 0.04383Error 12 3.01612 0.25134

Test of H0 is Fcalc =MS(Between)

MS(Error) = 4.10.

If H0 true (and assumptions met!), Fcalc ∼ F2,12

Between 2 2.06337 1.03169 4.1047 0.04383Error 12 3.01612 0.25134

MS(Error) = 4.10.

Between 2 2.06337 1.03169 4.1047 0.04383Error 12 3.01612 0.25134

MS(Error) = 4.10.

Some F distributions

0 1 2 3 4 5

df=2,12df=3,12df=8,12df=8,100

Dr. Crescendo’s Fcalc

0 1 2 3 4 5 6

F dist, 2 and 12 df5% cutpt = 3.89

Generally: only large valuesof Fcalc considered unusual(1-tail)Reject H0 ifFcalc > F2,12,.95 = 3.89.Dr. Crescendo:Fcalc = 1.03/0.25 = 4.10,p = 0.044. Reject H0.

Outline

3 M&M data

4 t-test

5 2-sample t-test

7 1-way ANOVA

8 Summary remarks

Interacting with statistician

“What questions should I ask the statistician?”

Who asks the first questions?I ask: “Tell me about your experiment and your question.”We need some background

If designing an experiment:I ask: “What is your research question (explain in terms ofdata you plan to collect)?”We can help you

Decide what data to collect to answer questionHow to design experimentFigure out reasonable sample size (but need furtherinformation from you: estimated effect)

For data already collectedI ask: “What is your research question (explained in termsof data)?”Plots of data helpful

Interacting with statistician

“What questions should I ask the statistician?”Who asks the first questions?

I ask: “Tell me about your experiment and your question.”We need some background

If designing an experiment:I ask: “What is your research question (explain in terms ofdata you plan to collect)?”We can help you

Decide what data to collect to answer questionHow to design experimentFigure out reasonable sample size (but need furtherinformation from you: estimated effect)

For data already collectedI ask: “What is your research question (explained in termsof data)?”Plots of data helpful

How a statistician can help

Design of experimentStatistician must understand study aims and backgroundCan help investigator decide

what data to collecthow to design data collection processhow to use data to answer research questions

AnalysisDetermine appropriate modelUnderstand and check assumptions (use different model ifviolated)Carry out analysis

What does Biostatistics Dept offer?

CoursesBST463 = Introduction to Biostatistics (fall semester)

Consulting serviceRotation system: faculty and PhD student or postdocFree assistance with grant preparation if appropriate %effort for BiostatisticsFree 1-hour consultation on anythingIf related to CTSI, voucher possible for up to 10 free hoursStudent projects: faculty mentor must attend initial meetingand most other meetings with BiostatisticiansFaculty advisor can contact Susan Messing(Susan_Messing@urmc.rochester.edu) for appointment,see http://www.urmc.rochester.edu/biostat/consulting/

Parting remarks

Statistics is not just “crunching numbers”Requires understanding of model and assumptions.

Good experimental design and data collection is keyPoor design: may be impossible to make valid inferenceabout population.Biased data collection: may never be able to adjust for bias.Good design: enables statistical inference about populationof inference based on data collected.

Basic Statistics: MBI-540 · Basic Statistics: MBI-540 Sally W. Thurston, PhD Associate Professor...

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