summary statistics iv

8/12/2019 Summary Statistics IV

1/54

Compare parameters of the

original pop. of all individual scores

and the parameters of thesampling dist. of all possible xs

mean =m& std. dev. =sOriginal population:

mxsx

=m (same mean as

individual values)

=sn

(different std. dev.,

but related!)


2/54

s

m

XZ

Z Formula for X - value

Z = the number of standard deviations

that an X - value is from the mean.


3/54

n

X

Zs

m

Z Formula for Sample Means,X


4/54

Anytime the original pop.is Normal (true for any n).

Anytime the original pop.is not Normal, but

n is BIG (n > 30).

Reminder

When is the population ofall possible X values Normal?


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Anytime the original population

is not NormalAND

n is NOT BIG.

Reminder

When is the population ofall possible X values NOT Normal?


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Confidence Interval

and

Hypothesis


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The Margin of Error of

a sample mean in estimatingthe true mean of the population.

MOE at 95% confidence =

The maximum amount by which 95%

of ALL X-bars miss the population mean mWith 95% confidence, the most by which

MY X-bar misses m


8/54

m= true population mean

Where are the 95% of all possible X-bars

that are closest to the population mean m?

m X

.95

? ?

-1.96 0 +1.96 Z

.025.025

Illustration of Margin of Error


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? - m

Z = s n

+1.96 =

? -m

sn

? = m + 1.96 s n

Margin of Error

Convert to the X-bar axis:


10/54

1.96 l sn

sxstandard error

of the mean

Margin of Error for 95% confidence:

MOE =

The confidence

multiplier

More confidence?

Less confidence?Larger sample size?

Smaller sample size?

Larger MOE

Smaller MOESmaller MOE

Larger MOE


11/54

General form for margin of error

when sis known:

MOE = Z l sna

2

Z a2

appropriate percentile from thestandard normal distribution,

i.e., the Z table.

where is the


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Explanation of symbol:

Z ~ N(0,1)0

Za/2

Za/2 cuts off the top tail at area = a/2a/ 21 a/2


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Amount ofconfidence

Area ineach tail

Table value

.95

.90

.80

.98

.0250

.0500

.1000

.0100

1.96

1.645

1.28

2.33

1 - a a/2 Z a/ 2

Examples


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Confidence Interval:

Point estimate MOE


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(1-a)100% Confidence Interval:

Point estimate MOE(1-a)100% is the amount of

confidence desired.

.95 confidence a= .05 risk


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Example: Exam 1 scores

X=

X~

individual score on Exam 1

Population: Exam 1 scores

?( m= , s= ) slightly skewed-left? 14.0

What do we know?

Is the population of X-bars Normal?

Maybe. The CLT might apply for n =

25.

Estimate with 98% confidence the mean

score on Exam 1 if the pop. std. dev. is 14.0.The mean of a sample of 25 exams is 71.04.


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MOE = Z l sna 2

= 2.33 l14

5= 6.52pts

= Z.01 l14 25

Note: Were assuming that n = 25 islarge enough for the CLT to apply.



score on Exam 1 if the pop. std. dev. is 14.0.The mean of a sample of 25 exams is 71.04.


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71.04 6.52

Point estimate MOE

or64.52 to 77.56 pts

98% confidence interval:

Example: Exam 1 scores,con tinued . . .


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I am 98% confident that

the true mean Exam 1 score (m)in the population of all test-takers

is within the interval 64.52 to 77.56 points.

A statement in the L.O.P. contains four parts:

1. the level of confidence.

2. the parameter estimated in the L.O.P.

3. the population to which we generalize

in the L.O.P.

4. the calculated interval.

Statement in the L.O.P.


20/54

Before sampling, there is a 98% chance that well

pick an X-bar within MOE of m.

m X

.98

m- MOE m+ MOE

.01.01

Think about this:

MOE | MOE| |

If my X-bar is one of the 98% closest, then

X-bar +/- MOE will contain the true mean m.

X[ ] CIMOE + MOE


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Replace s with s.

Replace z with t.

What do we do if the

true population standarddeviation sis unknown?


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Estimated margin of error (MOE)

when sis UN-known:

MOE = tl

sna 2 ,n1

sx

estimated

standard

error of

themean

ta2

,n1where is the

appropriate percentile from

the t-distribution.


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Explanation of symbol:

t-distribution0 ta/ 2 , n1

ta/2, n-1cuts off the top tail at area= a/2

Use the t-tableto find the value.

a/ 21 a/2


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0 t

Table gives right-tail area.

(e.g., for a right-tail area

of 0.025 and d.f. = 15,

the t value is 2.131.)

0.1 0.05 0.025 0.01 0.005

d.f. = 1 3.078 6.314 12.706 31.821 63.6562 1.886 2.920 4.303 6.965 9.925

3 1.638 2.353 3.182 4.541 5.841

4 1.533 2.132 2.776 3.747 4.604

5 1.476 2.015 2.571 3.365 4.032

6 1.440 1.943 2.447 3.143 3.7077 1.415 1.895 2.365 2.998 3.499

8 1.397 1.860 2.306 2.896 3.355

9 1.383 1.833 2.262 2.821 3.250

10 1.372 1.812 2.228 2.764 3.16911 1.363 1.796 2.201 2.718 3.106

12 1.356 1.782 2.179 2.681 3.055

13 1.350 1.771 2.160 2.650 3.012

14 1.345 1.761 2.145 2.624 2.977

15 1.341 1.753 2.131 2.602 2.947

16 1.337 1.746 2.120 2.583 2.921

17 1.333 1.740 2.110 2.567 2.898

18 1.330 1.734 2.101 2.552 2.878

19 1.328 1.729 2.093 2.539 2.861

20 1.325 1.725 2.086 2.528 2.845

21 1.323 1.721 2.080 2.518 2.831

22 1.321 1.717 2.074 2.508 2.819

23 1.319 1.714 2.069 2.500 2.807

24 1.318 1.711 2.064 2.492 2.797

25 1.316 1.708 2.060 2.485 2.78726 1.315 1.706 2.056 2.479 2.779

27 1.314 1.703 2.052 2.473 2.77128 1.313 1.701 2.048 2.467 2.763

29 1.311 1.699 2.045 2.462 2.756

30 1.310 1.697 2.042 2.457 2.750

31 1.309 1.696 2.040 2.453 2.744

32 1.309 1.694 2.037 2.449 2.738

33 1.308 1.692 2.035 2.445 2.733

34 1.307 1.691 2.032 2.441 2.728

35 1.306 1.690 2.030 2.438 2.724

0.1 0.05 0.025 0.01 0.005

36 1.306 1.688 2.028 2.434 2.719

37 1.305 1.687 2.026 2.431 2.71538 1.304 1.686 2.024 2.429 2.712

39 1.304 1.685 2.023 2.426 2.708

40 1.303 1.684 2.021 2.423 2.704

41 1.303 1.683 2.020 2.421 2.70142 1.302 1.682 2.018 2.418 2.698

43 1.302 1.681 2.017 2.416 2.695

44 1.301 1.680 2.015 2.414 2.692

45 1.301 1.679 2.014 2.412 2.690

97 1.290 1.661 1.985 2.365 2.627

98 1.290 1.661 1.984 2.365 2.627

99 1.290 1.660 1.984 2.365 2.626

100 1.290 1.660 1.984 2.364 2.626

1.282 1.645 1.960 2.326 2.576

Want 95% CI,

n = 20,

a/2 = .025d.f. = n-1 = 19

t.025, 19

= 2.093


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0 t

0.1 0.05 0.025 0.01 0.005

d.f. = 1 3.078 6.314 12.706 31.821 63.6562 1.886 2.920 4.303 6.965 9.925

3 1.638 2.353 3.182 4.541 5.841

4 1.533 2.132 2.776 3.747 4.604

5 1.476 2.015 2.571 3.365 4.032

6 1.440 1.943 2.447 3.143 3.7077 1.415 1.895 2.365 2.998 3.499

8 1.397 1.860 2.306 2.896 3.355

9 1.383 1.833 2.262 2.821 3.250

10 1.372 1.812 2.228 2.764 3.16911 1.363 1.796 2.201 2.718 3.106

12 1.356 1.782 2.179 2.681 3.055

13 1.350 1.771 2.160 2.650 3.012

14 1.345 1.761 2.145 2.624 2.977

15 1.341 1.753 2.131 2.602 2.947

16 1.337 1.746 2.120 2.583 2.921

17 1.333 1.740 2.110 2.567 2.898

18 1.330 1.734 2.101 2.552 2.878

19 1.328 1.729 2.093 2.539 2.861

20 1.325 1.725 2.086 2.528 2.845

21 1.323 1.721 2.080 2.518 2.831

22 1.321 1.717 2.074 2.508 2.819

23 1.319 1.714 2.069 2.500 2.807

24 1.318 1.711 2.064 2.492 2.797

25 1.316 1.708 2.060 2.485 2.78726 1.315 1.706 2.056 2.479 2.779

27 1.314 1.703 2.052 2.473 2.77128 1.313 1.701 2.048 2.467 2.763

29 1.311 1.699 2.045 2.462 2.756

30 1.310 1.697 2.042 2.457 2.750

31 1.309 1.696 2.040 2.453 2.744

32 1.309 1.694 2.037 2.449 2.738

33 1.308 1.692 2.035 2.445 2.733

34 1.307 1.691 2.032 2.441 2.728

35 1.306 1.690 2.030 2.438 2.724

0.1 0.05 0.025 0.01 0.005

36 1.306 1.688 2.028 2.434 2.719

37 1.305 1.687 2.026 2.431 2.71538 1.304 1.686 2.024 2.429 2.712

39 1.304 1.685 2.023 2.426 2.708

40 1.303 1.684 2.021 2.423 2.704

41 1.303 1.683 2.020 2.421 2.70142 1.302 1.682 2.018 2.418 2.698

43 1.302 1.681 2.017 2.416 2.695

44 1.301 1.680 2.015 2.414 2.692

45 1.301 1.679 2.014 2.412 2.690

97 1.290 1.661 1.985 2.365 2.627

98 1.290 1.661 1.984 2.365 2.627

99 1.290 1.660 1.984 2.365 2.626

100 1.290 1.660 1.984 2.364 2.626

1.282 1.645 1.960 2.326 2.576

Want 98% CI,

n = 33,

a/2 = .01d.f. = n-1 = 32

t.01, 32= 2.449



of 0.025 and d.f. = 15,



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0 t

0.1 0.05 0.025 0.01 0.005

d.f. = 1 3.078 6.314 12.706 31.821 63.6562 1.886 2.920 4.303 6.965 9.925

3 1.638 2.353 3.182 4.541 5.841

4 1.533 2.132 2.776 3.747 4.604

5 1.476 2.015 2.571 3.365 4.032

6 1.440 1.943 2.447 3.143 3.7077 1.415 1.895 2.365 2.998 3.499

8 1.397 1.860 2.306 2.896 3.355

9 1.383 1.833 2.262 2.821 3.250

10 1.372 1.812 2.228 2.764 3.16911 1.363 1.796 2.201 2.718 3.106

12 1.356 1.782 2.179 2.681 3.055

13 1.350 1.771 2.160 2.650 3.012

14 1.345 1.761 2.145 2.624 2.977

15 1.341 1.753 2.131 2.602 2.947

16 1.337 1.746 2.120 2.583 2.921

17 1.333 1.740 2.110 2.567 2.898

18 1.330 1.734 2.101 2.552 2.878

19 1.328 1.729 2.093 2.539 2.861

20 1.325 1.725 2.086 2.528 2.845

21 1.323 1.721 2.080 2.518 2.831

22 1.321 1.717 2.074 2.508 2.819

23 1.319 1.714 2.069 2.500 2.807

24 1.318 1.711 2.064 2.492 2.797

25 1.316 1.708 2.060 2.485 2.78726 1.315 1.706 2.056 2.479 2.779

27 1.314 1.703 2.052 2.473 2.77128 1.313 1.701 2.048 2.467 2.763

29 1.311 1.699 2.045 2.462 2.756

30 1.310 1.697 2.042 2.457 2.750

31 1.309 1.696 2.040 2.453 2.744

32 1.309 1.694 2.037 2.449 2.738

33 1.308 1.692 2.035 2.445 2.733

34 1.307 1.691 2.032 2.441 2.728

35 1.306 1.690 2.030 2.438 2.724

0.1 0.05 0.025 0.01 0.005

36 1.306 1.688 2.028 2.434 2.719

37 1.305 1.687 2.026 2.431 2.71538 1.304 1.686 2.024 2.429 2.712

39 1.304 1.685 2.023 2.426 2.708

40 1.303 1.684 2.021 2.423 2.704

41 1.303 1.683 2.020 2.421 2.70142 1.302 1.682 2.018 2.418 2.698

43 1.302 1.681 2.017 2.416 2.695

44 1.301 1.680 2.015 2.414 2.692

45 1.301 1.679 2.014 2.412 2.690

97 1.290 1.661 1.985 2.365 2.627

98 1.290 1.661 1.984 2.365 2.627

99 1.290 1.660 1.984 2.365 2.626

100 1.290 1.660 1.984 2.364 2.626

1.282 1.645 1.960 2.326 2.576

Want 90% CI,

n = 600,

a/2 = .05d.f. = n-1 = 599

t.05, = 1.645Same as Z



of 0.025 and d.f. = 15,



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t-dist.Z

Bell shaped,

symmetric

Mean

Std. dev.

As n1 increases, tn1 approaches Z

Degrees of freedom

= 0

> 1

Yes Yes

= 0

= 1

n-1

Comparison of Z and t


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= 2.492 l16.8

5= 8.37pts

= t.01, 24 l16.8 25

MOE = sn

t la2

, n-1



score on Exam 1.In a sample of 25 exams, the mean is 71.04and the std dev is 16.8.

(1-a)100%n

X

s


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71.04 8.37

Point estimate MOE

or62.67 to 79.41 pts

Example,con tinued . . .

98% confidence interval:


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I am 98% confident that

the true mean Exam 1 score (m)in the population of all test-takers

is within the interval 62.67 to 79.41 points.

A statement in the L.O.P. contains four parts:

1. the level of confidence.

2. the parameter estimated in the L.O.P.3. the population to which we generalize

in the L.O.P.

4. the calculated interval.

Statement in the L.O.P.


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Compare the two 98% MOE values.

Sample 2. True swas NOT known:

Sample 1. True swas known:= 2.33 l

14

5 = 6.52ptsMOE

= 2.492 l16.8

5= 8.37ptsMOE


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To get a smaller margin of error:q Increase n.

q Decrease the amount ofconfidence desired.

Margin of Error for 95% confidence:

= 1.96 l s nMOE


33/54

What sample size is needed to estimate

the mean mpg of Toyota Camrys with an

MOE of 0.2 mpg at 90% confidence if the

pop. std. dev. is 0.88 mpg?

MOE = Z l s na 2

0.2 = 1.645 l 0.88

n

n =1.6452l0.882

0.22 = 52.39

Round

UP,

use 53Camrys


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Sample Size Determinations

Problem: What sample size nis needed to havea margin of error of MOE with

(1a)100% confidence?

MOE za/2

s

n

n za/2

s

MOE

2


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Confidence Interval

for a Parameter

point estimate margin of error

Choose the appropriate statistic

(point estimate)and its MOE based on the problem that is tobe solved.

E ti ti f P t


36/54

Mean, m

ifsis known:

PopulationParameter

Point EstimateMargin of Error

at (1-a)100% confidence

Mean, m

ifsis unknown:

MOE Za2

s

n

MOE ta2,n

1

s

n

X

X

A (1-a)100% confidence interval estimate of a parameter is

point estimate MOE

Estimation of Parameters

E ti ti f P t


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Diff. of twoproportions,p1 - p2:

A (1-a)100% confidence interval estimate of a parameter is

21 pp

1 1 2 2

21 2

p (1-p p (1-pm.o.e. = Z +n n

) )

21 xx 2 2

1 2

21 2

s sm.o.e. = Z +

n n

Proportion,p: 2

m.o.e. = Z p(1-p) n

MOE = Z 2

gsn

Mean, m

ifsis known:

( , n-1)2

sm.o.e. = t

n

PopulationParameter

Point Estimate Margin of Errorat (1-a)100% confidence

Mean, m

ifsis unknown:

/ ,p X n

X

X

point estimate MOE

Estimation of Parameters

Diff. of two

means, m1 - m2 :

(for large sample sizes only)

( , n-2)2

sm.o.e. = tEqu.2

Slope of regression

line,b:b

* 2

( , n-2)2

1 (x -x )m.o.e. =t s +

n Equ. 2

Mean from a

regression

when X =x*:

* bxay

where s MSE


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Hypothesized mean

Making a decision using a CI.

A value of the parameter that

we believe is,or ought to be,

the true value of the mean.

We gather evidence and make a

decision about this hypothesis.


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Question of interest: Is there evidence

that the true meanis different from the

hypothesized mean?

Making a decision using a CI.

q If the hypothesized value is insidethe CI,

then it ISa plausible value.

Reach a vagueconclusion..

q If the hypothesized value is notin the CI,then it IS NOTa plausible value.

Reject it! Reach a strongconclusion.

Take appropriate action!


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5 8 11 14 17 20

The true population

mean is hypothesized

to be 13.0.

X-axis

Population of

all possible

X-bar values,

assuming . . . .

My ONE

sample mean.

My ONE

Confidence Interval.

Conclusion:

The hypothesis is

wrong. The true

mean not 13.0!

13.0 does NOT fall in

my confidence interval;

it is not a plausible value

for the true mean.

ll

l10.25.6 7.9

Middle

95%

The dataconvince me

the truemeanis

smallerthan 13.0.

I am 95%confident

that . . . .

A more likely location


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5 8 11 14 17 20

The true population

mean is hypothesized

to be 13.0.

Conclusion:

The hypothesis is

wrong. The true

mean not 13.0!

ll

l10.25.6 7.9

X-axis

A more likely location

of the population.

13.0 does NOT fall in

my confidence interval;

it is not a plausible value

for the true mean.

The dataconvince me

the truemeanis

smallerthan 13.0.

I am 95%confident

that . . . .


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Net weight of potato chip bags

should be 16.00 oz.

FDA inspector takes a sample.

If the 95% CI is (15.81 to 15.95),

If the 95% CI is (15.71 to 16.05),

then 16.00 is NOTin the interval.

Therefore, reject16.00 as a plausible

value. Take action against the company.

X = 15.88

then 16.00 ISin the interval.

Therefore, 16.00 is a plausible value.

Take no action.

X = 15.88


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Net weight of potato chip bags

should be 16.00 oz.

FDA inspector takes a sample.

then 16.00 is NOTin the interval.

Therefore, reject16.00 as a plausible

value. But, the FDA does not care that

the company is giving away potato chips.

The FDA would obviously take no action

against the company.

X = 16.10If the 95% CI is (16.05 to 16.15),


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Hypothesis Testing

for Means

C t


45/54

Court case

Hypothesis: Defendant is innocent.

Alternative: Defendant is guilty.

Decisions: Based on the sample data.

Reject

Innocence

Declare

Guilty

Persongoes to

jail!

Do not Reject

Innocence

Declare

Not

Guilty

Person

goes

free!

C t


46/54

Court case

H0

: Defendant is innocent.

H1: Defendant is guilty.

Decisions: Based on the sample data.

Reject

Innocence

Declare

Guilty

Persongoes to

jail!

Do not Reject

Innocence

Declare

Not

Guilty

Person

goes

free!

Types of Errors in a Court Case


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Type I: Send an innocent person to jail.

Type II: Let a guilty person go free.

a= level of risk deemed reasonablefor the occurrence of a Type I error.

= the point of reasonable doubt.

Types of Errors in a Court Case


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Reject the hypothesized value if:

1. it is outside the confidence interval.

2. the p-value is less than theuser specified a-level. (p-value < a-level)

Statistical Inference Methods:

Two methods; both give the same result.


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Testing the mean MPG of a car population

Ho: m= 40.0H1: m40.0

Ho: The population mean is 40.0

H1: The population mean is NOT 40.0

Decision:

Do NOT reject Ho.

Conclusion:

Not sufficient evidence

to say that H1 is correct.

Reject Ho!

The true population mean

is NOT 40.0. Take action!

X is in the vicinity of 40.0.

Its too close to call.

X is far enough away

from 40.0 to convince me.

then 40.0 is a plausible for m then 40.0 is NOT plausible.p-value < a-level, or40.0 is outsidethe CI,

Ifp-value > a-level, or40.0 is insidethe CI,

If

1 Confidence Inter al Method


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1. Confidence Interval Method

Two tailed test:Is the mean something

other than 40.0?

Hypothesized mean: 40.0

.05 .95

.10 .90

.01 .99

Desired

a-level:

Size of

CI to use:

1 - a

Result: Each tail has half of a.

2 l M th d


51/54

2. p-value Method

p-value = the probability of observingastatistic valuethat is more extreme(more contradictory to hypothesis)thanthe value observed, assuming that thehypothesized parameter value is correct.

Calculate p-value using the

appropriate distribution.

Decision rule:If p-value < a-level,

reject the hypothesized value.

X = 42 6Hypo mean: 40 0p Value:


52/54

X = 42.6

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

40 X42.6

Hypo. mean: 40.0,p-Value:

37.4

Two tailed test:

Is the mean something

other than 40.0?

p-value

/

2p-value

/

22.62.6

p-value is a measure of how far the observed estimateof the true parameter is from the hypothesized value.

Small

p-value

Far away!

(inconsistent)

Reject

hypothesized

value

Large

p-value

Fairly close

(consistent)

Do not reject

hypothesized

value

P bl 1 i V l


53/54

Sample results:X = 43.0

s = 7.2

-4.0 -3.0 -2.0 -1.0 0.0 1.0 2.0 3.0 4.0

0 Z1.50

Z =43.040.0

2.0= 1.50

.0668

40 X43.0

Hypothesized mean: 40.0.Change ads if MPG is off

in eitherdirection.

More extreme

Alsomore extreme

-1.50

p-value = .0668 2= .1336

than 3.0 unitsthan 3.0 units

Problem 1, using p-Value

What distribution

should be used?

Pick a= .05

(p-value=.1336)> (a=.05);Do not reject 40.0.

X~ N(m= ?, s= 8.0)n= 16X~ N(m X= ?, s X= 2.0)

P bl 1 ith C fid I t l


54/54

Sample results:X = 43.0

s = 7.2

Hypothesized mean: 40.0.Change ads if MPG is off

in eitherdirection.

Problem 1 withConfidence Interval

Which CI formula

should be used?

Pick a= .05

40.0 falls inside the CI. Do not reject 40.0.

X Z025sn43.01.96 2.043.0 3.92(39.08, 46.92)

X~ N(m= ?, s= 8.0)n= 16X~ N(m X= ?, s X= 2.0)

summary statistics iv

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