7 Hypothesis Testing1

7/28/2019 7 Hypothesis Testing1

1/25

Chapter

7

Elementary Statistics Larson Farber

Hypothesis Testing


2/25

A Statistical Hypothesis

Alternativehypothesis H a contains a statement

of inequality , such as.

Null hypothesis H 0 tatistical hypothesis

hat contains a

tatement of equality ,uch as , = or .

If I am false,you are true If I am false,

you are true

H 0 H a

Complementary Statements

A claim about a population.


3/25

Write the claim about the population. Then,write its complement. Either hypothesis, the

ull or the alternative, can represent the claim.

A hospital claims its ambulance responsetime is less than 10 minutes .

H0 : 10 min

Ha : 10 min claim

Ha : 60.0 p

H0 : 60.0 p claim

Writing Hypotheses

A consumer magazine claims theproportion of cell phone calls made duringevenings and weekends is at most 60%.


4/25

A type I error: Null hypothesis is actuallyrue but the decision is to reject it.

Level of significance, a Maximum probability of committinga type I error.

D e c

i s i o n

Actual Truth of H 0

Errors and Level of Significance

H0 True H0 False

Do not

reject H 0

Reject H 0

Correct

Decision

CorrectDecision

Type II

Error

Type IError


5/25

Sampling distribution for x

The rejection region is the range of values for which the null hypothesis is not

probable . It is always in the direction of the alternative hypothesis. Its area is equalto a .

Rejection Region

0 z z0

A critical value separates the rejectionregion from the non-rejection region

Critical Value z0

Rejection Regions


6/25

z00

-z0

0 z0

Right- tail test Ha: >valu Reject H 0 if z > z 0 otherwise fail to reject H 0.

Two- tail test Ha: value

Reject H 0 if z > z 0 or z


7/25

Claim is H 0

There is notenoughevidence toreject the claim

There isenoughevidence toreject the claim

Interpreting the Decision


8/25

Claim is H a

There is notenoughevidence tosupport theclaim

There isenoughevidence tosupport theclaim

Interpreting the Decision


9/25

1 . Write the null and alternative hypothesis

2 . State the level of significance

3 . Identify the sampling distribution

Write H 0 and H a as mathematical statements.Remember H 0 always contains the = symbol.

This is the maximum probability of rejecting thenull hypothesis when it is actually true. (Makinga type I error.)

The sampling distribution is the distribution for the test statistic assuming that H 0 is true andthat the experiment is repeated an infinitenumber of times.

8 Steps in a Hypothesis Test


10/25

7 . Make your decision

6. Find the test statistic

5. Find therejection region

4 . Find thecritical value

8 . Interpret your decision

The critical value separates the rejection regionof the sampling distribution from the non-rejection region.

Perform the calculations to standardize your sample statistic.

If the test statistic falls in the critical region,reject H 0. Otherwise, fail to reject H 0.


11/25

he critical value z 0 separates the rejection regionrom the non-rejection region. The area of theejection region is equal to a .

z0 0

ejectiongion

z00

Rejectionregion

-z0 0 z 0

Reject ionregion

Rejectionregion

ind z 0 for a left-tailest with a =.01

Find z 0 for a right-tailtest with a =.05

Find - z0

and z0

for a two-tail test with a =.01

z0=-2.33

-z0=-2.575 and z 0 =2.575

z0=1.645

Critical Values


12/25

A cereal company claims the meansodium content in one serving of its cereal isno more than 230 mg. You work for anational health service and are asked to test

this claim. You find that a random sample of 52 servings has a mean sodium content of 232 milligrams and a standard deviation of 10 mg. At a = 0.05, do you have enough

evidence to reject the companys claim? 1. Write the null and alternative hypothesis

H0: 230 mg.(claim) Ha: > 230 mg.

2. State the level of significancea = 0.05

3. Determine the sampling distribution

Since the sample size is at least 30, the samplingdistribution is normal.

The z-test for a Mean


13/25

7. Make your decision


8. Interpret your decision

5. Find the rejectionregion

Rejectionregion

Since H a contains the > symbol, this is a right tail test.

n=52x = 232s=10

44.1387.1

2

5210

230232 z

= 1.44 does not fall in the rejection region, soail to reject H 0

here is not enough evidence to reject the

ompanys claim that there is at most 230mg of odium in one serving of its cereal.

z00

1.645

4. Find the criticalvalue


14/25

The P-value is the probability of obtaininga sample statistic with a value as extremeor more extreme than the one determinedby the sample data. Reject H 0 if P > a .

z 0

Area inleft tail

For a left tail test

z 0

Area in right tail

For a right tail test

z 0 z

If z is

negative,twice thearea inthe lefttail

If z is positivetwice the

area in theright tail

For a two-tail test

P-Values

P-value = indica ted area


15/25

The standardized z -score for the hypothesistest on the sodium content of cereal was z =1.44. Find the P-value and make your decisionat the 0.05 level of significance.

The standardized z -score for the hypothesisest on the sodium content of cereal was z =.44. Find the P-value and make your decisiont the 0.05 level of significance.

ince this was a right-tail test, the P-value is therea to the right of the z score. The area to theght of z = 1.44 is 0.0749 so the P-value is .0749.ince 0.0749 > 0.05, fail to reject H 0.

If P = 0.0246, what is your decision if 1) a =0.052) a =0.01

If P = 0.0246, what is your decision if 1) a =0.052) a =0.01

1) Since 0.0246 < 0.05, reject H 0.

2) Since 0.0246 > 0.01, fail to reject H 0 .

Using P-Values


16/25

For a P-value decision, compare areas.If P < , reject H 0. P , fail to reject H 0.

0

For = 0.05, z 0 = -1.645. If z = -1.23

the P-value is 0.1093. The actual area to theleft of the z -score is 0.1093. With astandardized z -score of -1.23, fail to rejectthe null hypothesis at the 0.05 level of

significance.

z0

Rejection area

0.05

z

If = 0.05 and z = -1.23

Using the P-value of a testcompare areas


17/25

Find the critical value t 0 for a left-tailedtest given a = 0.01 and n = 18.

Find the critical value t 0 for a left-tailedtest given a = 0.01 and n = 18.

0

Area inleft tail

Find the critical values - t 0 and t 0 for atwo-tailed test given a = 0.05 and n = 11.

Find the critical values - t 0 and t 0 for atwo-tailed test given a = 0.05 and n = 11.

d.f.= 18-1 = 17

to

t0 = - 2.567

0

d.f.= 11-1 = 10

-t0

t0

-t0 =-2.228 and t 0=2.228

The t SamplingDistribution


18/25

A university says the mean number of classroomhours per week for full-time faculty is 11.0. Aandom sample of the number of classroom hoursor full-time faculty for one week is listed below.

You work for a student organization and are askedo test this claim At a = 0.01, do you have enough

evidence to reject the university s claim?11.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1

A university says the mean number of classroomours per week for full-time faculty is 11.0. Aandom sample of the number of classroom hoursor full-time faculty for one week is listed below.

You work for a student organization and are askedo test this claim At a = 0.01, do you have enoughvidence to reject the universitys claim? 1.8 8.6 12.6 7.9 6.4 10.4 13.6 9.1

. Write the null and alternative hypothesis

H0: = 11.0 (claim) Ha: 11.0

. State the level of significance

a = 0.01


ince the sample size is 8, the samplingistribution is a t-distribution with 8 - 1 = 7 d.f.

Testing -small Sample


19/25




n=8x = 10.050s=2.485

08.1878.0

95.0

8485.2

0.11050.10t

= -1.08 does not fall in the rejection region, soail to reject H 0 at the 0.01 level of significance.

here is not enough evidence to reject the

niversitys claim that faculty spend a mean of 11lassroom hours.


Rejectionregion

Rejectionregion

ince H a contains the symbol, this is a two- tail test.

-t0 0 t0

4. Find the criticalvalues

-3.499 3.499


20/25

T-Test of the Mean

Test of mu = 11.000 vs mu not = 11.000

Variable N Mean StDev SE Mean T PHours 8 0.050 2.485 0.879 -1.08 0.32

Enter the data in C1, Hours.Choose t-test in the STAT menu .

Minitab reports the t-statistic and the P-value.

Since the P-value is greater than the level of significance (0.32 > 0.01), reject the nullhypothesis at the 0.01 level of significance.

Minitab Solution


21/25

Test for Proportions

A communications industry spokespersonlaims that over 40% of Americans either owncellular phone or have a family member whooes. In a random survey of 1036 Americans,56 said they or a family member owned aellular phone. Test the spokespersons claim

t a = 0.05. What can you conclude?


H0: p 0.40 Ha: p > 0.40 (claim)

. State the level of significance a = 0.05

The standardized test statistic is:n

pq

p p z

is the population proportion of successes.The test statistic isf np 5 and nq 5 the sampling distribution for

normal.

n x p

p


22/25





6.201522.0

04.0

1036)60)(.40(.

40.44. z

z = 2.63 falls in the rejection region, so reject H 0

here is enough evidence to support the claim that

ver 40% of Americans own a cell phone or have aamily member who does.

0

036(.40) > 5 and 1036(.60)>5 . The s amplingistribution is normal.

n =1036x = 456


1.645

5. Find the rejection

region

Rejectionregion


23/25

Find a c 20 critical value for a left-tail test when=17 and a = 0.05.

2 is the test statistic for the population variance. Itsampling distribution is a c 2 distribution with n-1 d.f.

Find critical values c 20 for a two-tailed test when= 12 and a = 0.01.

he standardized test statistic is2

22 )1(

c

sn

c 20 =7.962

c 2l =2.603 and c 2R =26.757

0 1 0 2 0 3 0 4 0

Critical Values for 2


24/25

A state school administrator says that the standarddeviation of test scores for 8th grade students whotook a life-science assessment test is less than 30.You work for the administrator and are asked to test

this claim. You find that a random sample of 10 testshas a standard deviation of 28.8. At a = 0.01, do youhave enough evidence to support the administratorsclaim? Assume test scores are normally distributed.


H0: 30 Ha: < 30 (claim)

. State the level of significance a = 0.01


he sampling distribution is c 2 with 10 - 1 = 9 d.f.

Test for


25/25




n=10s = 28.8

c 2 = 8.2944 does not fall in the rejection region,so fail to reject H 0

here is not enough evidence to support the

2944.8

30

8.28)110()1(2

2

2

22

c sn

0 1 0 2 0 3 0 4 0



2.088

7 Hypothesis Testing1

Documents

Transcript of 7 Hypothesis Testing1