modular 17 ch 10.2 part ii and ch 10.3 part ii. ch 10.3 hypothesis test about a population mean with...

Modular 17Ch 10.2 Part II

andCh 10.3 Part II

Ch 10.3 Hypothesis Test about a Population Mean with unknown

Ch 10.2 Hypothesis Test for a Population Mean

Objective B : – Value ApproachP

Objective B : – Value ApproachP

Ch 10.2 Hypothesis Test for a Population Mean Objective B : – Value ApproachP

3) the sampled values are independent of each other ( ).

2) to guarantee that a normal distribution can be used to test hypotheses for ;

1) simple random sampling ;

sample proportion, , provided the sample is obtained by

– Test for a Proportion A hypothesis test involving a population proportion can be considered as a binomial experiment. As we learned from Ch 8.2, the best point estimate of , the population proportion, is a

Z

p

n

xp ˆ

10)1( 00 pnp

00 : ppH

Nn 05.0

Testing Hypotheses Regarding a Population Proportion, .Use the following steps to perform a – Test for a Proportion.Z

p

Example 1: The percentage of physicians who are women is 27.9%. In a survey of physicians employed by a large university health system, 45 of 120 randomly selected physicians were women. Use the - Value Approach to determine whether there is sufficient evidence at the 0.05 level of significance to conclude that the proportion of women physicians at the university health system exceeds 27.9%?

05.0,45,120 xn

279.0:0 pH

279.0:1 pH (right-tailed test)

Step 1 :

Step 2 : 05.0

P

Sample : 45,120 xn

375.0120

45ˆ n

xp

npp

ppz

)1(

ˆ

00

00

120)279.01)(279.0(

279.0375.0

120)721.0)(279.0(

279.0375.00

z 34.2 * (test statistic)

Step 3 :

– value approach-- > Let alone. -- > Convert to a tailed area. 34.20 z

P

– value = 1 – 0.9904P= 0.0096

From Table V

9904.03.2

04.0

34.20 z

9904.0

Is – value < ?P Is 0.0096 < 0.05 ? Yes, reject .

There is enough evidence to support the claim that the proportion of female physicians exceeds 27.9%.

0H

Step 4 :

Step 5 :

Example 2: In 1997, 4% of mothers smoked more than 21 cigarettes during their pregnancy. An obstetrician believes that the percentage of mothers who smoke 21 cigarettes or more is less than 4% today. She randomly selects 120 pregnant mothers and finds that 3 of them smoked 21 or more cigarettes during pregnancy. Does the sample data support the obstetrician’s belief? Use the - Value Approach with level of significant.

P01.0

04.0:0 pH

04.0:1 pH (left-tailed test)

025.0120

3ˆ n

xp

01.0,3,120 xn

npp

ppz

)1(

ˆ

00

00

120)04.01)(04.0(

04.0025.0

120)96.0)(04.0(

04.0025.00

z 84.0 * (test statistic)

– value approach-- > Let alone. -- > Convert to a tailed area. 84.00 z

P

– value = 0.2005P

From Table V

2005.08.0

04.0

84.00 z

2005.0

Is – value < ?P Is 0.2005 < 0.01 ? No, do not reject .

There is not enough evidence to support the claim that the percentages of mothers who smoke 21 cigarettes or more is less than 4%.

0H

Example 3: In 2000, 58% of females aged 15 years of age and older lived

alone, according to the U.S. Census Bureau. A sociologist tests whether this percentage is different today by conducting a random sample of 500 females aged 15 years of age and older and finds that 285 are living alone. Use the -Value Approach to determine whether there is sufficient evidence at the level of significance to conclude the proportion has changed since 2000?

1.0P

1.0,285,500 xn

58.0:0 pH

58.0:1 pH (Two-tailed test)

57.0500

285ˆ n

xp

npp

ppz

)1(

ˆ

00

00

500)58.01)(58.0(

58.057.0

500)42.0)(58.0(

58.057.00

z 45.0 * (test statistic)

– value approach-- > Let alone. -- > Convert to a two-tailed area. 45.00 z

P

– value = Double (0.3264)P

From Table V

3264.04.0

05.0

45.00 z

3264.0

45.00 z

3264.0

= 0.6528

Is – value < ?P

Is 0.6528 < 0.1 ? No, do not reject .

There is not enough evidence to support the claim.

0H

Ch 10.3 Hypothesis Test about a Population Mean with unknown

Ch 10.2 Hypothesis Test for a Population Mean

Objective B : – Value ApproachP

Objective B : – Value ApproachP

Objective B : – Value ApproachP

Ch 10.3 Hypothesis Test about a Population Mean with unknown

The – test procedure requires either that the sample was drawn from a normally distributed population or the sample size was greater than 30. Minor departures from normality will not adversely affect the results of the test. However, if the data include outliers, the – test procedure should not be used. A normality plot is used to test whether the sample was drawn from a normally distributed population. A boxplot is used to detect outliers.

t

t

Example 1: Find the – value interval for each test value.P(a) right-tailed,15,624.20 nt

141151 ndfFrom Table VI

01.0

624.214

df

624.20 t14df

01.0valueP

(b) right-tailed,28,703.10 nt

271281 ndfFrom Table VI

05.0

703.127

df

703.10 t27df

05.0valueP

(c) left-tailed,23,321.20 nt

221231 ndf From Table VI

02.0

183.2

22

df

321.20 t22df

02.0value01.0 P321.2

508.2

01.0

– value is between 0.01 and 0.02.P02.0value01.0 P

(d) two-tailed ,17,562.10 nt

161171 ndf From Table VI

10.0

337.1

16

df

562.10 t17df

562.1

746.1

05.0

is between 0.05 and 0.10.)value(2

1P

10.0)value(2

105.0 P

20.0value10.0 P

20.0value10.0 P

reject .

fail to reject .

Example 2: Use the – value approach to determine whether to reject

or fail to reject .

P0H

(a) – value versus for a right-tailed test126.0 05.0P

Since – value is not smaller than , P 0H)05.0()126.0(

(b) – value versus for a left-tailed test001.0 025.0P

Since – value < , P 0H)025.0()001.0(

Example 3: A survey of 15 large U.S. cities finds that the average commute time one way is 25.4 minutes. A chamber of commerce executive feels that the commute in his city is less and wants to publicize this. He randomly selects 25 commuters and finds the average is 22.1 minutes with a standard deviation of 5.3 minutes. At , is there enough evidence to support the claim? Use the – value approach.

P01.0

(left-tailed test)

3.5,1.22,25,4.250 sxn

ns

xt 00

253.5

4.251.22113.3 *

4.25:0 H

4.25:0 H

01.0

Setup :Step 1 :

Step 2 :

Step 3 :

Is – value < ?

241251 ndf From Table VI

0025.0

091.3

24

df

113.30 t24df

0025.0value001.0 P

113.3

467.3

001.0

– value is between 0.001 and 0.0025.P0025.0value001.0 P

– value ApproachP

Yes , reject .P )01.0(001.0Between(

There is sufficient evidence to support the executive’s claim that the commute time in his city is less than 25.4 minutes.

Step 4 :0H

Step 5 :

)0025.0and

As we have experienced from example 1 to 3, we were unable to find exact – values using the table because the – table provides

– values only for certain area. To find exact – values, we use statistical software.

P tPt

Example 4: Use StatCrunch to calculate the following question.