Chapter 9 Tests of Hypothesis

Single Sample TestsThe Middle Game –

applications to the real world

Chapter 9B

Recall?

If Z1, Z2, ..., Zn are independent standard normal random variables, then

2 2 2 2 21 2 ... ( )nZ Z Z n chi-square distribution

with n degrees freedom

2( )

/

ZT t n

n

t-distribution with n degrees freedom

2

2

( ) /( , )

( ) /

n nF F n m

m m

F-distribution with n degrees freedom inthe numerator and m degreesof freedom in the denominator

9-3 Tests on the Mean of a Normal Distribution, Variance Unknown

One-Sample t-Test

2( )

/

ZT t n

n

9-3.1 Hypothesis Tests on the Mean

Figure 9-9 The reference distribution for H0: = 0 with critical region for (a) H1: 0 , (b) H1: > 0, and (c) H1: < 0.

Tests on the mean - variance unknown

Peter Perk is concerned about the amount of caffeine in his pop (soft drink). It appears that his midday soda is keeping him awake during the boring statistics lectures. The label on the can states that it contains only 20 mg of caffeine. Peter Perk doesn’t believe this.

A random sample and testing of 25 cans of soda resulted in the following caffeine levels:

Test at a 2 percent level of significance.

H0: = 20

H1: > 20

22mg, 5mgX s

The t-test

One can of soda contains:• About 10 teaspoons of sugar • 150 calories • 30-55 mg of caffeine • Artificial food colors and sulphites

00

, 1 .02,24

0

22 202

/ 5 / 252.1715

2 2.1715; cannot rejectn

XT

s nt t

T

0 .02,24 /

520 2.1715 22.1715

25

critX t s n

.05,24at 5% : 1.711;

520 1.711

25

21.711; reject

crit

t

X

The Prob-Value

22mg, 5mg; n=25X s H0: = 20

H1: > 20

conclusion: Peter should be able to sleep during the boring lectures.

24 24

22 2022 | 20 2 .0285

5 / 25P X P t P t

Non-Central t Distribution – a Small Complication

•If the alternative hypothesis is true, the statistic T0 does not have a mean of zero.

•It has what is called the non-central t distribution.

•Many of the operations performed for the central t must now be done using numerical techniques, e.g. integrating the non-central t distribution.

•No sweat for us. The results we need are tabulated – we need the O.C. curves!

0

000

if

/ /

XX nT

ss n s n

estimatednoncentralityparameter

Non-Central t Distribution – a Small Complication

•O.C. curves relate , n, and d, where d=|0|/. and is the Type II error probability.

•Since variance is unknown, estimate – either from previous experiments where we measured s, or after the current measurement set is collected.

•Alternatively we can think of d as a certain number of standard deviations (if a relative measure is satisfactory).

O.C. Curves for t-test

0d

two-sided t-testwith = .05

More on Peter Perk’s Pop

If the actual caffeine level is 22 mg per can, what is the probability of not rejecting at the 5 percent level (i.e. the probability of a Type II error)?

d = |22-20|/5 = .4n = 25

Pr{do not reject| = 22} .5

9-4.1 Hypothesis Test on the Variance

9-4.1 Hypothesis Test on the Variance

More on Variance Tests

Much more on Variance Tests

A Variance Test Example Professor Vera Vance asserts that the variability of the

IQ’s among the Engineering Management (ENM) students is significantly less than for the population at large.

It is well know that the distribution of IQ’s is normal with a mean of 100 and a standard deviation of 15.

Thirty ENM students were forced to take an exhaustive IQ test in which the sample standard deviation was computed to be 12.78.

Professor Vance is willing to test her assertion at the 10 percent level.

Add a little variance to your lifeH0: 2 = 152

H1: 2 < 152

2220 2 2

0

2 21 , 1 .90,29

1 29 12.7821.05

15

19.7677; cannot rejectn

n s

Professor Vera Vance lecturingthe students on their excessive variability.

Prob-value and so much more…

Prob-Value 229 21.05 .1427P

9-4.2 Type II Error and Choice of Sample Size:

Operating characteristic curves are provided in Charts VIIi through VIIn

0

O.C. Curves for Variance

0

Problem 9-75

1) The parameter of interest is the true variance of sugar content, 2.

2) H0 : 2 = 18 3) H1 : 2 18 4) = 0.05

5) 02 = ( )n s 1 2

2

6) Reject H0 if 02

1 2 12 / ,n where 70.22

9,975.0 or 02

2 12 , ,n where

02.1929,025.0

7) n = 10, s = 4.8

02 = 52.11

18

)8.4(9)1( 2

2

2

sn

8) Since 2.70 < 11.52 < 19.02 do not reject H0 and conclude there is insuffi cient evidence to indicate the true variance of sugar content is significantly diff erent f rom 18 at = 0.05.

More sugar?

milligrams2

Problem 9-75 P-Value

29Pr 11.52 .2417

P-value 2 .2417 .4834

Problem 9-75 Confidence Interval

Problem 9-75 Assume 2 = 40what sample size to detect difference with prob = .9?

Using the chart in the Appendix, with 49.118

40 and = 0.10, we fi nd

n = 50 at .01 alpha level

This is the O.C. graphic for alpha = .01

9-5.1 Large-Sample Tests on a Proportion

Many engineering decision problems include hypothesis testing about p.

An appropriate test statistic is

9-5 Tests on a Population Proportion

Another form of the test statistic Z0 is

or

ˆ XP

n the sample proportion

Test on a Proportion - Example

A cable news commentator has asserted that more than half the population believe that the United States should not have taken military action against Iraq. Test this assertion at the 5 percent level.

"Looking back, do you think the United States did the right thing in taking military action against Iraq, or should the U.S. have stayed out?"

Right Thing

Stayed Out Unsure    

% % %    

9/14-16/07 39 53 8

CBS News Poll. Sept. 14-16, 2007. N=706 adults nationwide. MoE ± 4 (for all adults).

An Example Continued

0

0 .05 0

.53 .501.594

.50 1 .50 / 706

1.594 1.6449; cannot reject

z

z z H

H0: p .5H1: p > .5 = .05

p-Value Pr 1.594 .0555z

ˆ .53, 706p n

9-5.2 Type II Error and Choice of Sample Size

For a two-sided alternative where p is the true value

If the alternative is p < p0

If the alternative is p > p0

Type II Error

H0: p .5H1: p > .5 = .05

assume .55, 706p n

.5 .55 1.645 .5 1 .5 / 706

1.017 .1546.55 1 .55 / 706

9-5.3 Type II Error and Choice of Sample SizeFor a two-sided alternative

For a one-sided alternative

Sample Size Calculations

H0: p .5H1: p > .5 = .01, =.05 when p = .55

2

2.33 .5 1 .5 1.645 .55 1 .551573.5 1574

.55 .50n

Goodness-of-fit (GOF) Tests

Testing for the distribution of the underlying population

9-7 The Chi-Square GOF Test

• Assume there is a sample of size n from a population whose probability distribution is unknown.

• Let Oi be the observed frequency in the ith class interval.

• Let Ei be the expected frequency in the ith class interval.

The test statistic is chi-square with

df = k – p – 1 where p = number of estimated parameters

Example 9-12

More of Example 9-12

Much More of Example 9-12

Still Example 9-12

Yes, Example 9-12

The last of Example 9-12

Next Week – Chapter 10

How to do it with two samples!