“Students” t-test

Recall: The z-test for means

ns

x

n

xxz

x

000

The Test Statistic

Comments

• The sampling distribution of this statistic is the standard Normal distribution

• The replacement of by s leaves this distribution unchanged only if the sample size n is large.

For small sample sizes:

ns

xt 0

The sampling distribution of

is called “students” t distribution with n –1 degrees of freedom

Properties of Student’s t distribution

• Similar to Standard normal distribution– Symmetric– unimodal– Centred at zero

• Larger spread about zero.– The reason for this is the increased variability introduced

by replacing by s.

• As the sample size increases (degrees of freedom increases) the t distribution approaches the standard normal distribution

-4 -2 2 4

0.1

0.2

0.3

0.4

t distribution

standard normal distribution

The Situation

• Let x1, x2, x3 , … , xn denote a sample from a normal population with mean and standard deviation . Both and are unknown.

• Let

• we want to test if the mean, , is equal to some given value 0.

mean sample the1

n

xx

n

ii

deviation standard sample the

11

2

n

xxs

n

ii

The Test Statistic

ns

xt 0

The sampling distribution of the test statistic is the t distribution with n-1 degrees of freedom

The Alternative Hypothesis HA

The Critical Region

0: AH

0: AH

0: AH

2/2/ or tttt

tt

tt

t and t/2 are critical values under the t distribution with n – 1 degrees of freedom

Critical values for the t-distribution

or /2

0 t

tt or 2/

Critical values for the t-distribution are provided in tables. A link to these tables are given with today’s lecture

Look up df

Look up

Note: the values tabled for df = ∞ are the same values for the standard normal distribution, z

…

Example

• Let x1, x2, x3 , x4, x5, x6 denote weight loss from a new diet for n = 6 cases.

• Assume that x1, x2, x3 , x4, x5, x6 is a sample from a normal population with mean and standard deviation . Both and are unknown.

• we want to test:

0: AH

0:0 H

versus

New diet is not effective

New diet is effective

The Test Statistic

ns

xt 0

The Critical region:

tt Reject if

The Data

The summary statistics:

462418.1 and 96667.0 sx

1 2 3 4 5 6

2.0 1.0 1.4 -1.8 0.9 2.3

The Test Statistic

619.1

6462418.1

096667.00

ns

xt

The Critical Region (using = 0.05)

d.f. 5for 0152050 .tt . Reject if

Conclusion: Accept H0:

Confidence Intervals

Confidence Intervals for the mean of a Normal Population, , using the Standard Normal distribution

nzx

2/

Confidence Intervals for the mean of a Normal Population, , using the t distribution

n

stx 2/

The Data

The summary statistics:

462418.1 and 96667.0 sx

1 2 3 4 5 6

2.0 1.0 1.4 -1.8 0.9 2.3

Example

• Let x1, x2, x3 , x4, x5, x6 denote weight loss from a new diet for n = 6 cases.

The Data:

The summary statistics:

462418.1 and 96667.0 sx

1 2 3 4 5 6

2.0 1.0 1.4 -1.8 0.9 2.3

Confidence Intervals (use = 0.05)

n

stx 025.0

6

462418.1571.296667.0

535.196667.0

50.2 to57.0

Summary

Statistical Inference

Estimation by Confidence Intervals

Confidence Interval for a Proportion

pzp ˆ2/ˆ

n

pp

n

ppp

ˆ1ˆ1ˆ

point critical 2/upper 2/ z

ndistribtio normal standard theof

ˆ/ 2 / 2 / 2

ˆ ˆ1 1p

p p p pB z z z

n n

Error Bound

The sample size that will estimate p with an Error Bound B and level of confidence P = 1 – is:

where:• B is the desired Error Bound• z is the /2 critical value for the standard normal

distribution• p* is some preliminary estimate of p.

2

22/ *1*

B

ppzn a

Determination of Sample Size

Confidence Intervals for the mean of a Normal Population,

/ 2 xx z

/ 2or x zn

/ 2or s

x zn

sample meanx point critical 2/upper 2/ z

ndistribtio normal standard theof sample standard deviation s

The sample size that will estimate with an Error Bound B and level of confidence P = 1 – is:

where:• B is the desired Error Bound• z is the /2 critical value for the standard normal

distribution• s* is some preliminary estimate of s.

2

222/

2

222/ *

B

sz

B

zn aa

Determination of Sample Size

Confidence Intervals for the mean of a Normal Population, , using the t distribution

n

stx 2/

Hypothesis Testing

An important area of statistical inference

To define a statistical Test we

1. Choose a statistic (called the test statistic)

2. Divide the range of possible values for the test statistic into two parts

• The Acceptance Region

• The Critical Region

To perform a statistical Test we

1. Collect the data.

2. Compute the value of the test statistic.

3. Make the Decision:

• If the value of the test statistic is in the Acceptance Region we decide to accept H0 .

• If the value of the test statistic is in the Critical Region we decide to reject H0 .

Determining the Critical Region

1. The Critical Region should consist of values of the test statistic that indicate that HA is true. (hence H0 should be rejected).

2. The size of the Critical Region is determined so that the probability of making a type I error, , is at some pre-determined level. (usually 0.05 or 0.01). This value is called the significance level of the test.

Significance level = P[test makes type I error]

To find the Critical Region

1. Find the sampling distribution of the test statistic when is H0 true.

2. Locate the Critical Region in the tails (either left or right or both) of the sampling distribution of the test statistic when is H0 true.

Whether you locate the critical region in the left tail or right tail or both tails depends on which values indicate HA is true.

The tails chosen = values indicating HA.

3. the size of the Critical Region is chosen so that the area over the critical region and under the sampling distribution of the test statistic when is H0 true is the desired level of =P[type I error]

Sampling distribution of test statistic when H0

is true

Critical Region - Area =

The z-test for Proportions

Testing the probability of success in a binomial experiment

Situation• A success-failure experiment has been

repeated n times

• The probability of success p is unknown. We want to test either

0 0 01. : versus :AH p p H p p

0 0 0

or

2. : versus :AH p p H p p

0 0 0

or

3. : versus :AH p p H p p

The Test Statistic

n

pp

ppppz

p 00

0

ˆ

0

1

ˆ

ˆ

Critical Region (dependent on HA)

Alternative Hypothesis Critical Region

0 :AH p p

0 :AH p p

0 :AH p p

or z z z z

z z

z z

The z-test for the mean of a Normal population (large samples)

Situation• A sample of n is selected from a normal

population with mean (unknown) and standard deviation . We want to test either

0 0 01. : versus :AH H

0 0 0

or

2. : versus :AH H

0 0 0

or

3. : versus :AH H

The Test Statistic

0 0 0 x

x x xz

s

n n

if is large.n

Critical Region (dependent on HA)

Alternative Hypothesis Critical Region

0 :AH

0 :AH

0 :AH

or z z z z

z z

z z

The t-test for the mean of a Normal population (small samples)

Situation• A sample of n is selected from a normal

population with mean (unknown) and standard deviation (unknown). We want to test either

0 0 01. : versus :AH H

0 0 0

or

2. : versus :AH H

0 0 0

or

3. : versus :AH H

The Test Statistic

0 0 x

x xt

ssn

Critical Region (dependent on HA)

Alternative Hypothesis Critical Region

0 :AH

0 :AH

0 :AH

or t t t t

t t

t t

Testing and Estimation of Variances

Let x1, x2, x3, … xn, denote a sample from a Normal distribution with mean and standard deviation (variance 2)

The point estimator of the variance 2 is:

The point estimator of the standard deviation is:

2

2 1

1

n

ii

x xs

n

2

1

1

n

ii

x xs

n

The statistic

has a 2 distribution with n – 1 degrees of freedom

2

21

2 2

1

n

ii

x xn s

U

Sampling Theory

Critical Points of the 2 distribution

0

0.1

0.2

0 5 10 15 202

Confidence intervals for 2 and .

0

0.1

0.2

0 5 10 15 202

/ 2

/2

22 21 / 2 / 22

11

n sP

21 / 2

/2

Confidence intervals for 2 and .

22 21 / 2 / 22

11

n sP

2 2 22 2

/ 2 1 / 2

1 11

n nP s s

2 2

/ 2 1 / 2

1 11

n nP s s

It is true that

from which we can show

and

Hence (1 – )100% confidence limits for 2 are:

2 22 2

/ 2 1 / 2

1 1 to

n ns s

and (1 – )100% confidence limits for are:

2 2

/ 2 1 / 2

1 1 to

n ns s

Example• In this example the subject is asked to type his

computer password n = 6 times.

• Each time xi = time to type the password is recorded. The data are tabulated below:

i 1 2 3 4 5 6 Sx iSx i

2

x i 6.63 8.51 9.01 8.69 8.71 8.83 50.38 426.9062

50.388.3967

6

ii

xx

n

2

22 1

1

50.38426.9062

6 0.8811511 5

n

ini

ii

x

xn

sn

95% confidence limits for the mean

.025

sx t

n

0.881151or 8.3967 2.571

6

.025 2.571 for 5 . .t d f

8.3967 0.9249

7.472 to 9.322

95% confidence limits for

97

1 1 to

n ns s

2 2.975 .0250.8312, 12.83 for 5 . .d f

95% confidence limits for 2

2 2

97

1 1 to

n s n s

5 5(0.881151) to (0.881151)

12.83 0.8312

2 25(0.881151) 5(0.881151) to

12.83 0.8312

0.550 to 2.161

0.303 to 4.671

Testing Hypotheses for 2 and .

2

20

1n sU

Suppose we want to test:

The test statistic:

2 2 2 20 0 0: against :AH H

If H 0 is true the test statistic, U, has a 2 distribution with n – 1 degrees of freedom:

2 2

1 / 2 / 22 20 0

1 1 or

n s n s

Thus we reject H0 if

0

0.1

0.2

0 5 10 15 202

/ 2

/2

21 / 2

/2

Accept RejectReject

One-tailed Tests for 2 and .

2

20

1n sU

Suppose we want to test:

The test statistic:

2 2 2 20 0 0: against :AH H

2

20

1n s

We reject H0 if

0

0.1

0.2

0 5 10 15 202

Accept Reject

2

20

1n sU

Or suppose we want to test:

The test statistic:

2 2 2 20 0 0: against :AH H

2

120

1n s

We reject H0 if

0

0.1

0.2

0 5 10 15 2021

AcceptReject

Example

• The current method for measuring blood alcohol content has the following properties– Measurements are

1. Normally distributed

2. Mean = true blood alcohol content

3. standard deviation 1.2 units

• A new method is proposed that has the first two properties and it is believed that the measurements will have a smaller standard deviation.

• We want to collect data to test this hypothesis.

• The experiment will be to collect n = 10 observations on a case were the true blood alcohol content is 6.0

• The data are tabulated below:

6.0550i

i

xx

n

2

2 1

21 0.692359, 0.4793611

n

ini

ii

x

xn

s sn

i 1 2 3 4 5 6 7 8 9 10 Sx iSx i

2

x i 5.21 6.90 5.69 5.05 5.75 5.90 6.92 6.48 6.85 5.80 60.55 370.9445

2

20

1n sU

To test:

The test statistic:

2 2 2 20 : 1.2 against : 1.2AH H

1 0.95 3.325 for 9 . .U d f

We reject H0 if

2

9 0.4793612.996

1.2U

Thus we reject H0 if = 0.05.

Two sample Tests