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Write My Essay For MeChapter 9 HYPOTHESIS TESTS ABOUT ONE PARAMETER

9.2 Z TEST for a MEAN: testing claim about one mean and is known and the

population is normal OR for any population when is known and n 30.

Sample Problem: GIVEN SAMPLE STATISTICS

*A survey claims that the average cost of a hotel room in Atlanta is $69.21. To test

the claim, a researcher selects a sample of 30 hotel rooms and finds that the average

cost is $68.43. The population standard deviation is $3.72. Test the above claim

using =.05. Assume population normal.

Test Statistic (TS) Z = -1.15

Test Statistic Z = -1.15

p=.2508 > =.05 do not reject H

There is not enough evidence to reject the claim.

Ho might be true.

σ

σ ≥

a

0

1

: 69.21

: 69.21

: 69.21

Claim

H

H

µ

µ

µ

=

=

¹

a 0

Entry Display

[STAT]

[>][>]TESTS

[1] for 1:Z-Test

Calculate

[ENTER]

P=.2508

z= -1.15 (TS)

2

Sample Problem: GIVEN DATA

*The average 1 –year old (both sexes) is 29 inches tall. A random sample of 30 one-year olds in

a large day care franchise resulted in the following heights. At =.05 , test the claim that the

average height differs from 29 inches. The population standard deviation is 2.6108. Assume

the population is normal.

25 32 35 25 30 26.5 26 25.5 29.5 32

30 28.5 30 32 28 31.5 29 29.5 30 34

29 32 27 28 33 28 27 32 29 29.5

Entry Display

Enter data into list

one.

[STAT]

[>][>] TESTS

[1] for 1:Z-Test

Z-Test

Inpt: Data [Stats]

:29

: 2.6108

List:

Freq:1

Calculate

[ENTER]

z= .944 (TS)

p=.3451

Test Statistic Z = .944

p=.3451 > =.05 do not reject H

There is not enough evidence to support the claim.

Ho might be true.

a

σ

0

1

: 29

: 29

: 29

Claim

H

H

µ

µ

µ

¹

=

¹

µο

σ

1L

:µ µo¹

a 0

3

9.3 T TEST for a MEAN: testing a claim about one mean and is not

known and the population is normal OR for any population when is not

known and n 30.

Sample Problem: GIVEN STATISTICS

*The average amount of rainfall during the summer months for the northeast part of

the United States is less than 11.52 inches. A researcher selects a random sample of

10 cities in the northeast and finds that the average amount of rainfall for 1995 was

7.42 inches and the standard deviation was 1.3 inches. At =.05 test the above

claim. Assume the population is normal.

Test Statistic (TS) t = -9.97

Entry Display

[STAT]

[>][>] TESTS

[2] for 2:T -Test

Calculate

[ENTER]

t= -9.97 (TS)

p= 1.8E-6

Test Statistic t =-9.97

p=1.8E-6 < =.05 reject H

There is enough evidence to support the claim.

H1 is true.

σ

σ

≥

a

Claim:µ <11.52

H0 :µ =11.52

H1 :µ <11.52

a 0

4

Sample Problem: GIVEN DATA

In one part of a test developed by a psychiatrist, the test subject is asked to

form a word by unscrambling letters. Given are the times (in seconds)

required by 15 randomly selected persons to unscramble letters (50, 16, 17,

50, 25, 52, 60, 40, 33, 30, 76, 16, 30, 74, 59) At the = .05 level of

significance, test the claim that the mean time is equal to 50 seconds.

Assume the population is normal.

Input the times into a list, say L1. Press [STAT] [>] [TESTS] [T-Test]

Highlight DATA and press [ENTER]

Input the following:

Highlight calculate and press enter.

Test Statistic t= -1.57

p=.1395 > do not reject Ho

There is not enough evidence to reject the claim.

Ho might be true.

a

Claim:µ = 50

H0 :µ = 50

H1 :µ ≠ 50

a

5

9.4 Z Test for Proportion: 1Prop Z Test: To test a claim about one proportion.

Sample Problem: GIVEN X AND N

*A telephone company wants to advertise that more than 30% of all its customers

have more than two telephones. To support this ad, the company selects a sample of

200 customers and finds that 72 have more than two telephones. Test the claim at

=.05 . Assume the population is normal.

Test Statistic (TS)

Entry Display

[STAT]

[>][>] TESTS

[5] for

5:1-PropZTest

1-PropZTest

po: .3

x: 72

n: 200

prop: > po

Calculate

[ENTER]

z=1.85 (TS)

p=.0320

Test Statistic Z = 1.85

p=.0320< =.05 reject H

There is enough evidence to support the claim.

H1 is true.

Note: In some problems (such as the next problem) instead of being given x and n,

you are given n and the sample proportion. To find x, simply multiply n and the

sample proportion.

a

Claim:p > .30

H0 :p = .30

H1 :p > .30

a 0

6

Sample Problem: GIVEN SAMPLE PROPORTION AND N

*The American Automobile Association (AAA) claims that 54% of fatal

car/truck accidents are caused by driver error. A researcher studies 30

randomly selected accidents and finds 47% were caused by driver error. Test

the above claim at =.05 . Assume the population is normal.

n = 30 =.47 (round to the nearest

whole number: .5-.9 ROUND UP, and .1 – .4 ROUND DOWN)

Test Statistic (TS)

Entry Display

[STAT]

[>][>] TESTS

[5] for

5:1-PropZTest

1-PropZTest

po: .54

x: 14

n: 30

prop: po

Calculate

[ENTER]

z=-.806 (TS)

p=.420

Test Statistic Z = -.806

p=.4203 > =.05 do not reject H

There is not enough evidence to reject the claim.

Ho might be true.

a

p̂ x = p̂ ⋅n = (.47)(30) =14.1→14

Claim:p = .54

H0 :p = .54

H1 :p ≠ .54

¹

a 0

7

Ch 10 &11 HYPOTHESIS TESTS/CONFIDENCE INTERVALS ABOUT TWO

PARAMETERS

11.1 and 10.1 2 SAMPLE Z TEST: Testing a Claim about Two Means:

Independent Samples, known, Both populations normal OR

both samples are at least 30.

Sample Problem: GIVEN STATISTICS

*In a study of women science majors, the following data were obtained on two

groups, those who left their profession within a few months after graduation

(leavers) and those who remained in their profession after they graduated (stayers).

Test the claim that those who stayed had a higher science grade-point average than

those who left. Use =.05. Assume populations are normal.

[STAT][>][>]TESTS

[3] for 3: 2-SampZTest

2-SampZTest

Inpt: Stats

TS Z = -2.01

p = .0222 < =.05

Reject the null hypothesis.

There is enough evidence to support the claim.

H1 is true.

σ1 and σ 2

a

Leavers Stayers

n1 =103 n2 = 225

x1 = 3.16 x2 = 3.28

σ1 = .52 σ2 = .46

claim:µ1 < µ2

H0 :µ1 = µ2

H1 :µ1 < µ2

1

2

1 2

:.52

:.46

1:3.16

1:103

2:3.28

2:225

:

X

n

X

n

Calculate

s

s

µ µ<

a

8

Note: To do a confidence interval, simply choose 2SampZInt.

Sample Problem: GIVEN DATA

*A researcher claims that the average length of the major rivers in the United States

is the same as the average length of the major rivers in Europe. The data (in miles)

of a sample of rivers are shown. At =.01, test the above claim. Given

and . Assume populations are normal.

United States Europe

729 560 434 481 724 820

329 332 360 532 357 505

450 2315 865 1776 1122 496

330 410 1036 1224 634 230

329 800 447 1420 326 626

600 1310 652 877 580 210

1243 605 360 447 567 252

525 926 722 824 932 600

850 310 430 634 1124 1575

532 375 1979 565 405 2290

710 545 259 675 454

300 470 425

Enter US data in L1 and Europe data in L2

[STAT][>][>]TESTS

[3] for 3: 2-SampZTest

2-SampZTest

Inpt: Data

TS Z = -.856

p=.3920 > =.01

Do not reject the null hypothesis.

a σ1 = 449.8703

σ 2 = 474.1258

1 2

0 1 2

1 1 2

:

:

:

claim

H

H

µ µ

µ µ

µ µ

=

=

¹

1

2

1 2

:449.8703

:474.1258

1: 1

2: 2

1:1

2:1

:

List L

List L

Freq

Freq

Calculate

s

s

µ µ¹

a

9

There is not enough evidence to reject the claim.

Ho might be true.

_________________________________________________________________________________________________

2 SAMPLE T TEST: Testing the Claim about Two Means with

Independent Samples and not known, both populations

normal OR both sample sizes are at least 30.

Sample Problem: GIVEN STATISTICS

*A real estate agent wishes to determine whether tax assessors and real estate

appraisers agree on the values of homes. Summary statistics for a random sample of

the two groups is below. Test the claim that there is a significant difference in the

values of homes for each group. Use = .05. Assume both populations are normal

and the variances are the same. Assume populations are normal.

[STAT][>][>] TESTS [4] for 4: 2 SampTTest

2-SampTTest

Inpt: Stats

__________________________

t = -4.02 (TS)

p = 8.03 E-4 < = .05

Reject the null hypothesis.

There is enough evidence to support the claim. H1 is true.

σ1 and σ 2

a

REstate Tax Assessors

n1 =10 n2 =10

x1 = $83,256 x2 = $88,354

S1 = $3256 s2 = $2341

1 2

0 1 2

1 1 2

:

:

:

claim

H

H

µ µ

µ µ

µ µ

¹

=

¹

1 2

1:83256

1:3256

1:10

2:88354

2:2341

2:10

:

:

X

sx

n

X

sx

n

Pooled Yes

Calculate

µ µ¹

a

10

NOTE: Since the variances are given to be equal, we say YES to POOLED.

NOTE: If the variances are given to be unequal, we say NO to POOLED.

Example) To construct a 95% Confidence Interval using the above problem:

[STAT] [>][>] TESTS [O] for O: 2-Samp TInt

2-SampTInt

Inpt: Stats

Enter data same as above.

C-level: .95

Pooled: Yes

Calculate

(-7762, -2434)

Sample Problem: GIVEN DATA

*A health care worker wishes to see if the average number of family day care homes

per county is greater than the average number of day care centers per county. The

number of centers for a selected sample of counties is shown. At = .01 test the

above claim. Assume both populations are normal and the variances are equal.

Number of Family day care homes Number of day care centers

25 57 34 5 28 37

42 21 44 16 16 48

Enter data for homes in L1 and centers in L2

[STAT][>][>] TESTS [4] for 4: 2 SampTTest

2-SampTTest

Inpt: Data

t = 1.44 (TS)

p = .0895 > = .01

Do not reject the null hypothesis.

There is not enough evidence to support the claim. Ho might be true.

a

claim:µ1 > µ2

H0 :µ1 = µ2

H1 :µ1 > µ2

1 2

1: 1

2: 2

1:1

2:1

:

:

List L

List L

Freq

Freq

Pooled Yes

Calculate

µ µ>

a

11

11.2 and 10.2 Test Difference Between Two Means Dependent Samples

BEFORE/AFTER means Dependent samples. Both populations normal

OR both sample sizes are at least 30.

In a before/after test, this two-sample test reduces to a one-sample t test on

the differences.

Sample Problem: claim after scores less

*A new composition teacher wishes to see whether a new grammar program will

reduce the number of grammatical errors her students make when writing a two-

page essay. The data is shown below. Can it be concluded that the number of errors

has been reduced? Assume both populations are normal. Use = .025

Student 1 2 3 4 5 6

Errors

before

12 9 0 5 4 3

Errors

After

9 6 1 3 2 3

Since claiming that the errors has reduced, assuming that is true, such as 10 before

and 5 after. Then after – before = 5 – 10 = – 5, negative five which is less than zero.

That is why the claim is stated as the mean of the differences less than zero.

Enter Before data in L1 and After data in L2

Let L3 = L2 –L1 (AFTER – BEFORE)

In list mode, using arrows on keypad, highlight L3, press enter, type L2-L1 ENTER.

a

claim:µd < 0

H0 :µd = 0

H1 :µd < 0

12

[STAT] [>][>]TESTS [2] for 2:T-Test

T-Test

Inpt: Data

TS t = -2.24

p = .0378 > = .025

Do not reject the null hypothesis

There is not enough evidence to support the claim.

Ho might be true.

_________________________________________________________________

Alternative way to do the previous problem

Note: If I Let L3 = L1 – L2 (BEFORE – AFTER) then

Since claiming that the errors has reduced, assuming that is true, such as 10 before

and 5 after. Then before – after = 10 – 5 = 5, positive five which is greater than zero.

That is why the claim is stated as the mean of the differences greater than zero.

Enter Before data in L1 and After data in L2

Let L3 = L1 –L2

In list mode, using arrows on keypad, highlight L3, press enter, type L1-L2

ENTER.

[STAT] [>][>]TESTS [2] for 2:T-Test

T-Test

Inpt: Data

TS t = 2.24

p = .0378 > = .025

Do not reject the null hypothesis There is not enough evidence to support the claim.

Ho might be true.

µ0 :0

List :L3

Freq :1

µ :< µ0

Calculate

a

claim:µd > 0

H0 :µd = 0

H1 :µd > 0

0

3

0

:0

:

:1

:

List L

Freq

Calculate

µ

µ µ>

a

13

IMPORTANT SUMMARY OF THE 3 CASES in Before/After Hypothesis Tests

In a before/after test, this two-sample test reduces to a one-sample t test on

the differences. There are 3 cases explained below.

L1 = before data

L2 = after data

1) Test the claim that the number of errors has been increased.

For ex: before 7 and after 9. L3= L2 – L1 =after – before = 9 – 7 = +2 positive or > 0

Claim greater than zero

Claim: >0. (The mean of the differences greater than zero)

OR———————–

L3 = L1 – L2= before – after = 7 – 9 = -2 negative or < 0

Claim less than zero

Claim: <0 (The mean of the differences less than zero.)

2) Test the claim that the number of errors has been reduced.

For ex: before 9 and after 7. L3= L2 – L1 =after – before = 7 – 9 = -2 negative or < 0

Claim less than zero

Claim: <0 (The mean of the differences less than zero.)

OR———————–

L3 = L1 – L2= before – after=9 – 7 = + 2 positive or > 0

Claim greater than zero

Claim: >0. (The mean of the differences greater than zero)

3) Test the claim that the number of errors is the same before vs. after.

ex: before 9 and after 9. L3= L2 – L1 =after – before = L3 = L1 – L2= before – after=

9 – 9 = 0

Claim equals zero

Claim: = 0. (The mean of the differences equals zero.)

NOTE: To construct a confidence interval for the above, simply use TInterval

with the above data.

——————————————-

µd

µd

µd

µd

µd

14

11.3 and 10.3 Testing the Claim about Two Proportions: 2-Prop Z Test

Sample Problem: GIVEN THE SAMPLE

PROPORTION AND N

*In a sample of 80 Americans, 55% wished that they were rich. In a sample of 90

Europeans, 45% wished that they were rich. At = .01 test the claim that there is a

difference between the two groups. Assume populations are normal.

Note:

X1 = .55 *80 = 44

X2 = .45*90= 40.5 or 41

_________________________________

[STAT][>][>]TESTS [6] for 6:2-PropZTest

2-PropZTest

TS Z = 1.23

P = .2190 > = .01

Do not reject the null hypothesis

There is not enough evidence to support the claim.

Ho might be true.

To calculate the 99% Confidence Interval:

[STAT][>][>]TESTS [ALPHA][B] for B: 2-PropZInt

2-PropZInt

Enter data same as above.

C-level: .99

Calculate

(-.1026, .29145)

a

1 2

0 1 2

1 1 2

:

:

:

claim p p

H p p

H p p

¹

=

¹

ˆx p n= ×

1 2

1:44

1:80

2:41

2:90

:

X

n

X

n

p p

Calculate

¹

a

15

Sample Problem: GIVEN X AND N

*A survey of 80 homes in a Washington, D.C., suburb showed that 45 were air-

conditioned. A sample of 120 homes in a Pittsburgh suburb showed that 63 had air-

conditioning. At = .05 test the claim that there is a difference in the two

proportions. Assume populations are normal.

[STAT][>][>]TESTS [6] for

6:2-PropZTest

2-PropZTest

Test Statistic Z = .521

P = .6022 > = .05

Do not reject the null hypothesis

There is not enough evidence to support the claim.

Ho might be true.

a

1 2

0 1 2

1 1 2

:

:

:

claim p p

H p p

H p p

¹

=

¹

1 2

1:45

1:80

2:63

2:120

:

X

n

X

n

p p

Calculate

¹

a

16

11.4 Testing the Claim about Two Variances: 2 Sample F Test

Sample Problem: GIVEN DATA

* A tax collector wishes to see if the variances of the values of the tax-exempt

properties are different for two large cities. The values of the tax-exempt properties

are shown below. The data are given in millions of dollars. At = .05, is there

enough evidence to support the tax collector’s claim that the variances are different?

Assume populations are normal.

City A City B

113 22 14 8 82 11 5 15

25 23 23 30 295 50 12 9

44 11 19 7 12 68 81 2

31 19 5 2 20 16 4 5

Enter City A data in L1 and City B data in L2

TI83 [STAT][>]CALC [1] for 1:1-Var Stats [2nd] [L1] [ENTER]

see

TI83 [STAT][>]CALC [1] for 1:1-Var Stats [2nd] [L2] [ENTER]

see (It is customary to let larger standard deviation be s )

TI84

[STAT][>][>]TESTS

[APLHA] [D] or [E] for 2-SampFTest

2-Samp F Test

Inpt: Stats

a

2 2

1 2

2 2

0 1 2

2 2

1 1 2

:

:

:

claim

H

H

s s

s s

s s

¹

=

¹

225.97xs s= =

172.74xs s= = 1

17

TS F = 7.85

P=2.67E -4 < = .05

Reject the null hypothesis.

There is enough evidence to support the claim. H1 is true.

1 2

1:72.74

1:16

2:25.97

2:16

sx

n

sx

n

Calculate

s s¹

a

18

4.1 & 4.2 & 13.1 LINEAR CORRELATION (linear relationship

between 2 variables) and REGRESSION(finding the line of best fit)

Table of Total Points and Personal Fouls in basketball for players of a team

Question: Can we predict Total Points (Y) by knowing Personal Fouls (X)?

Note: The variable that you are predicting is Y. Assume populations are normal.

Let’s enter the data into lists.

[2ND] [MEM] [4] for 4:ClrAllLists [ENTER] clears all lists from the home screen

[STAT] [1] for 1:Edit enter fouls (x) data into L1 and total points (y) data into L2

To do a scatter plot (a graph of the ordered pairs):

Press [Y=] [CLEAR] To clear the y= editor.

Press [2ND] [STATPLOT] [4] for 4:Plots Off [ENTER] (to turn plots off)

Press [2ND] [STATPLOT] [1] for 1:Plot 1 [ENTER]

Player x( L1) PersonalFouls Y ( L2)TotalPo int s

1 2 3

2 25 76

3 2 1

4 19 60

5 10 10

6 4 8

7 6 36

8 21 47

9 2 1

10 4 3

11 23 58

19

[ZOOM] [9] for 9: ZoomStat

Note: There appears to be a positive correlation between x and y.

_____________________________________________________________________________________________

*r linear correlation coefficient gives strength and direction of a linear

relationship. Notice when r is positive, slope of the line is positive, and when r is

negative, slope of the line is negative.

_____________________________________________________________________________________________

To compute r the linear correlation coefficient:

need to turn diagnostics on (You only need to do this one time.)

[2ND] [CATALOG]

arrow down to DIAGNOSTICS ON

[ENTER]

[ENTER]

20

TI83

[STAT] [>] [CALC] [4] for 4:LinReg(ax +b) [2ND] [ ] [,] [2ND] [, ] [ ] [ENTER]

Note: To get above,

[VARS] [>] [YVARS] [1] for 1:Function] [1] for 1:

TI84

[STAT] [>] [CALC] [4] for 4:LinReg(ax +b)

Note: To get above,

[VARS] [>] [YVARS] [1] for 1:Function] [1] for 1:

r = .935 (looks significant, however we need to do a formal hypothesis test)

= .875 is the coefficient of determination

87.5 % of the total variation in total points can be explained by the variation in the

number of personal fouls. The other 12.5% is attributable to other factors.

Note: r and are displayed because the diagnostics are on.

Note: -1 r 1, when r = -1 perfect negative linear correlation, when r = +1

perfect positive linear correlation, when r = 0 no correlation.

To see the equation of the regression line:

Simply press [Y=]

y= 2.85x – 3.03

To graph the regression line on the scatter plot: Simply press GRAPH

1L [ ]2L 1Y

1Y

1Y

1Y

1Y

r2 r2

r2

≤ ≤

21

y = mx + b Sample Linear Model; m is Sample Slope

Population Linear Model ; is the Population Slope.

To graph the regression line on the scatter plot: Simply press GRAPH

IMPORTANT: The population correlation coefficient ”rho” and the population

slope ”Beta sub one” always have the same sign, both will be positive or both will

be negative. Additionally, whenever one of them is equal to zero, the other is equal

to zero as well. When the slope of a line is zero, we have a horizontal line and

changes in x do not result in change in y; this is NOT a linear relationship, and hence

the correlation is zero as well.

Test the claim of no linear correlation. Use a level of significance of .05.

Claim: = 0

(r is the sample statistic and the population parameter is

NOTE: This problem could also be worded as: Test the claim of zero slope.

( is the population slope and m is the sample slope.)

[STAT][>][>][TESTS][ALPHA] [E] or [F] for LinRegT-Test

Enter in the following:

Note: To get above,

[VARS] [>] [YVARS] [1] for 1:Function] [1] for 1:

Highlight Calculate and press [ENTER]

TS t = 7.93

p= 2.37E-5 <

reject

There is enough evidence to reject the claim. H1 is true.

Do a 95% confidence interval for or , we do a LinRegTInterval.

(2.0379, 3.6635)

y = β1x + β0 β1

ρ

β1

ρ

H

H

0

1

0

0

:

:

ρ

ρ

=

≠

ρ

Claim :β1 = 0

H0 : β1 = 0

H1 : β1 ≠ 0

β1

1Y

1Y

α =.05

H0

β1 ρ

22

Interpreting this interval: We are 95% confident that the true value of or lies in

the interval. Note: LinReg T interval is located on my TI84 Plus CE under TESTS, but

my TI83 Plus does not have it. This is NOT a topic that I will be testing you on.

Predictions

Since there is significant linear correlation between number of personal fouls and

the number of total points, we can use the regression equation for predictions.

Example: Find the best predicted total points given 10 personal fouls.

[VARS] [>] [Y-VARS] [1] for 1:Function [1] for 1:

(10) = 25.47 total points

———————————————————–

13. 3 Multiple Regression

In many cases, a better prediction model can be found for a dependent (response) Y

variable by using more than one independent (explanatory) X variables.

Just substitute the x values into the regression equation to compute the

predicted value of y.

β1 ρ

1Y

Y1

23

12.1 GOODNESS of FIT TEST

1) Under TESTS, check to see if your calculator has the Chi-Square GOF Test already. My

TI84 Plus C has this test under TESTS, option D: Chi-Square GOF TEST (best option)

2) If you do not have this test, you can either:

– follow my handout on the GOF PROGRAM from Canvas. (recommended)

– follow the directions below for the “long way”

————————————————————————————————————

Ex) Mars Inc. claims that it’s M&M candies are distributed with the color

percentages of 30% brown, 20% yellow, 20% red, 10% orange, 10% green, and

10% blue. A classroom exercise resulted in the observed frequencies listed below.

At the .05 level, test the claim that the color distribution is as claimed by Mars Inc.

Assume populations are normal.

Claim= Ho: The distribution is as stated above.

H1: The distribution differs from the claimed distribution.

L1 L2 L3

Enter the observed frequencies in List 1 and the experimental probabilities in List 2

to figure out the exp f (expected frequency) column Recall E = np

To generate List 3:

= sum( [ENTER]

Note: the SUM command is from 2ND STAT à à MATH and choose 5)SUM

Using the Chi-Square GOF Test, enter data in L1 and L2, and generate L3 as

explained above. You do not need L4.

color obsf P f

brown

yellow

red

blue

green

orange

exp exp

.

.

.

.

.

.

17 30

39 20

8 20

26 10

12 10

19 10

L3 L L1 2)*

24

STAT—->—–> TESTS

Choose: D: Chi-Square GOF Test

Observed: L1 (observed f)

Expected: L3 (expected f)

Df: 5 (#categories – 1)

CALCULATE

You will see………

Chi-Square TS = 50.06

P value = 1.34 E-9 < alpha = .05

Reject the null hypothesis.

There is enough evidence to reject the claim.

H1 is true.

____________________________________________________________________________________________

OR “GOF Program” from Canvas Module

Follow my directions to type program in your calculator (easy to do!), then

Enter the data in L1 and L2 and generate L3 as explained above.

Then press PRGM, 1: GOF, ENTER

__________________________________________________________________________________________

OR “long way”

To generate List 4:

[2ND] [QUIT]

To find the Test Statistic =50.06

Now, find the p-value:

[2nd][DISTR] [7] or [8] for : cdf( TS, E99, df) [ENTER]

df= #categories – 1

(Note: to get ‘E’ press [2nd] [EE] )

(50.06, E99, 5) [ENTER]

p = 1.35 E-9 < = .05

Reject the null hypothesis.

There is enough evidence to reject the claim.

H1 is true.

L L L L4 1 3

2

3= −( ) /

2

4( )sum Lχ =

2χ

2χ

α

25

Ex) Mars Inc. claims that it’s M&M candies are distributed with the same frequency.

The colors are: Brown, yellow, red, orange, green, and blue. A classroom exercise

resulted in the observed frequencies listed below. At the .05 significance level, test

the claim that the color distribution is as claimed by Mars Inc. Assume populations

are normal.

WE ARE ASKED TO TEST THE CLAIM THAT THE COLORS OCCUR WITH THE SAME

FREQUENCY. SINCE WE HAVE 6 COLORS, EACH MUST OCCUR 1 DIVIDED BY 6

or .1667.

Claim= Ho: The distribution is as stated above.

H1: The distribution differs from the claimed distribution.

L1 L2 L3

Enter the observed frequencies in List 1 and the experimental probabilities in List 2

To figure out the exp f (expected frequency) column Recall E = np

To generate List 3: = sum( [ENTER]

Note: the SUM command is from 2ND STAT à à MATH and choose 5)SUM

Continue the problem either using the Chi-Square GOF Test or GOF Program or the

“long way” as explained in the last example.

Chi-Square TS = 30.48

P value = 1.18E-5 < alpha = .05

Reject Ho

There is enough evidence to reject the claim.

H1 is true.

———————————————————————————————————–

NOTE: For some of the problems at ALEKS, you are given the Observed Frequencies

(L1) and the Expected Frequencies (L3). In my notes, we are given the Observed

Frequencies (L1) and the Expected Probabilities (L2), and then we calculate the

Expected Frequencies (L3). For problems like these at ALEKS, put the Observed

Frequencies in L1 and the Expected Frequencies in L3 and nothing in L2.

color obsf expP exp f

brown 17 .1667

yellow 39 .1667

red 8 .1667

blue 26 .1667

green 12 .1667

orange 19 .1667

L3 L L1 2)*

26

12.2 TESTS OF INDEPENDENCE

Ex) Responses to a survey question are broken down according to employment and

the sample results are given below. At the.10 significance level, test the claim that

the response and employment status are independent. Assume populations are

normal.

Claim=Ho: Response and employment status are independent.

H1: Response and employment status are dependent.

enter the data in Matrix [A]:

[2ND] [MATRIX] [>] [>] [EDIT] press [1] for matrix A

Note: We have a 2 (rows) x 3 (columns) matrix.

[STAT] [>][>] [TESTS] [ALPHA][C] for C: Test

Highlight Calculate and press [ENTER]

TS

p=.0512 < =.10

reject the null hypothesis

There is enough evidence to reject the claim.

H1 is true.

Yes no undecided

Employed

Unemployed

30 15 5

20 25 10

2χ

2 5.94χ =

α

27

14.1 One Way ANOVA: Testing claims about 3 or more population

means.

Ex) At the .05 significance level, test the claim that the mean distance traveled by

each type of golf ball is the same. Assume populations are normal.

____________________________________________________________________________________

Note other ways of stating the alternative hypothesis:

H1: At least one mean is different from the others.

H1: Means are not all equal.

_________________________________________________________________

Enter data into lists 1,2 and 3

[STAT][>] [>][TESTS] [ALPHA][F] or [H] for ANOVA [2ND] [ ][,] [2ND] [ ][,]

[2ND][ ] [ENTER]

TS F = .653

p=.531 > = .05

Do not reject the null hypothesis

There is not enough evidence to reject the claim.

Ho might be true.

…

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