Review:Large Sample Confidence Intervals

• 1- confidence interval for a mean:x +/- z/2 s/sqrt(n)

• 1- confidence interval for a proportion:

p +/- z/2 p(1-p)/sqrt(n)

• 1- confidence interval for the difference between two means:

x1 – x2 +/- z/2 sqrt(s21/n1+s2

2/n2)

n >30 or so for means, np and n(1-p) both > 5 for proportions

In General:

Estimate (that is normally distributed via the Central

Limit Theorem)+/- standard deviation

Z/2 of estimate ( )This gives an interval: (Lower Bound , Upper Bound)

Interpretation: This is a plausible range for the true value of the number that we’re estimating. is a tuning parameter

for level of plausibility: smaller = more conservative estimate.

Large Sample Confidence Intervals

• 1- confidence interval for difference between two proportions:

p1-p2 +/- z/2 sqrt[(p1(1-p1)/n1)+(p2(1-p2)/n2)]

np and n(1-p) > 5 for all p’s…

Designing an Experiment and Choosing a Sample Size

• Example: Compare the shrinkage in a tumor due to a “new” cancer treatment relative to standard treatment

• 100 patients randomly assigned to “new” treatment or standard treatment

xinew = reduction in tumor size for person i under new treatment

xjstd = reduction in tumor size for person j under std treatment

xnew and s2new

xstd and s2std

Mean and sample variance of the changes in size for the new and standard treatments

Suppose the data are:

xnew = 25.3

snew = 2.0

xstd = 24.8

sstd = 2.3

95% Confidence Interval for difference:

x1 – x2 +/- z/2 sqrt(s21/n1+s2

2/n2)

= 0.5 +/- 0.84

What can we conclude?

• There’s no difference?

• Can’t see a difference?

• There’s a difference, but it’s too small to care about?

There is a difference between:

• Can’t see a difference

• There’s no difference

Situation forCancer example

(In cancer experiment, we can assume we care about small differences.)

• Can’t see a difference (that is big enough to care about) = wasted experiment

• AVOID / PREVENT THE WASTE AND ASSOCIATED TEARS

USE SAMPLE SIZE PLANNING

Sample Size Planning

• Length of a 1- level confidence interval is:

“2 z/2 std deviation of estimate”

2z/2s/sqrt(n)

2z/2p(1-p)/sqrt(n)

2z/2sqrt((s21/n1)+(s2

2/n2))

2z/2sqrt[(p1(1-p1)/n1)+(p2(1-p2)/n2)]

1. Suppose we want a 95% confidence interval no wider than W units.

2. is fixed. Assume a value for the standard deviation (or variance) of the estimator.

3. Solve for an n (or n1 and n2) so that the width is less than W units.

4. When there are two sample sizes (n1 and n2), we often assume that n1 = n2.

Cancer example

• Let W = 0.1. Want 95% CI for difference between means with width less than W.

• Suppose s2new = s2

std = 6 (conservative

guess)

W > 2z/2sqrt((s2new/n1)+(s2

std/n2))

0.1 > 2(1.96)sqrt(6/n + 6/n)

0.1 > 3.92sqrt(12/n)

0.01 > (3.922)12/n

n > 18439.68 (each group…)

Book’s B = our W/2

Hypothesis testing and p-values (Chapter 9)

We used confidence intervals in two ways:1. To determine an interval of plausible values for

the quantity that we estimate. Level of plausibility is determined by 1-. 90% (=0.1) is less conservative than 95% (=0.05) is less conservative than 99% (=0.01)...

2. To see if a certain value is plausible in light of the data:

If that value was not in the interval, it is not plausible (at certain level of confidence). Zero is a common certain value to test, but not the only one.

Hypothesis tests address the second use directly

Example: Dietary Folate• Data from the Framingham Heart Study

0 200 400 600 800 1000 1200

0

20

40

60

80

100

Dietary Folate (micrograms / day, calorie adjusted to 2000 calorie diet)

Count

n = 333 Elderly Men

Mean = x = 336.4

Std Dev = s = 193.4

Can we conclude that the mean is greater than 300 at 5% significance? (same as 95% confidence)

Five Components of the Hypothesis test:

1. Null Hypothesis = “What we want to disprove”= “H0” = “H not”= Mean dietary folate in the population

represented by these data is <= 300.

= <= 3002. Alternative Hypothesis

= “What we want to prove”= “HA”= Mean dietary folate in the population

represented by these data is > 300.

= > 300

3. Test Statistic

To test about a mean with a large sample test, the statistic is z = (x – )/(s/sqrt(n))

(i.e. How many standard deviations (of X) away from the hypothesized mean is the observed x?)

4. Significance Level of Test, Rejection Region, and P-value

5. Conclusion

Reject H0 and conclude HA if test stat is in rejection region. Otherwise, “fail to reject” (not same as concluding H0 – can only cite a “lack of evidence”

(think “innocent until proven guilty”)

(Equivalently, reject H0 if p-value is less than .)

Next page

• Significance Level: =1% or 5% or 10%... (smaller is more conservative) (Significance = 1-Confidence)

• Rejection Region:– Reject if test statistic in rejection region.– Rejection region is set by:

• Assume H0 is true “at the boundary”.• Rejection region is set so that the probability of seeing the observed test

statistic or something further from the null hypothesis is less than or equal to

• P-value– Assume H0 is true “at the boundary”.– P-value is the probability of seeing the observed test statistic or

something further from the null hypothesis.– = “observed level of significance”

Note that you reject if the p-value is less than .(Small p-values mean “more observed significance”)

Example:• H0: <=300, HA: >300• z (x-)/(s/sqrt(n))

= (336.4 – 300)/(193.4/sqrt(333))= 3.43

• Significance level = 0.05• When H0 is true, Z~N(0,1). As a result, the cutoff

is z0.05=1.645. (Pr(Z>1.645) = 0.05.)• P-value = Pr(Z>3.43 when true mean is 300) =

0.0003 • Reject. Mean is greater than 300.• Would you reject at significance level 0.0001?

Picture

1.645Test Statisistic

De

nsi

ty

-4 -2 0 2 4

0.0

0.1

0.2

0.3

0.4

Rejection region

Distribution ofZ = (X – 300)/(193.4/sqrt(333))when true mean is 300.

Area to right of 1.645=0.05 = sig level

3.43Area to right of 3.43=0.0003 = p-value

Test statistic

ObservedTest Statistic

One Sided versus Two Sided Tests

• Previous test was “one sided” since we’d only reject if the test statistic is far enough to “one side” (ie. If z > z0.05)

• Two sided tests are more common (my opinion):

H0: =0, HA: does not equal 0

Two Sided Tests (cntd)Test Statistic (large sample test of mean)

z = (x – )/(s/sqrt(n))

Rejection Region:

reject H0 at signficance level if |z|>z/2

i.e. if z>z/2 or z<-z/2

Note that this “doubles” p-values. See next example.

Example:• H0: =300, HA: doesn’t equal 300• z=(x-)/(s/sqrt(n))

= (336.4 – 300)/(193.4/sqrt(333))= 3.43

• Significance level = 0.05• When H0 is true, Z~N(0,1). As a result, the cutoff

is z0.025=1.96. (Pr(|Z|>1.96)=2*Pr(Z>1.96)=0.05• P-value = Pr(|Z|>3.43 when true mean is 300) =

Pr(Z>3.43) + Pr(Z<-3.43) = 2(0.0003)=0.0006• Reject. Mean is not equal to 300.• Would you reject at significance level 0.0005?

Picture

1.96Test Statisistic

De

nsi

ty

-4 -2 0 2 4

0.0

0.1

0.2

0.3

0.4

Rejection region

Distribution ofZ = (X – 300)/(193.4/sqrt(333))when true mean is 300.

3.43Area to right of 3.43=0.0003

Test statistic

-3.43

Area to left of -3.43=0.0003

1.96

Rejection region

Sig level = area to right of 1.96 + area to the left of -1.96 = 0.05=

Pvalue=0.0006=Pr(|Z|>3.43)

Power and 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 )

• Assuming H0 is true, what’s the probability of making a type I error?

• H0 is true means true mean is 0.• This means that the test statistic has a

N(0,1) distribution. • Type I error means reject which means |

test statistic| is greater than z/2.• This has probability .