Stat 528 (Autumn 2008)

Inference for the mean of a population

(One sample t procedures)

Reading: Section 7.1.

• Inference for the mean of a population.

• The t distribution for a normal population.

• Small sample CI for µ in a normal population.

• Robustness of the t procedures.

• Testing hypotheses about a single mean

(the one sample t-test).

• Methods for matched pairs

– The paired t-test

– The sign test for matched pairs

• The power of the one sample t-test.

1

Inference for the mean of a population

• So far we have based inference for the population mean

on the Z statistic

Z =X − µ

σ/√

n.

For large n, Z is approximately N(0,1).

• Problem: in practice we do not know the population stan-

dard deviation, σ.

– Instead we use the sample standard deviation, s, as

an estimate for σ.

2

The distribution of t for a normal population

• Let X1, X2, . . . Xn be a SRS from a normal population

with population mean µ. Then the standardized variable

t =X − µ

s/√

n,

has a t distribution with n − 1 degrees of freedom (df).

• The impact of estimating σ is to add uncertainty about our

standardization.

• Smaller n leads to fewer degrees of freedom and less certainty.

• We say that t has a tn−1 distribution

• The quantity, s/√

n is the (estimated) standard error for

the sample mean.

– It is denoted SE mean in MINITAB.

3

Properties of the t distribution

−4 −2 0 2 4

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0.2

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standard normalt with 5 dft with 2 dft with 1 df

• The density curve is symmetric with mean zero and is bell-

shaped like the normal distribution.

• The t distribution has heavier tails than the normal dis-

tribution (more spread out about zero).

• As the degrees of freedom increase the tails become thinner,

and more of the density is concentrated in the center of the

distribution. t∞ = standard normal distribution.

4

A small sample CI for µ

(The normal population case)

• For one random sample of normal data, a C = 100(1−α)%

level confidence interval for µ is given by

x̄ ± tn−1,α/2s√n,

where tn−1,α/2 is the critical value of the t distribution with

n − 1 degrees of freedom.

• The tn−1,α/2 value is tabulated in Table D.

1. Look at the bottom of the table for the confidence

level C of the two sided interval, OR

2. Look up α/2 as the upper tail probability p.

• Recall that the CI for µ comes from a family of hypothesis

tests about µ.

5

Robustness of the t-procedures

• What if the population is not normal – can we still use the

t distribution?

• Practical guidelines from the textbook:

1. n < 15: Use t procedures if data are close to normal.

If data are clearly non-normal or if outliers are present,

do not use the t procedure.

2. n ≥ 15: Use t procedures except in presence of strong

skewness or outliers.

3. Roughly n ≥ 40: The t procedures are valid even for

clearly skewed distributions.

• Use plots of the data to help you decide!

6

Polymerization example

The article “Measuring and understanding the aging of craft in-

sulating paper in power transformers” contained the following

observations on the degree of polymerization for paper specimens

for which viscosity times concentration fell in a certain middle

range.

418 421 421 422 425 427 431 434 437

439 446 447 448 453 454 463 465

Plots of the data show that a normality assumption for the data

is reasonable. (Note that x̄ = 438.29, s = 15.14, n = 17).

Form a 95% confidence interval for the true average degree

of polymerization (as did the authors of the article).

Does the interval suggest that 440 is a plausible value for the

true average degree of polymerization? What about 450?

7

Testing hypotheses about a single mean

The one sample t test

• Data: We assume x1, x2, . . . xn is a random sample from a

normal population with mean µ.

• We state our hypotheses:

H0: µ = µ0, for some constant value µ0

Ha: µ < µ0, µ 6= µ0, OR µ > µ0

(remember to define what µ is (in words) for your problem).

• We calculate the test statistic,

t =x̄ − µ0

s/√

n.

• Under H0, the test statistic follows a tn−1 distribution.

• Decision: Compare the observed t-statistic to the critical

value found in Table D.

8

Drawing conclusions in the one-sample t-test

• For a test of significance at the level α

– If the observed t-statistic is in the tail, we reject H0 (in

favor of HA).

– If the observed t-statistic is not in the tail, we do not

reject H0.

• Alternatives and tails

– For a two-tailed alternative, reject if |t| ≥ tα/2.

– For an upper-tailed alternative, reject if t ≥ tα.

– For a lower-tailed alternative, reject if t ≤ tα.

• As always, write your conclusion(s) in words.

• It is important to think about the assumptions that you made

to carry out the t-test.

– Remember that some assumptions can be validated using

plots of the data.

9

Example

The one-sample t statistic from a sample of n = 50 observations

for the two-sided test of

H0 : µ = 50 versus Ha : µ 6= 50,

has the value t = 1.65.

• What are the degrees of freedom for the test statistic, t?

• Is the value t = 1.65 statistically significant at the 10% level?

At the 5% level?

• Locate the two critical values, t∗ from Table D that bracket

t. What are the right-tail probabilities for these two values?

How would you report the P-value for this test?

10

Matched pairs (revision and analysis)

• Suppose we have two treatments.

• In the matched pairs design we try to gain precision in the

response by matching pairs of similar individuals.

– we assign each treatment randomly to each subject (each

subject only receives one treatment).

• Or an individual serves as his/her own partner.

– the individual receives both treatments.

• Each pair of subjects (individual) form their own block.

• To analyze the results of this type of experiment, we com-

pare the responses across the pairs (individuals).

– We usually take differences, and carry out the

statistical inference using the paired t-test.

11

Football example

Two identical footballs, one air-filled and one helium-filled, were

used outdoors on a windless day at The Ohio State University’s

athletic complex. The kicker was a novice punter and was not

informed which football contained the helium. Each football was

kicked 39 times. The kicker changed footballs after each kick so

that his leg would play no favorites if he tired or improved with

practice.

(Source: Lafferty, M. B. (1993), ”OSU scientists get a kick out of sports

controversy, ”The Columbus Dispatch (21 Nov 1993), B7.)

12

The data

(all distances are in yards)

Trial Air Helium

1 25 25

2 23 16

3 18 25

4 16 14

5 35 23

6 15 29

7 26 25

8 24 26

9 24 22

10 28 26

11 25 12

12 19 28

13 27 28

Trial Air Helium

14 25 31

15 34 22

16 26 29

17 20 23

18 22 26

19 33 35

20 29 24

21 31 31

22 27 34

23 22 39

24 29 32

25 28 14

26 29 28

Trial Air Helium

27 22 30

28 31 27

29 25 33

30 20 11

31 27 26

32 26 32

33 28 30

34 32 29

35 28 30

36 25 29

37 31 29

38 28 30

39 28 26

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A scatterplot

14

The paired t procedure – the setup

• Suppose we have pairs of data values

(x1, y1), (x2, y2), . . . (xn, yn).

e.g., In our example the pairs of values are the

(helium-filled, air-filled) distances for each kick.

• Clearly the x and y values are not independent.

• Instead, we calculate the differences

di = yi − xi,

for each i = 1, . . . , n.

• We assume d1, d2, . . . dn is a random sample from a normal

population with mean µd and stdev σd.

– µd is the population mean of the differences between the

x and y values.

– σd is the population stdev of the differences.

15

The paired t procedure

• We want to test:

H0: µd = µ0, for some constant value µ0

Ha: µd < µ0, µd 6= µ0, OR µd > µ0

• We compute the test statistic,

t =d̄ − µ0

sd/√

n,

where d̄ is the sample average of the differences, and

sd is the sample stdev of the differences.

• Under H0, the test statistic follows a tn−1 distribution.

• We make our decision in the same way that we did for the

one-sample t-test.

– if the observed t-statistic is in tail, we reject H0,

– if the observed t-statistic is not in the tail, we do not

reject H0.

16

Identifying the hypotheses

• There is a belief that on average a helium-filled ball travels

further than the air-filled ball. State the appropriate H0 and

Ha. Be sure to identify the parameters appearing in the

hypotheses.

17

Summary figures

18

Performing the test

• Carry out a test. Can you reject H0 at the 5% significance

level? At the 1% significance level? Write down you conclu-

sion in words.

Variable N N* Mean SE Mean StDev

Air-Helium 39 0 -0.462 1.10 6.87

Variable Minimum Q1 Median Q3 Maximum

Air-Helium -17.00 -4.00 -1.00 2.00 14.00

• Provide a 90% confidence interval for the mean difference in

the distances (air-filled minus helium-filled).

19

Inference for non-normal populations

• If the data do not seem to be drawn from a normal popula-

tion, then the t procedures may not be valid.

• Three possible strategies:

1. Learn about other probability distributions. For exam-

ple, there plenty of skewed distributions (e.g, exponential,

gamma, Weibull). Use methods for these distributions in-

stead of the methods for the normal distribution.

2. Transform your data to make it look “as normal as pos-

sible” (recall the ladder of power transformations). Can

be hard to interpret the results when using a transforma-

tion.

3. Use distribution-free tests. These tests do not assume

a particular distribution for the population. Often these

test are based on other parameters of the distribution

such as the median (rather than the mean). These tests

can be less powerful in practice.

20

The sign test for matched pairs

• Example of a distribution-free test.

• As before, consider pairs of data values:

(x1, y1), (x2, y2), . . . (xn, yn).

• We will test

H0 : population median of differences = 0, versus

Ha : population median of differences 6= 0.

• Let di = yi − xi (i = 1, . . . , n) be the differences.

• Exclude the differences that are zero. Let X denote the count

out of the remaining m differences that are positive.

• Then under H0, X is Binomial(m,0.5).

(If the median is zero, then half the nonzero differences are

above zero, and the other half are below zero).

• If x is the observed X value, then the P-value is

2 × P (X ≤ x) or 2 × P (X ≥ x).

21

The sign test for matched pairs (cont.)

• For the football example:

– Out of n = 39 differences, m = 37 differences are nonzero.

– Thus under H0, X is Binomial(37, 0.5).

– Out of the 37, we observe 17 that are above zero.

– P-value = 2 × P (X ≤ 17) = 2 × 0.3714 = 0.7428.

– No evidence to reject H0.

• See the textbook for the one-sided test.

• Note: If the population of differences is normally (or ap-

proximately normally) distributed then this test will be less

powerful at detecting differences than the paired t-test.

22

The power of the one sample t-test

• The power calculation for the one sample t-test is simi-

lar to the power calculation for the z-test.

• But, the math is much harder!

– Instead we use MINITAB.

• Stat → Power and Sample Size → 1-Sample t.

• Under Options select the Alternative Hypothesis and

Significance Level

• Then enter any two of the following three items:

1. Sample sizes:

2. Differences:

3. Power values:

• Enter the Standard deviation (the sample stdev in this

case) and click OK.

23

A value for σ

• There are four main ways to obtain a value for σ.

– Literature search. Use historical data from similar

studies.

– Pilot study. Use the results of a pilot study. The

estimate of σ will often need to be adjusted.

– Elicit σ. Two useful methods are the Range/4 method

and the Range/6 method.

– Construct a value for σ. Some probability mod-

els yield a value for σ. (e.g. For a Bernoulli RV, σ =√

p(1 − p)).

• Be conservative. Use several methods and consider a slightly

larger value of σ than these methods suggest.

24

An agricultural field trial example

An agricultural field trial compares the yield of two varieties of

tomatoes for commercial use. The researchers divide in half each

of 10 small plots of land and plant each tomato variety on one half

of each plot. After harvest, they compare the yields in pounds per

plant at each location. The ten differences (Variety A - Variety

B) give the following statistics: x̄ = 0.46 and s = 0.92. Is there

convincing evidence that Variety A has the higher mean yield?

Let µd denote the population mean of the difference in the yields.

We test: H0: µd = 0 versus Ha: µd > 0.

The MINITAB output for the paired t test is:

One-Sample T, Test of mu = 0 vs > 0

95% Lower

N Mean StDev SE Mean Bound T P

10 0.460000 0.920000 0.290930 -0.073307 1.58 0.074

25

Agricultural trial (cont.)

The tomato experts who carried out the field trial suspect that

the relative lack of significance is due to low power. They would

like to detect a mean difference in yields of 0.6 pounds per plant

at the 0.05 significance level. Based on the previous study, use

0.92 as an estimate of the population σ.

• What is the power of the test with n = 12 against the alter-

native of µ = 0.6?

• If the sample size is increased to n = 30 plots of land, what

will be the power against the same alternative?

26