Define tomorrow.universityof south africa

Tutorial letter 104/2/2017

Applied Statistics II

STA2601

Semester 2

Department of Statistics

TRIAL EXAMINATION PAPER

STA2601/104/2/2017

Dear Student

Congratulations if you obtained examination admission by submitting assignment 1. I would like to

take the opportunity of wishing you well in the coming examinations. I hope you found the module

stimulating.

The examination

Please note the following with regard to the examination:

* The duration of the examination paper is two-hours. You will be able to complete the

set paper in 2 hours, but there will be no time for dreaming or sitting on questions you

are unsure about. Make sure that you take along a functional scientific calculator that

you can operate with ease as it can save you some time. My advice to you would be to

do those questions you find easy first; then go back to the ones that need more thinking.

I do not mind to mark questions in whatever order you do them, just make sure that you

number them clearly!

* A copy of the list of formulae is attached to the trial examination paper. Please ensure

that you know how to test the various hypotheses.

* All the necessary statistical tables will be supplied (see the trial paper).

* Pocket calculators are necessary for doing the calculations.

* Working through (and understanding!) ALL the examples and exercises in the study

guide, workbook and in the assignments as well as the trial paper will provide beneficial

supplementary preparation.

* Make sure that you know all the theory as well as the practical applications.

* All the chapters in the study guide are equally important and don’t try to spot!

* Start preparing early and don’t hesitate to call or email me if something is unclear.

The enclosed trial examination papers should give you a good indication of what to expect in the

examination.

Best wishes with your preparation for the examination and do not hesitate to contact me if you have

any questions about STA2601.

Ms S. Muchengetwa

GJ GERWEL (C-Block), Floor 6, Office 6-05

Tel: (011) 670-9253

Cel: 074 065 9020

e-mail: muches@unisa.ac.za

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Trial paper 1

Reserve two hours for yourself and do the trial paper under exam conditions on your own!

Duration: 2 hours 100 Marks

INSTRUCTIONS

1. Answer ALL questions.

2. Marks will not be given for answers only. Show clearly how you solve each problem.

3. For all hypothesis-testing problems always give

(i) the null and alternative hypothesis to be tested;

(ii) the test statistic to be used; and

(iii) the critical region for rejecting the null hypothesis.

4. Justify your answer completely if you make use of JMP output to answer a question.

3

May/June 2017 Paper One Final Examination

QUESTION 1

Complete the following statements in your answer book (i.e. give the missing words and do not

waste time rewriting everything):

(a) The statistic T is called an unbiased estimator for the parameter θ if ..................... (1)

(b) The efficiency of two estimators of the same parameter is a function of their .......................

(1)

(c) If V and W are random variables such that V ∼ χ25 and W ∼ χ2

9 then, U =V/5

W/9has a

......................... distribution with ......................... degree(s) of freedom. (2)

(d) When multiple measurements or observations are made on each of the individuals or units in

a sample the assumption of ....................... is violated. (1)

(e) If you reject the null hypothesis on the basis of sample data, when in fact no difference exists,

you have made a .......................................... error. (1)

[6]

QUESTION 2

(a) Let X1, X2, ..., Xn be a random sample of size n from a discrete distribution with probability

function

P (X = r) =λre−λ(

1− e−λ)r !

for r = 1; 2; ...

[Please note:

In a real life situation a random sample will result in, for example, X1 = 2; X2 = 3; X3 =3; X4 = 10; X5 = 2 etc ... . Do not fall into the trap to argue that X1 = 1; X2 = 2; ...; Xn =n because this is only one very specific outcome out of the millions of other possibilities.

Denote the sample outcome by X1 = r1; X2 = r2; ...; Xn = rn.]

(i) Find the likelihood function for the sample. (4)

(ii) Show that∂lnL (λ)

∂λ= −n −

ne−λ(1− e−λ

) +n∑

i=1

ri

λ(4)

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(b) Let X1, X2, ..., Xn be independent random variables from a distribution with expected value

7θ. Find the least squares estimator for θ. (6)

[14]

QUESTION 3

The following data gives the number of orders received by a company each week over a period of

40 weeks.

8 9 10 11 12 13 13 14 14 15

15 16 17 17 17 18 18 18 19 19

20 20 21 21 22 22 23 23 24 24

25 26 26 26 27 27 28 29 31 32

(a) The following summary statistics were obtained.

N = 40∑

X i = 790 X = 19.75∑(X i − X)2 = 1 485.5

∑(X i − X)3 = 234.75

∑(X i − X)4 = 120 961.5313

Test whether this distribution is symmetric. (Use α = 0.10.) (7)

(b) The 40 observations were classified into five classes with equal probability for each class

interval and the observations were assumed to be from a n(20; 62

)distribution. The following

table of observed frequencies was obtained:

Table of observed and expected frequencies

Class interval Oi ei = nπ i

X < 14.95 9 ......14.95 ≤ X < 18.48 9 ......18.48 ≤ X < 21.52 6 ......21.52 ≤ X < 25.05 7 ......

X ≥ 25.05 9 ......

Totals 40 40

(i) Show how intervals 1 is derived. (4)

(ii) Calculate the missing expected frequencies in the above table. (1)

5

(iii) At the 0.05 level, use a chi-square goodness-of-fit test to test if the 40 observations in

the sample come from a normal distribution with mean 20 and standard deviation 6. (7)

(c) The following SAS JMP outputs in Figure 1 and Figure 2 were obtained.

Figure 1

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Figure 2

(i) One would like to test whether the average number of orders is less than 18. What

assumption(s) is/are necessary in order to conduct the statistical test specified? (2)

(ii) Test whether the average number of orders is less than 18 using the p-value approach.

Use α = 0.05 and the output given. (3)

(iii) Is there any reason to reject the null hypothesis H0 : σ = 5? Test at the 5% level of

significance using the p-value approach. (3)

7

(iv) Interpret the 95% confidence interval for µ. (1)

(d) Suppose another sample of 30 weeks is taken and the following statistics are obtained:

Y = 22 S2Y = 42.25

Would you say that the mean number of orders of population B is more than the mean number

of orders for A?

Test the hypothesis H0 : µX = µY

H1 : µX < µY

against

sat the 5% level of significance.

[Hint: assume that both variances are unknown but equal. Assume that S2X = 38.09 for the

sample with n1 = 40 from population A.] (8)

[36]

QUESTION 4

(a) The operations manager of a company that manufactures shirts wants to determine whether

there are differences in the quality of workmanship among the three daily shifts. She ran-

domly selects 600 recently made shirts and carefully inspects them. Each shirt is classified

as either perfect or flawed, and the shift that produced it is also recorded. The accompanying

table summarises the number of shirts that fell into each cell.

Shirt condition Shift

1 2 3

Perfect 240 191 139

Flawed 10 9 11

The operations manager wanted to find out whether the data provide sufficient evidence to in-

fer that there are differences in quality between the three shifts at the 5% level of significance.

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The following SAS JMP output in Figure 3 was obtained.

Figure 3

(i) Interpret the Mosaic Plot. (3)

(ii) State the appropriate null and alternative hypothesis for this test. (2)

(iii) What test statistic is used to test these hypotheses and what is the value of the test

statistic? (2)

(iv) What is your final conclusion? (2)

9

(b) To study telephone calls at a business office, the office manager times incoming calls and

outgoing calls for 1 day. She finds that 17 incoming calls last an average of 5.16 minutes with

a standard deviation of 1.12 minutes and 12 outgoing calls last an average of 4.13 minutes

with a standard deviation of 2.36 minutes. Determine a 95% confidence interval forσ 2

1

σ 22

and

interpret the results. (6)

[15]

QUESTION 5

The extent to which a person’s attitude can be changed depends in part on how big a change you

are trying to produce. In a classic study on persuasion, Aronson, Turner, and Carlsmith (1963)

obtained three groups of participants. One group listened to a persuasive message that differed

only slightly from the participants’ original attitudes. For the second group, there was a moderate

discrepancy between the message and the original attitudes. For the third group, there was a large

discrepancy between the message and the original attitudes. For each participant, the amount of

attitude change was measured. The following data was obtained.

Size of discrepancyAmount of attitude

changex i

6∑j=1

(xi j − x i

)2Short 1 0 0 2 3 0 x1 = 1 8

Moderate 3 4 6 3 5 3 x2 = 4 8

Large 0 2 0 4 0 0 x3 = 1 14

X = 2

[You may also accept that3∑

i=1

6∑j=1

(xi j − x

)2= 66.]

(Regard the data as random samples from normal populations.)

(a) What are the values of S21 , S2

2 , and S23 . (3)

(b) (i) Compute the “ordinary” average of the three variances computed in (a). (2)

(ii) Compute the MSE according to the definition in the study guide. What do you notice?(4)

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(c) Suppose you are given the following output in Figure 4.

Figure 4

Test at the 5% level of significance whether the amount of discrepancy between the origi-

nal attitude and the persuasive argument has a significant effect on the amount of attitude

change.

(i) State the null and alternative hypotheses.

(ii) State the rejection region and conclusion.

11

(4)

(d) For these data, describe how the effectiveness of a persuasive treatment is related to the

discrepancy between the argument and a person’s original attitude. (2)

(e) The Tukey-Kramer HSD method of multiple comparisons test was done to determine which

means differ. The following output in Figure 5 was obtained.

Figure 5

Using the above output discuss your results. (4)

[19]

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QUESTION 6

The following table shows the observations of transportation time and distance for a sample of ten

rail shipments made by a motor parts supplier.

Delivery time (days) 5 7 6 11 8 11 12 8 15 12

Distance (kilometres) 210 290 350 480 490 730 780 850 920 1 010

The following output in Figure 6 was obtained.

Figure 6

Assume that a linear relationship Yi = β0 + β1xi + εi where the εi ’s are independent n(0; σ 2

)random variables, is meaningful. Using Figure 6:

13

(a) Does the assumption of linearity appear to be reasonable and why? (2)

(b) Give an interpretation of the numerical value of the regression coefficient β1. (1)

(c) Use the estimated regression equation that can be used to predict delivery time for a cus-

tomer situated 600 kilometres from the company. (1)

(d) Derive a 95% confidence interval for the slope β1. (4)

(e) Give the value of R2 and interpret it. (2)

[10]

[100]

Trial paper 2

Reserve two hours for yourself and do the trial paper under exam conditions on your own!

Duration: 2 hours 100 Marks

INSTRUCTIONS

1. Answer ALL questions.

2. Marks will not be given for answers only. Show clearly how you solve each problem.

3. For all hypothesis-testing problems always give

(i) the null and alternative hypothesis to be tested;

(ii) the test statistic to be used; and

(iii) the critical region for rejecting the null hypothesis.

4. Justify your answer completely if you make use of JMP output to answer a question.

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May/June 2017 Paper Two Final Examination

QUESTION 1

(a) Complete the following statement:

If Y ∼ n(0; 1), then Y 2 has a .............................distribution. (1)

(b) What is the connection between random variables being uncorrelated and independent? (3)

(c) What is the relationship between a Type II error and the power of the test? (2)

[6]

QUESTION 2

(a) Write down, in general terms, the method of obtaining a least squares estimator. (4)

(b) Let X1, X2, ..., Xn be independent random variables from a distribution with expected value

θ. Find the least squares estimator for θ. (4)

(c) The probability distribution function (p.d.f.) of the two-parameter gamma distribution (with

parameters α > 0 and β > 0) is given by

f (x;α;β) =1

0 (α) βαxα−1e−x/β for x > 0

= 0 for x ≤ 0

Let X1, X2, ..., Xn be a random sample from this distribution and assume that the parameter

α is known, but that the parameter β is unknown.

(i) Show that the likelihood function L (β) in given by

L (β) = 0 (α)−n β−αn

(n∏

i=1

X i

)α−1

e−∑

X i/β

(4)

(ii) Find Log L (β). (3)

15

(iii) Show that the maximum likelihood estimator (m.l.e) of β equals∑

X i/nα. (4)

[19]

QUESTION 3

(a) Let X1, X2, ..., X100 be a random sample that yields the following:

100∑i=1

X i = 50,1

100

100∑i=1

(X i − X

)2= 25

1

100

100∑i=1

(X i − X

)3= 9.6 and

1

100

100∑i=1

(X i − X

)4= 832.2

(i) Test at the 10% level of significance whether this sample comes from a symmetrical

distribution. (7)

(ii) Does this sample have the kurtosis of a normal distribution? Test at the 10% level of

significance. (7)

(iii) Would you say that this is a sample from a normal distribution? (1)

(b) Starting in 2008 an increasing number of people found themselves facing mortgages that

were worth more than the value of their homes. A fund manager who had invested in debt

obligations involving grouped mortgages was interested in determining the group most likely

to default on their mortgages. He speculates that older people are less likely to default on

their mortgage and thinks the average age of those who do is less than 55 years. To test this,

a random sample of 30 who had defaulted was selected; the following sample data reflect the

ages of the sampled individuals:

40 55 78 27 55 33 51 76 54 67

40 31 60 61 50 42 78 80 25 38

74 46 48 57 30 65 80 26 46 49

Let µ denote the mean average age, and assume that σ 2 is unknown.

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Figure 1

17

Figure 2

You make use of the JMP outputs in Figure 1 and Figure 2.

(i) What assumption(s) is/are necessary in order to conduct the statistical test specified in

(b) below? Are they met? Give a brief discussion. (4)

(ii) Test H0 : µ ≥ 55 against H1 : µ < 55 at the α = 0.05 level of significance. (4)

(iii) How will the test procedure change, if in fact you know that σ = 18? (2)

(iv) Interpret the 95% confidence interval for µ. Can you use this interval to confirm your

conclusion in (a)? Justify your answer. (4)

[29]

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∈

QUESTION 4

(a) Suppose that the temperament of people (i.e., how good-or ill-tempered they are) can be

measured by a psychological scale and classified into three distinct groups. A random sample

of 1000 people from a certain nationality was measured and classified by this test and the

results are as follows:

Bad-tempered N1 = 250

Even-tempered N2 = 480

Good-tempered N3 = 270

It is postulated that the population of this nationality is divided into the three temperament

groups in the following proportions:

π1 = 0.20; π2 = 0.50 and π3 = 0.30.

Test this hypothesis at the 5% level of significance. (8)

(b) In a study of human reaction time in response to a certain stimulus, psychologists used two

independent samples. Sample one was a random sample of 11 males between the ages

of 20 and 40 and sample two was a random sample of 13 females in the same age group.

The sample variances of the reaction times were 12m sec2 for the males and 4m sec2 for the

females. Can the psychologists conclude that the reaction times of males are more variable

than the reaction time of females? Use α = 0.05. (7)

(c) An educational psychologist was studying the influence of behavioural modification on nurs-

ery school behaviour. Children were praised for displaying a positive attitude towards a puppy

and were not praised for any negative behaviour (such as pinching or pulling its tail). The

number of negative behaviours for each of 20 children are listed below. [The first score is for

a period of a week before the praise period, and the second score is for the praise period

which duration was also one week].

Child Numbers of negative behaviours Child Numbers of negative behaviours

Before praise After praise Before praise After praise

A 4 2 K 7 4

B 2 2 L 4 2

C 3 1 M 3 2

D 5 1 N 5 2

E 1 0 O 2 1

F 8 3 P 6 3

G 2 3 Q 1 1

H 3 0 R 3 4

I 4 3 S 4 3

J 2 1 T 5 3

19

Test the hypothesis whether there was a significant decrease in negative behaviour? Use

α = 0.05.

Let Yi = After praise − Before praise

(i) Using the output in Figure 3, test the hypothesis whether there was a significant de-

crease in negative behaviour, that is, H0 : µ = 0 against H1 : µ < 0. (4)

(ii) Show that the 95% confidence interval for the difference in the two means is −0.9174 to

−2.3826. (3)

Figure 3

[22]

QUESTION 5

To test the effect of certain additives in petrol, 24 identical engines were randomly divided into four

groups of six each, Each engine was then filled with 1.0 litre of petrol and 0.1 litre of additive. As

a control, group 1 received 1,1 litre of petrol only. All the engines were switched on and the total

running times were recorded in hours. The following results were obtained:

Group 1 (Control) 1.1 1.2 1.0 1.3 1.2 1.1

Group 2 (Additive A) 0.9 0.8 0.85 0.9 0.95 1.0

Group 3 (Additive B) 0.8 0.9 1.1 1.0 0.8 1.0

Group 4 (Additive C) 1.2 1.1 1.2 1.3 1.1 1.2

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Consider the data as random samples from n(µi ; σ

2)

- populations for i = 1, ..., 4.

Study the following JMP output and answer the questions given below:

Figure 4

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Figure 5

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Figure 6

(a) Use Bartlett’s test to determine if the four groups have equal population variances? Use

α = 0.05. (3)

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(b) Do these results indicate that the engines gave the same result at the 5% level of signif-

icance?

Justify your answer by giving attention to the following detail:

(i) State the appropriate null and alternative hypothesis for this test.

(ii) What test statistic is used to test these hypotheses?

(iii) What is the value of the test statistic? (4)

(c) Discuss the results of the multiple comparisons in Figure 6 on all pairs. (5)

[12]

QUESTION 6

The following table gives the rate (Y ) of flow of blood through the kidney for individuals of X years

of age:

Individual Age (X) Rate of flow (Y )

1 40 467

2 40 573

3 45 430

4 45 476

5 50 466

6 50 375

7 55 352

8 55 426

9 60 340

10 60 405

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The following output was obtained:

Figure 7

(a) What are the estimates of β0 and β1? Hence, give the least squares regression line. (3)

(b) Test for the significance of the slope, β1 at the 5% level of significance.

Justify your answer by giving attention to the following detail

(i) State the appropriate null and alternative hypothesis for this test.

(ii) What test statistic is used to test these hypotheses?

(iii) What is the value of the test statistic? (4)

25

(c) Derive a 95% confidence interval for the slope β1. (4)

(d) Find the predicted rate of flow of blood in the kidney of a 70-year-old person. (1)

[12]

[100]

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Formulae / Formules

B1=

1n

n∑i=1

(X i − X)3

[ 1n

n∑i=1

(X i − X)2]32

B2=

1n

n∑i=1

(X i − X)4

[ 1n

n∑i=1

(X i − X)2]2

A=

1n

n∑i=1

∣∣X i − X∣∣√

1n

n∑i=1

(X i − X)2

ρ =eη − e−η

eη + e−η

T =√

n − 2U11 −U22

2

√U11U22 −U 2

12

T =(X1 − X2)− (µ1 − µ2)

S

√1n1+ 1

n2

υ =

[S2

1

n1+

S22

n2

]2

S41

n21(n1−1)

+S4

2

n22(n2−1)

F =

nk∑

i=1

(X i − X)2/(k − 1)

k∑i=1

n∑j=1

(X i j − X i )2/(kn − k)

β1 =

n∑i=1

Yi (X i − X)

d2Note: d2 =

n∑i=1

(X i − X)2 and β0 =

n∑i=1

Yi − β1

n∑i=1

X i

n= Y − β1X

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Table A. Percentage points for the distribution of B1

Lower percentage point = − (tabulated upper percentage point)

Size of sample Percentage points Size of sample Percentage points

n 5% n 5%

25 0, 711 200 0, 280

30 0, 662 250 0, 251

35 0, 621 300 0, 230

40 0, 587 350 0, 213

45 0, 558 400 0, 200

50 0, 534 450 0, 188

500 0, 179

60 0, 492 550 0, 171

70 0, 459 600 0, 163

80 0, 432 650 0, 157

90 0, 409 700 0, 151

100 0, 389 750 0, 146

800 0, 142

125 0, 350 850 0, 138

150 0, 321 900 0, 134

175 0, 298 950 0, 130

200 0, 280 1000 0, 127

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Table B. Percentage points of the distribution of B2

Size of Percentage points

sample n Upper 5% Lower 5%

50 3, 99 2, 15

75 3, 87 2, 27

100 3, 77 2, 35

125 3, 71 2, 40

150 3, 65 2, 45

200 3, 57 2, 51

250 3, 52 2, 55

300 3, 47 2, 59

350 3, 44 2, 62

400 3, 41 2, 64

450 3, 39 2, 66

500 3, 37 2, 67

550 3, 35 2, 69

600 3, 34 2, 70

650 3, 33 2, 71

700 3, 31 2, 72

800 3, 29 2, 74

900 3, 28 2, 75

1000 3, 26 2, 76

37

Table C. Percentage points for the distribution of A =mean deviation

standard deviation

Size of Percentage points

sample n n − 1 Upper 5% Upper 10% Lower 10% Lower 5%

11 10 0,9073 0,8899 0,7409 0,7153

16 15 0,8884 0,8733 0,7452 0,7236

21 20 0,8768 0,8631 0,7495 0,7304

26 25 0,8686 0,8570 0,7530 0,7360

31 30 0,8625 0,8511 0,7559 0,7404

36 35 0,8578 0,8468 0,7583 0,7440

41 40 0,8540 0,8436 0,7604 0,7470

46 45 0,8508 0,8409 0,7621 0,7496

51 50 0,8481 0,8385 0,7636 0,7518

61 60 0,8434 0,8349 0,7662 0,7554

71 70 0,8403 0,8321 0,7683 0,7583

81 80 0,8376 0,8298 0,7700 0,7607

91 90 0,8353 0,8279 0,7714 0,7626

101 100 0,8344 0,8264 0,7726 0,7644

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Table D

Tabel D

The hypergeometric probability distribution: P (X ≤ x) for N = 12

Die hipergeometriese verdeling: P (X ≤ x) vir N = 12

n k x P n k x P n k x P

1 1 0 0,917 4 4 0 0,141 6 2 0 0,227

1 1 1 1,000 4 4 1 0,594 6 2 1 0,773

4 4 2 0,933 6 2 2 1,000

2 1 0 0,833 4 4 3 0,998

2 1 1 1,000 4 4 4 1,000 6 3 0 0,091

6 3 1 0,500

2 2 0 0,682 5 1 0 0,583 6 3 2 0,909

2 2 1 0,985 5 1 1 1,000 6 3 3 1,000

2 2 2 1,000

5 2 0 0,318 6 4 0 0,030

3 1 0 0,750 5 2 1 0,848 6 4 1 0,273

3 1 1 1,000 5 2 2 1,000 6 4 2 0,727

6 4 3 0,970

3 2 0 0,545 5 3 0 0,159 6 4 4 1,000

3 2 1 0,955 5 3 1 0,636

3 2 2 1,000 5 3 2 0,955 6 5 0 0,008

5 3 3 1,000 6 5 1 0,121

3 3 0 0,382 6 5 2 0,500

3 3 1 0,873 5 4 0 0,071 6 5 3 0,879

3 3 2 0,995 5 4 1 0,424 6 5 4 0,992

3 3 3 1,000 5 4 2 0,848 6 5 5 1,000

5 4 3 0,990

4 1 0 0,667 5 4 4 1,000 6 6 0 0,001

4 1 1 1,000 6 6 1 0,040

5 5 0 0,027 6 6 2 0,284

4 2 0 0,424 5 5 1 0,247 6 6 3 0,716

4 2 1 0,909 5 5 2 0,689 6 6 4 0,960

4 2 2 1,000 5 5 3 0,955 6 6 5 0,999

5 5 4 0,999 6 6 6 1,000

4 3 0 0,255 5 5 5 1,000

4 3 1 0,764

4 3 2 0,982 6 1 0 0,500

4 3 3 1,000 6 1 1 1,000

39

Table E

Upper 5% percentage points of the ratio, S2max/S

2min

v k = 2 3 4 5 6

2 39, 0 87, 5 142 202 266

3 15, 4 27, 8 39, 2 50, 7 62, 04 9, 60 15, 5 20, 6 25, 2 29, 55 7, 15 10, 8 13, 7 16, 3 18, 7

6 5, 82 8, 38 10, 4 12, 1 13, 77 4, 99 6, 94 8, 44 9, 70 10, 88 4, 43 6, 00 7, 18 8, 12 9, 03

9 4, 03 5, 34 6, 31 7, 11 7, 80

10 3, 72 4, 85 5, 67 6, 34 6, 92

12 3, 28 4, 16 4, 79 5, 30 5, 72

15 2, 86 3, 54 4, 01 4, 37 4, 68

20 2, 46 2, 95 3, 29 3, 54 3, 76

30 2, 07 2, 40 2, 61 2, 78 2, 91

60 1, 67 1, 85 1, 96 2, 04 2, 11

∞ 1, 00 1, 00 1, 00 1, 00 1, 00

k = number of samples

v = degrees of freedom for each sample variance

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Table F:

100× (power) of the two-sided t-test with level α

φ 6 7 8 9 10 12 15 20 30 60 ∞ v = degrees of freedom

1.2 30 31 32 33 34 35 36 37 38 39 40

1.3 35 36 37 38 39 40 41 42 43 44 45

1.4 39 40 41 42 43 45 46 47 49 50 51

1.5 43 45 46 47 48 50 51 52 54 55 56

1.6 48 50 52 53 54 55 57 58 59 61 62

1.7 52 55 57 58 59 60 62 64 65 66 67

1.8 57 60 62 63 64 65 67 69 70 71 72

1.9 62 64 65 67 68 69 71 73 74 76 77

2.0 66 68 70 71 72 74 75 77 78 80 81

2.1 70 72 74 75 77 78 79 81 82 83 85

2.2 74 76 78 79 80 81 83 84 86 87 88

2.3 77 80 81 83 84 85 86 87 88 89 90

2.4 81 83 85 86 87 88 89 90 91 92 93

2.5 84 86 87 88 89 90 91 92 93 94 94

2.6 86 88 90 91 91 92 93 94 95 95 96

2.7 89 90 92 93 93 94 95 95 96 96 97

2.8 91 92 93 94 95 95 96 96 97 97 98

2.9 92 94 95 95 96 96 97 97 98 98 98

3.0 94 95 96 96 97 97 98 98 98 99 99

3.1 95 96 97 97 98 98 98 99 99 · ·3.2 96 97 98 98 98 99 99 · · · ·3.3 97 98 98 99 99 · · · · · ·3.4 98 98 99 · · · · · · · ·3.5 98 99 · · · · · · · · ·

α = 0.05

41

Table F (continued):

100× (power) of the two-sided t-test with level α

φ 6 7 8 9 10 12 15 20 30 60 ∞ v = degrees of freedom

2.0 31 33 37 40 42 45 48 50 54 57 60

2.2 39 42 46 49 51 54 58 61 64 67 70

2.4 47 51 55 58 60 63 67 70 74 77 80

2.6 55 60 63 67 69 72 76 79 82 85 87

2.8 62 68 71 74 77 80 83 86 88 90 92

3.0 69 75 78 81 83 86 89 91 92 94 95

3.2 75 81 84 87 88 90 93 94 96 97 97

3.4 81 86 88 91 92 94 95 97 98 98 99

3.6 86 90 92 94 95 96 97 98 99 99 ·3.8 90 93 95 96 97 98 99 99 · · ·4.0 93 95 97 98 98 99 · · · · ·4.2 95 97 98 99 99 · · · · · ·4.4 96 98 99 · · · · · · · ·4.6 97 99 · · · · · · · · ·4.8 98 · · · · · · · · · ·5.0 99 · · · · · · · · · ·

α = 0.01

42