• On Thursday, I’ll provide information about the project

• Due on Friday after last class.• Proposal will be due two weeks from today (April 15th)• You’re encouraged (but not required) to work in groups

of three people

• Homework:– Due next Tuesday– On web tonight

Announcements

Hypothesis Testing: 20,000 Foot View

1. Set up the hypothesis to test and collect data

Hypothesis to test: HO

Hypothesis Testing: 20,000 Foot View

1. Set up the hypothesis to testand collect data

2. Assuming that the hypothesis is true, are the observed data likely?

Data are deemed “unlikely” if the test statistic is in theextreme of its distribution when HO is true.

Hypothesis to test: HO

Hypothesis Testing: 20,000 Foot View

1. Set up the hypothesis to testand collect data

2. Assuming that the hypothesis is true, are the observed data likely?

3. If not, then the alternative to the hypothesis must be true.

Data are deemed “unlikely” if the test statistics is in theextreme of its distribution when HO is true.

Alternative to HO is HA

Hypothesis to test: HO

Hypothesis Testing: 20,000 Foot View

1. Set up the hypothesis to testand collect data

2. Assuming that the hypothesis is true, are the observed data likely?

3. If not, then the alternative to the hypothesis must be true.

4. P-value describes how likely the observed data are assuming HO is true. (i.e. answer to Q#2 above)

Data are deemed “unlikely” if the test statistics is in theextreme of its distribution when HO is true.

“Unlikely” if p-value <

Alternative to HO is HA

Hypothesis to test: HO

Large Sample Test for a Proportion:Taste Test Data

• 33 people drink two unlabeled cups of cola (1 is coke and 1 is pepsi)

• p = proportion who correctly identify drink

= 20/33 = 61%

• Question: is this statistically significantly different from 50% (random guessing) at = 10%?

Large Sample Test for a Proportion:Taste Test Data

• HO: p = 0.5HA: p does not equal 0.5

• Test statistic:z = | (p - .5)/sqrt( p(1-p)/n) |

= | (.61-.5)/sqrt(.61*.39/33) | = 1.25

• Reject if z > z0.10/2 = 1.645

• It’s not, so there’s not enough evidence to reject HO.

Large Sample Test for a Proportion:Taste Test Data

P-valuePr( |(P-p)/sqrt(P Q/n)| >

|(p-p)/sqrt(p q/n)| when H0 is true)=Pr( |(P-0.5)/sqrt(P Q/n) | > |1.25 | when H0 is true)=2*Pr( Z > 1.25) where Z~N(0,1)= 21%

i.e. “How likely is a test statistic of 1.25 when true p = 50%?”

Minitab

• Minitab computes the test statistic as:

z = | (p - .5)/sqrt( .5(1-.5)/n) |= | (.61-.5)/sqrt(.25/33) | = 1.22

Since .25 >= p(1-p) for any p, this is more conservative (larger denominator = smaller test statistic). Either way is fine.

Difference between two means

• PCB Data– Sample 1: Treatment = PCB 156– Sample 2: Treatment = PCB 156 + estradiol

• Response = estrogen produced by cells

• Question: Can we conclude that average estrogen produced in sample 1 is different from average by sample 2 (at = 0.05)?

• H0: 1 – 2 = 0HA: 1 – 2 does not = 0

• Test statistic:|(Estimate – value under H0)/Std Dev(Estimate)|

z = (x1 – x2)/sqrt(s12/n1 + s2

2/n2)

Reject if |z| > z/2

• P-value = 2*Pr[ Z > (x1 – x2)/sqrt(s1

2/n1 + s22/n2)] where

Z~N(0,1).

n x s

PCB156 96 1.93 1.00

PCB156+E 64 2.16 1.01

|z| = |-0.23/sqrt(1.002/96 + 1.012/64)| = |-1.42| = 1.42

z/2 = z0.05/2 = z0.025 = 1.96

So don’t reject.

P-value = 2*Pr(Z > 1.42) = 16%

Pr( Test statistic > 1.42 when HO is true)

• Test statistic:

|z| = |(Estimate – value under H0)/Std Dev(Estimate)|

Reject if |z| > z/2

P-value = 2*Pr( Z > z ) where Z~N(0,1).

In General, Large Sample 2 sided Tests:

Large Sample Hypothesis Tests: summary for means

Single meanHypotheses Test (level 0.05)

HO: = k Reject HO if |(x-k)/s/sqrt(n)|>1.96

HA: does not = k p-value: 2*Pr(Z>|(x-k)/s/sqrt(n)|)where Z~N(0,1)

Difference between two meansHypotheses Test (level 0.05)

HO: = D Let d = x1 – x2

HA: does not = D Let SE = sqrt(s12/n2 + s2

2/n2)Reject HO if |(d-D)/SE|>1.96

p-value: 2*Pr(Z>|(d-D)/SE|)where Z~N(0,1)

Large Sample Hypothesis Tests: summary for proportions

Single proportionHypotheses Test (level 0.05)

HO: true p = k Reject HO if |(p-k)/sqrt(p(1-p)/n)|>1.96

HA: p does not = k p-value: 2*Pr(Z>|(p-k)/sqrt(p(1-p)/n)|)where Z~N(0,1)

Difference between two proportionsHypotheses Test (level 0.05)

HO: p1-p2 = d Let d = p1 – p2

HA: p1-p2 does not = d Let p = total “success”/(n1+n2)Let SE = sqrt(p(1-p)/n1 + p(1-p)/n2)

Reject HO if |(p-d)/SE|>1.96p-value: 2*Pr(Z>|(d)/SE|)

where Z~N(0,1)

• A two sided level hypothesis test, H0: =k vs HA: does not equal k

is rejected if and only if k is not in a 1- confidence interval for the mean.

• A one sided level hypothesis test, H0: <=k vs HA: >k

is rejected if and only if a level 1-2 confidence interval is completely to the left of k.

Hypothesis tests versus confidence intervals

The following is discussed in the context of tests / CI’s for a single mean, but it’s true for all the confidence intervals / tests we have done.

• The previous slide said that confidence intervals can be used to do hypothesis tests.

• CI’s are “better” since they contain more information.

• Fact: Hypothesis tests and p-values are very commonly used by scientists who use statistics.

• Advice: 1. Use confidence intervals to do hypothesis testing2. know how to compute / and interpret p-values

Hypothesis tests versus confidence intervals

Type 1 and Type 2 Errors

Truth

H0 True

HA True

Action

Fail to Reject H0 Reject H0

correct

correct

Type 1error

Type 2error

Significance level = =Pr(Making type 1 error)

Power = 1–Pr(Making type 2 error)

In terms of our folate example, suppose we repeated the experiment

and sampled 333 new people

Pr( Type 1 error )= Pr( reject H0 when mean is 300 )= Pr( |Z| > z0.025 )= Pr( Z > 1.96 ) + Pr( Z < -1.96 ) = 0.05 =

When mean is 300, then Z, the test statistic, has a standard normal distribution.

Note that the test is designed to have type 1 error =

Power= Pr( reject H0 when mean is not 300 )= Pr( reject H0 when mean is 310)= Pr( |(X-300)/193.4/sqrt(333)| > 1.96)= Pr( (X-300)/10.6 > 1.96 )+Pr( (X-300)/10.6 < -1.96 )= Pr(X > 320.8) + Pr(X < 279.2)

= Pr( (X – 310)/10.6 > (320.8-310)/10.6 )+ Pr( (X – 310)/10.6 < (279.2-310)/10.6 )

= Pr( Z > 1.02 ) + Pr( Z < -2.90 ) where Z~N(0,1)= 0.15 + 0.00 = 0.15

In other words, if there true mean is 310, there’s an 85% chancethat we will not detect it. If 310 is scientifically significantly different from 300, then this means that our experiment was wasted in some sense.

As n increases, power goes up.As standard deviation of x decreses, power goes up.As increases, power goes up.

Picture for Power

True Mean

Po

we

r

260 280 300 320 340

0.2

0.4

0.6

0.8

1.0

Power forn=333 and = 0.05

“Pr(Reject HO when it’s false)”

As n increases and/or increases and/or stddev decreases, these

curves becomesteeper

Power calculations are a very important part of planning any

experiment:• Given:

– a certain level of – preliminary estimate of std dev (of x’s that go

into x)– difference that is of interest

• Compute required n in order for power to be at least 85% (or some other percentage...)

Power calculations are an integral part of planning any experiment:

• Bad News: Algebraically messy (but you should know how to do them)

• Good News: Minitab can be used to do them:• Stat: Power and Sample Size…

– Inputs:1. required power

2. difference of interest

– Output:Result = required sample size

– Options: Change , one sided versus 2 sided tests