EXERCISE 1

Table 9.4 shows the serum and bone magnesium levels of 14 patients are reported by Alfrey et

al.*. Can we conclude from these data that a relationship exist between serum magnesium and bone

magnesium in the sample population?

Serum Mg ( m Eq./L.) Bone Mg ( m Eq./kg ash )

3.60 672

2.85 610

2.80 621

2.70 567

2.60 570

2.55 638

2.55 612

2.45 552

2.25 524

1.80 400

1.45 277

1.35 294

1.40 338

0.90 230

*Alfrey, Allen C.,Nancy L. Miller, and Donald Butkus, “Evaluation Of Body Magnesium Stores,”J.

Lab. Clin. Med.,84 (1974), 153-1

SOLUTION

Step 1 : Hypothesis.

H0 : The serum and bone magnesium level are independent

H1 : There are either direct or inversely relationship between the serum and magnesium

level. (claim)

Serum Mg (Xi)

Bone Mg (Yi)

R(Xi) R(Yi) di = R(Xi) - R(Yi) (di)2

3.60 672 14 14 0 0

2.85 610 13 10 3 9

2.80 621 12 12 0 0

2.70 567 11 8 3 9

2.60 570 10 9 1 1

2.55 638 8.5 13 -4.5 20.25

2.55 612 8.5 11 -2.5 6.25

2.45 552 7 7 0 0

2.25 524 6 6 0 0

1.80 400 5 5 0 0

1.45 277 4 2 2 4

1.35 294 2 3 -1 1

1.40 338 3 4 -1 1

0.90 230 1 1 0 0

= 51.5

Step 2 : Compute Test Value.

Step 3 : Critical value.

From table A.21, when n=14 and α(2) = 0.05

Then, the critical value is 0.538

Step 4 : Decision.

Reject H0 since 0.538 < 0.8868

Step 5 : Conclusion.

There is enough evidence to support the claim that there are either direct or inversely

relationship between the serum and magnesium level.

EXERCISE 2

Ten seventh-grade children randomly selected from a certain public school system were ranked

according to the quality of their home environment and the quality of their performance in

school. The result is shown in table 9.45. Compute rs and determine whether one can conclude

that the two variables are directly related.

Ten seventh-grade children ranked according to quality of home environment and quality

of performance in school.

Child Home environment Performance in school

1 3 1

2 7 9

3 10 8

4 9 10

5 2 3

6 1 4

7 6 5

8 4 2

9 8 6

10 5 7

SOLUTION

Step 1 : Hypothesis.

H0 : The child’s home environment and their performance in school are independent

H1 : there are direct relationship between home environment and the performance in school

of the seventh-grade children ( claim )

ChildHome

environment ( Xi )

Performance in school

( Yi )R(Xi ) R(Yi ) di = R(Xi) -R(Yi) (di)2

1 3 1 3 1 2 4

2 7 9 7 9 -2 4

3 10 8 10 8 2 4

4 9 10 9 10 -1 1

5 2 3 2 3 -1 1

6 1 4 1 4 -3 9

7 6 5 6 5 1 1

8 4 2 4 2 2 4

9 8 6 8 6 2 4

10 5 7 5 7 -2 4

Step 2 : Compute Test Value.

Step 3 : Critical value.

From table A.21, when n= 10 and α(1) = 0.05

Then, the critical value is 0.564

Step 4 : Decision.

Reject H0 since 0.564 < 0.7818

Step 5 : Conclusion.

There is enough evidence to support the claim that there are direct relationship between home

environment and the performance in school of the seventh-grade children.

EXERCISE 3

The Spearman's Rank Correlation Coefficient is used to discover the strength of a link between

two sets of data. This example looks at the strength of the link between the price of a

convenience item (a 50cl bottle of water) and distance from the Contemporary Art Museum

(CAM ) in El Raval, Barcelona. compute rs and determine whether one can conclude that the

two variable are inversly related.

Convinence store Distance from CAM ( m) Price of 50cl bottle (€)

1 50 1.80

2 175 1.20

3 270 2.00

4 375 1.00

5 425 1.00

6 580 1.20

7 710 0.80

8 790 0.60

9 890 1.00

10 980 0.85

SOLUTION.

Step 1: Hypothesis

H0 : The price of a convenience item (a 50cl bottle of water) and distance from the

Contemporary Art Museum (CAM ) are independent

H1 : there are inverse relationship between the price of a convenience item (a 50cl bottle of

water) and distance from the Contemporary Art Museum (CAM ) ( claim )

Convinence store

Distance from CAM

( m) (Xi)

Price of 50cl bottle

(€)(Yi)

R(Xi) R(Yi) di = R(Xi) - R(Yi) (di)2

1 50 1.80 1 9 -8 64

2 175 1.20 2 7.5 -5.5 30.25

3 270 2.00 3 10 -7 49

4 375 1.00 4 5 -1 1

5 425 1.00 5 5 0 0

6 580 1.20 6 7.5 -1.5 2.25

7 710 0.80 7 2 5 25

8 790 0.60 8 1 7 49

9 890 1.00 9 5 4 16

10 980 0.85 10 3 7 49

=

285.5

Step 2 : Compute test value

Step 3 : Critical value.

From table A.21, when n=10 and α(1) = 0.05

Then, the critical value is -0.564

Step 4 : Decision.

Reject H0 since <

Step 5 : Conclusion.

There is enough evidence to support the claim that there is inverse relationship between the

price of a convenience item and distance from the Contemporary Art Museum (CAM).