Methods of Applied Statistics with Examples - Homework 1 | STAT 420, Assignments of Data Analysis & Statistical Methods

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STAT 420 Spring 2009

Homework

(10 points) (due Friday, January 30, by 3:00 p.m.)

From the textbook:

The salary of junior executives in a large retailing firm is normally distributed with

standard deviation σ = $1,500. If a random sample of 25 junior executives yields

an average salary of $16,400, what is the 95-percent confidence interval for μ, the

average salary of all junior executives.

σ = $1,500, n = 25, X = $16,400.

n

z σ

2

X± α

95% confidence, α = 0.05, z (^) 0.025 = 1.960.

16 , 400 ± 1. 960 ⋅^1 ,^500 16,400 ±±±± 588 ( 15,812 , 16,988 )

Redo Problem 3.5 without assuming that the population standard deviation is

known. You are given s = $1,575.

s = $1,575, n = 25, X = $16,400. ( )

n

t n 1 s

2

X ± α −

95% confidence, α = 0.05, t (^) 0.025 ( 24 ) = 2.064.

16 , 400 ± 2. 064 ⋅^1 ,^575 16,400 ±±±± 650.16 ( 15,749.84 , 17,050.16 )

3.18 (a), (b)

A cereal packaging machine is supposed to turn out boxes that contain 20 oz of cereal on the average. Past experience indicates that the standard deviation of the

content weights of the packages turned out by the machine is σ = 0.25 oz. Boxes

containing less than 19.8 oz of cereal, on the average, are considered unacceptable.

Thus the manufacturer wishes to test H 0 : μ = 20 and H a : μ = 19.8.

Hint: Treat this problem as a left-tailed test since 19.8 < 20. Also assume that the cereal weights are approximately normally distributed.

a) If samples of size n = 20 are drawn from the current production to test the

hypotheses, fund the power function of the test assuming a significance level

of α = 0.025.

Hint: Recall: Power = P ( Reject H 0 | H 0 is not true ) = P ( Reject H 0 | μ = 19.8 ).

Find the rejection region first and state it in terms of the sample mean X.

n = 20. α = 0.025.

Rejection Region:

Z =

n

σ

X − μ 0 < – z

α.^ Z^ =

X − 20 < – 1.96.

X < 20 − 1. 96 ⋅^0.^25 = 19.89.

Power = P ( X < 19.89 | μ = 19.8 ) =





















< −

P Z^19.^8919.^8

= P ( Z < 1.61 ) = 0..

b) If a sample of n = 20 items has mean weight 19.9, what conclusion is drawn about

μ at the 2.5-percent significance level? Compute the P value for this case, and

interpret it.

Z =

X 19. 9 20

σ

− μ 0 = −

n

= – 1.79. P( Z ≤ − 1.79 ) = 0..

Since 0.0367 > 0.025, Accept H 0 at αααα = 0..

If μ = 20, the probability of obtaining a random sample of size n = 20 with the

sample mean at or below 19.9 is 0.0367.

3.30 (a), (b)

Suppose an experiment yields m = 9, n = 12, X 1 = 68, X 2 = 60, s^21 = 25, and

s^22 = 30.

( Assume that the two populations are approximately normally distributed, and that

the overall variances are equal. )

a) Use the two-sample t statistic to test H 0 : μ 2 – μ 1 = 0 against H a : μ 2 – μ 1 ≠ 0

at a significance level of your choice.

( ) ( )

pooled (^) + −

= −^ ⋅ + − ⋅

m n

s m s n^ s = (^ )^ (^ )

+ −

− ⋅ + − ⋅ =

530 ≈ 27.

Test Statistic: t = (^ )^ (^ )





 



 +

− −





 



 +

− −

⋅

=

⋅

27.895^1

pooled

1 2 δ 0

m n

s

x x = 3.435.

Rejection Region:

Reject H 0 if t < – t 0.025 ( m + n – 2 = 19 d.f. ) OR t > t 0.025 ( 19 d.f. )

t 0.025 ( 19 ) = 2.093. Reject H 0 at αααα = 0.05.

OR

3.435 > 2.861 (^) 2 – tailed

t >^ t 0.005 ( 19 ) p-value < 0.005^ ×^ 2 = 0.01.

Reject H 0 at αααα = 0.01.

b) Construct 90- and 95-percent confidence intervals for μ 2 – μ 1.

( ) 



 



− ± ⋅ ⋅^ +

m n

x 2 x 1 t α 2 s pooled^211 m + n – 2 = 19 d.f.

90%: ( ) 



 



− ± ⋅ ⋅^ +

60 68 1. 729 27.895^1

– 8 ±±±± 4.03 ( – 12.03 , – 3.97 )

95%: ( ) 



 



− ± ⋅ ⋅^ +

60 68 2. 093 27.895^1

– 8 ±±±± 4.87 ( – 12.87 , – 3.13 )

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

Not from the textbook:

1. A coffee machine is regulated so that the amount of coffee dispensed is normally

distributed. A random sample of 17 cups had an average of 7.91 ounces per cup and a sample standard deviation of 0.175 ounces. The coffee machine manufacturer claims that the overall average amount of coffee dispensed is at least 8 ounces. Find

the p-value of the appropriate test. That is, find the p-value of the test H 0 : μ ≥ 8

vs. H a : μ < 8.

X = 7.91. s = 0.175. n = 17.

σ is unknown. Test Statistic:

T X^7918

s

μ 0

.

.

n

= − =^ − = −−−− 2..

P-value = left tail. n − 1 = 17 − 1 = 16 degrees of freedom.

… t 0.10 t 0.05 t 0.025 t 0.01 t 0.

16 d.f. …^ 1.337 1.746 2.120 2.583 2.

Area to the right of 2.120 is 0..

P-value = ( Area to the left of T = −−−− 2.120 ) = 0..

2. For a group of 12 men (35-40 years old) who did not exercise regularly, the average

blood cholesterol was 270, and the sample standard deviation was 37. For a group of 7 men (also 35-40 years old) who did exercise regularly, the average blood cholesterol was 220, with the sample standard deviation of 31. We wish to test the claim that regular exercise lowers cholesterol by more than 20 points, on average. What is the

p- value for the test H 0 : μ NE – μ E ≤ 20 vs. H a : μ NE – μ E > 20? ( Assume that

the two populations are approximately normally distributed, and that the overall

variances are equal. )

( ) ( )

pooled (^) + −

s = − ⋅ + − ⋅ = 1225

s pooled= 35

Test Statistic: T = (^ )^ (^ )

X Y

1 2 pooled

δ 0

+

= − −

+

− −

⋅ ⋅

n n

s

= 1.802.

n 1 + n 2 – 2 = 12 + 7 – 2 = 17 d.f.

0.025 < p-value ( right – tailed ) < 0.05 ( p-value ≈ 0.0446 )

_ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _ _

From the textbook:

Prove the algebraic identity

∑ ∑ (^ )^ ∑ (^ )^ ∑ ∑(^ )

= = = = =

− = − + −

J j

ij j

J j

j

J j

ij

n i

Y Y Y Y

n i

Y Y n

1 1

2 1

2 1 1

2

∑ ∑ (^ )^ ∑ ∑(^ )

= = = =

− = − + −

J j

n

i

ij j j

J j

n

i

ij

j j Y Y Y Y Y Y 1 1

2 1 1

2

= ∑ ∑ ( ) ∑ ∑ ( )( ) ∑ ∑( )

= = = = = =

− + − − + −

J j

n

i

j

J j

n

i

ij j j

J j

n

i

ij j

j j j Y Y Y Y Y Y Y Y 1 1

2 1 1 1 1

= ∑ ∑ ( ) ∑ ( ) ∑( ) ∑ ( )

= = = = =

− + − − + −

J j

j j

J j

n

i

j ij j

J j

n

i

Yij Yj Y Y Y Y n Y Y

j j 1

2 1 1 1 1

Since ∑( )

=

−

n j

i

Yij Yj 1

= 0, j = 1, … , J , ∑ ( ) ∑( )

= =

− −

J j

n

i

j ij j

j Y Y Y Y 1 1

2 = 0, and

∑ ∑ (^ )^ ∑ ∑(^ )^ ∑ (^ )

= = = = =

− = − + −

J j

j j

J j

n

i

ij j

J j

n

i

Yij Y Y Y n Y Y

j j 1

2 1 1

2 1 1

OR

( ) 2

1 1 1 1

2 1 1

2 2 1 1

Y Y^2 Y 2 Y Y Y Y 2 Y^ J Y JnY j

n i

ij

J j

n i

ij

J j

n i

ij ij

J j

n i

∑ ∑ ij −^ =∑ ∑^ − + =∑ ∑ − ∑ ∑ +

= = = = = = = =

= ( ) 2

1 1

(^22) 1 1

Y^2 2 Y JnY JnY^ J Y JnY j

n i

ij

J j

n i

∑ ∑ ij −^ + =∑ ∑ −

= = = =

.

( ) 2

1 1

2 1

2 2 1

nY Y^2 n Y 2 Y Y Y nY 2 Y^ J nY JnY j

j

J j

j

J j

j j

J j

∑ j −^ =∑ ^ − + =∑ − ∑ +

= = = =

= ( ) 2

1

2 2 1

nY^2 2 Y n JY JnY^ J nY JnY j

j

J j

∑ j −^ + =∑ −

= =

.

∑ ∑ (^ )^ ∑ ∑ ∑ ∑ ∑ ∑ ∑

= = = = = = = = =

= −^ +



 



− = ^ − +^ J

j

j

J j

n i

j ij

J j

n i

ij

J j

n i

ij ij j j

J j

n i

Yij Yj Y Y Y Y Y Y Y nY 1

2 1 1 1 1

2 1 1

2 2 1 1

= ∑ ∑ ∑ ( ) ∑ ∑ ∑ ∑

= = = = = = =

− + = −

J j

j

J j

n i

ij

J j

j

J j

j j

J j

n i

Yij Y nY nY Y nY 1

2 1 1

2 1

2 1 1 1

Therefore, ∑ ∑ (^ )^ ∑ (^ )^ ∑ ∑(^ )

= = = = =

− = − + −

J j

n i

ij j

J j

j

J j

n i

Yij Y nY Y Y Y 1 1

2 1

2 1 1

A transportation analyst wishes to compare the “degree of support” for a new rapid transit system among the following three groups of people in the Madison, Wisconsin, area: group A, University of Wisconsin faculty; group B, downtown businessmen; and group C, state government personnel. A questionnaire was designed from which a “rapid transit support index” could be computed for each respondent. The questionnaire was given to randomly selected people from each group. The results follow.

Rapid Transit Support Index Group A B C 66 59 70 65 76 84

Stating any assumptions you make, test for differences between the three populations

represented by the three samples using the F test. Specify the null and the alternative

hypotheses. Show the calculations leading to your conclusion in the form of an

ANOVA table. (You should be able to calculate easily all the ANOVA table entries

given the following summary measures: J = 3, n = 6, Y A = 70, YB = 60, YC = 62,

2

s A = 78.8,

2

s B = 169.2,

2

s C = 140.0.)

N = n A + n B + n C = 6 + 6 + 6 = 18.

Y = 6 ⋅^70 +^6 ⋅^60 +^6 ⋅^62 = = 64.

SSB = n A ⋅( YA − Y ) 2 + nB ⋅( YB − Y ) 2 + nC ⋅( YC − Y ) 2

= 6 ⋅ ( 70 – 64 ) 2 + 6 ⋅ ( 60 – 64 ) 2 + 6 ⋅ ( 62 – 64 ) 2 = 216 + 96 + 24 = 336.

MSB =

SSB =

J −

= 168.

SSW = ( nA − 1 ) ⋅ sA^2 +( nB − 1 )⋅ sB^2 +( nC − 1 ) ⋅ sC^2

= 5 ⋅ 78.8 + 5 ⋅ 169.2 + 5 ⋅ 140.0 = 394 + 846 + 700 = 1940.

MSW =

SSW = 1940

N − J

≈ 129.3333.

SSTot = SSB + SSW = 336 + 1940 = 2276.

F =

MSW

MSB ≈ ≈ 1.29897.

ANOVA table:

Source SS DF MS F

Between^336 2 168 1.

Within^1940 15 129.

Total^2276

F 0.05 ( 2, 15 ) = 3.68. Do NOT Reject H 0 : μ A = μ B = μ C at αααα = 0.05.

p-value = [ = FDIST ( 1.29897 , 2 , 15 ) ] = 0.3018.