One Sample Inf-1

),(~ 10N

n

y 0

t

n

sy

0

If sample came from a normal distribution, t has a

t-distribution with

n-1 degrees of freedom.

1) Symmetric about 0.

2) Looks like a standard normal density, only more spread out.

3) The spread of the distribution is indexed to a parameter called the degrees of freedom (df).

4) As the degrees of freedom increase, the t-distribution gets closer to the standard normal distribution. (Safe to use z instead of t when n>30.)

We know that:

What is the distribution of:

One Sample Means Test: What if is unknown? (sampling from normal population)

One Sample Inf-2

See Table 3 Ott & Longnecker

Df .1 .05 .025 .01 .005 .001

1 3.078 6.314 12.706 31.821 63.657 318.309

5 1.476 2.015 2.571 3.365 4.032 5.893

10 1.372 1.812 2.228 2.764 3.169 4.144

15 1.341 1.753 2.131 2.602 2.947 3.733

20 1.325 1.725 2.086 2.528 2.845 3.552

25 1.316 1.708 2.060 2.485 2.787 3.450

30 1.310 1.697 2.042 2.457 2.750 3.385

40 1.303 1.684 2.021 2.423 2.704 3.307

NORMAL(0,1)

1.282 1.645 1.960 2.326 2.576 3.090

Tail probabilities of the t-distribution

One Sample Inf-3

H0: = 0For Pr(Type I error) = , df = n - 1

HA: 1.

2.

3.

Reject H0 if t > t,n-1

t < -t,n-1

| t | > t/2,n-1

ns

yt 0:T.S.

Rejection Regions for hypothesis tests using t-distribution critical values

One Sample Inf-4

Degrees of Freedom

Why are the degrees of freedom only n - 1 and not n?

We start with n independent pieces of information with which we estimate the sample mean.

n1

ini 1

y y

Now consider the sample variance:

n21

in 1i 1

s (y y)

Because the sum of the deviations iy y are equal to zero, if weknow n-1 of these deviations, we can figure out the nth deviation. Hence there are only n-1 independent deviations that are available to estimate the variance (and standard deviation). That is, there are only n-1 pieces of information available to estimate the standard deviation after we “spend” one to estimate the sample mean. The t-distribution is a normal distribution adjusted for unknown standard deviation hence it is logical that it would have to accommodate the fact that only n-1 pieces of information are available.

One Sample Inf-5

with df = n - 1 and confidence coefficient (1 - ). (Can use z/2 if n>30.)n

sty n 1,2/

Example: Compute 95% CI for given

65931400252

30441409601z059

6530621403062t9n

65sn

sty81ndf140y

025

8025

8025

...

....

....

.

.

.

,.

,.

Confidence Interval for when unknown (samples are assumed to come from a normal population)

One Sample Inf-6

The Level of Significance of a Statistical Test (p-value)

• Suppose the result of a statistical test you carry out is to reject the Null.• Someone reading your conclusions might ask: “How close were you to not rejecting?”• Solution: Report a value that summarizes the weight of evidence in favor of Ho, on a scale of 0 to 1. This the p-value. The larger the p-value, the more evidence in favor of Ho.

Formal Definition: The p-value of a test is the probability of observing a

value of the test statistic that is as extreme or more extreme (toward Ha)

than the actually observed value of the test statistic, under the assumption

that Ho is true. (This is just the probability of a Type I error for the observed

test statistic.)

Rejection Rule: Having decided upon a Type I error probability , reject Ho if p-value .

One Sample Inf-7

Equivalence between confidence intervals and hypothesis tests

Rejecting the two-sided null Ho: = 0

is equivalent to 0 falling outside a (1-)100% C.I. for .

Rejecting the one-sided null Ho: 0

is equivalent to 0 being greater than the upper endpoint of a (1-2)100% C.I. for , or 0 falling outside a one-sided (1-)100% C.I. for with –infinity as

lower bound.

Rejecting the one-sided null Ho: 0

is equivalent to 0 being smaller than the lower endpoint of a (1-2)100% C.I. for , or 0 falling outside a one-sided (1-)100% C.I. for with +infinity as

upper bound.

One Sample Inf-8

Example: Practical Significance vs. Statistical Significance

Dr. Quick and Dr. Quack are both in the business of selling diets, and they have claims that appear contradictory. Dr. Quack studied 500 dieters and claims,

A statistical analysis of my dieters shows a significant weight loss for my Quack diet.

The Quick diet, by contrast, shows no significant weight loss by its dieters. Dr. Quick followed the progress of 20 dieters and claims,

A study shows that on average my dieters lose 3 times as much weight on the Quick diet as on the Quack diet.

So which claim is right? To decide which diets achieve a significant weight loss we should test:

Ho: 0 vs. Ha: < 0 where is the mean weight change (after minus before) achieved by dieters on each of the two diets. (Note: since we don’t know we should do a t-test.)

One Sample Inf-9

MTB output for Quack diet analysis (Stat Basic Stats 1 - Sample t)One-Sample T: QuackTest of mu = 0 vs mu < 0

Variable N Mean StDev SE MeanQuack 500 -0.913 9.744 0.436

Variable 95.0% Upper Bound T PQuack -0.194 -2.09 0.018

Difference Size Power 1 500 0.6295

R output for Quick diet analysis (Read 20 values into vector “quack”)> t.test(quick,alternative=c("less"),mu=0,conf.level=0.95) One Sample t-test, data: quick t = -1.0915, df = 19, p-value = 0.1443 alternative hypothesis: true mean is less than 0 95 percent confidence interval: -Inf 1.594617 sample estimate of mean of x: -2.73> power.t.test(n=20,delta=1,sd=11.185,type="one.sample",

alternative="one.sided") n = 20, delta = 1, power = 0.104

One Sample Inf-10

Summary

1. Quick’s average weight loss of 2.73 is over 3 times as much as the 0.91 weight loss reported by Quack.

2. However, Quack’s small weight loss was significant, whereas Quick’s larger weight loss was not! So Quack might not have a better diet, but he has more evidence, 500 cases compared to 20.

Remarks

1. Significance is about evidence, and having a large sample size can make up for having a small effect.

2. If you have a large enough sample size, even a small difference can be significant. If your sample size is small, even a large difference may not be significant.

3. Quick needs to collect more cases, and then he can easily dominate the Quack diet (though it seems like even a 2.7 pound loss may not be enough of a practical difference to a dieter).

4. Both the Quick & Quack statements are somewhat empty. It’s not enough to report an estimate without a measure of its variability. Its not enough to report a significance without an estimate of the difference. A confidence interval solves these problems.

One Sample Inf-11

A confidence interval shows both statistical and practical significance.

Quack two & one-sided 95% CIs

Quick two & one-sided 95% CIs

)194.0,(),(

)058.0,768.1()436(.96.1913.

05.

025.

nszy

n

szy

)595.1,(),(

)505.2,965.7(20

185.11093.273.2

19,05.

19,025.

nsty

n

sty

One-sided CI says mean is sig. less than zero.

One-sided CI says mean is NOT sig. less than zero.