LEARNING OUTCOMES

By the end of this section, students should be able to :

Find the lower limit, ML and upper limit, MU by using table A.1 for small sample size and using the formula for large sample size and find the confidence interval that the population median lies on.

TWO METHODS FOR CONSTRUCTING INTERVAL ESTIMATES OF AN

UNKNOWN POPULATION

1. Based on sign test.

2. Based on Wilcoxon test.

Thompson and Savur describe this procedure.

David gives the mathematical derivation of the interval.

Hollander and Wolfe give the approximation of large sample.

SMALL SAMPLE SIZE

1. Arrange the order of data given according to the values.

2. Find the cumulatives probabilities from Table A.1 or build your own table.

3. Find K’.

4. Calculate both confidence interval by using 100(1-2α)%.

5. Determine the (K’+1)th value in the ordered array as ML and (n-K’)th value in the ordered array as MU.

6. Make the conclusion.

LARGE SAMPLE APPROXIMATION

1. Arrange the order of data given according to the values.

2. Find K’+1 by using :

3. Determine the (K’+1)th value in the ordered array as ML and (n-K’)th value in the ordered array as MU.

4. Make the conclusion.

42)1'(

2

nz

nK

12n

EXAMPLE 1(SMALL SAMPLE)

Question

In a study of side effects of drug, reported the data shown in table below. Let us construct a 95% confidence interval for the median of the population from which these sample data have been drawn.

Percentage of patients who experienced side

effects from a drug used to lower a person’s

cholesterol level.

4.0 3.0 7.0 5.4 3.8 7.3 10.0 7.0 2.6 1.8

SOLUTION

Step 1

Arrange the data in ordered array.

Step 2

Determine the amount of sample.

Small sample size

1.8 2.6 3.0 3.8 4.0 5.4 7.0 7.3 7.5 10.0

12n10n

Step 3

When we refer to the question, we want to construct a 95% confidence interval. Thus, we need to find the value of K’ as observation value.

Value of K’ will be refer by cumulative of probability in Table A.1 as below;

95%2.5% 2.5%

K’ K’

K’ Probabilities (from table A.1) Cumulative Probabilities

0 0.0010 0.0010

1 0.0098 0.0108

2 0.0439 0.0547

3 0.1172

4 0.2051

5 0.2461

6 0.2051

7 0.1172

8 0.0439 0.0547

9 0.0098 0.0108

10 0.0010 0.0010

0547.0)50.0,102(

0108.0)50.0,101(

KP

KP

Now, we choose K’=1 and K’=2 as observation value. This is because we need to construct the first case and the second case.

Step 4

First case: The wider interval and the higher confidence.

Let, K’=1

When K’=1 cumulative probability is 0.0108.

Thus, 100[1-2(0.0108)]= 97.84% of confidence interval.

The amount of cumulative probability should be less than 0.05 as significant value because of the chosen interval is wider and higher of confidence.

Step 5

Second case: The narrower interval and lower confidence.

Let, K’=2

When K’= 2 cumulative probability is 0.0547.

Thus, 100[1-2(0.0547)]=89.06% of confidence interval.

The amount of cumulative probability should be more than 0.05 as significant value because of the chosen interval is narrower and lower of confidence interval.

K’ Probabilities (from table A.1) Cumulative Probabilities

0 0.0010 0.0010

1 0.0098 0.0108

2 0.0439 0.0547

3 0.1172

4 0.2051

5 0.2461

6 0.2051

7 0.1172

8 0.0439 0.0547

9 0.0098 0.0108

10 0.0010 0.0010

0547.0)50.0,102(

0108.0)50.0,101(

KP

KP

Step 6

Therefore, for the first and second case, we will find the upper and lower limit as well.

Case 1:

K’=1, K’+1=2

Thus, the observation is second observation from each end of the ordered array yields as table below.

Rank Values

1 1.8

2 2.6

3 3.0

4 3.8

5 4.0

6 5.4

7 7.0

8 7.3

9 7.5

10 10.0

2nd observationRank of lower

limit

2nd observationRank of upper

limit

Thus, we can say that we are 97.84 % confident that

the population median is between 2.6 and 7.5

Case 2:

K’=2, K’+1=3

Thus, the observation is third observation from

each end of the ordered array yields as table below.

Rank Value

1 1.8

2 2.6

3 3.0

4 3.8

5 4.0

6 5.4

7 7.0

8 7.3

9 7.5

10 10.0

3th observationRank of lower

limit

3th observationRank of upper

limit

Thus, we can say that we are 89.06%

confident that the population median is between

3.0 and 7.3

EXAMPLE 2(LARGE SAMPLE)

QUESTION

Agnow et(E8)*, in study of sleep patterns, reported the data shown below. Let us construct a 95% confidence interval for the median of the population from which these sample data have been drown.

Percentage of total sleep time in stage 0 sleep

by 16 months mentally and physically healthy males

between ages of 50 and 60

*E8 refer to Stitt, John T., James D.Hardy, and Ethan R. Nadel, “Surface Area of the Squirrel Monkey in Relation to Body Weigth,” J. Appl. Physiol., 31 (1971), 140-141.

1.90 3.08 9.10 3.53 1.99 3.10 10.16 0.69

1.74 2.41 4.01 3.71 8.11 8.23 0.07 3.07

SOLUTION

Step 1

Arrange the data in ordered array

Step 2

Determine the amount of sample.

n=16

In this case n>12 which is large sample of

Approximation.

Formula of Large Sample Approximation

4)2/()1'(2

nznK

0.07 0.69 1.74 1.90 1.99 2.41 3.07 3.08

3.10 3.53 3.71 8.11 8.23 8.23 9.10 10.16

Step 3

Now, we count the fourth observation from each

end of the ordered array.

441696.1)2/16()1'( K

96.12

05.0

2

zz

Rank Values

1 0.07

2 0.69

3 1.74

4 1.90

5 1.99

6 2.41

7 3.07

8 3.08

9 3.10

10 3.53

11 3.71

12 4.01

13 8.11

14 8.23

15 9.10

16 10.16

4th observationRank of lower

limit

4th observationRank of upper

limit

Conclusion

Counting to the fourth observation from each end of the ordered array yields ML = 1.90 and MU =8.11

Hence, one can be 95% confident that the population median is between 1.90 and 8.11

EXERCISE

QUESTION 1

A meteorologist study the number of death per years from tornadoes in the United States. The number of death for a sample of 11 years is shown. Construct the possible confidence interval for 95% confidence interval for the population median.

DEATH DUE TO SEVERE WEATHER

53 39 39 67 69 40

25 33 30 130 94

Answer: 1.Confidence Interval is 93.44%, that the population median is between 33 and 692.Confidence Interval is 98.82%, that the population median is between 30 and 94

QUESTION 2

A game commissioner study the number of hunting in counties in Western Pennsylvania. A sample of counties is selected and the number of hunting accident are shown. Construct the possible confidence interval for 95% confidence interval for the population median.

HUNTING ACCIDENTS IN COUNTIES IN WESTERN PENNSYLVANIA

10 11 9 13 17

21 11 17 8 15

Answer: 1.Confidence Interval is 97.84% that the population median is between 9 and 172.Confidence Interval is 89.06% that the population median is between 10 and 17

QUESTION 3

Abu-Ayyash (E10) found that the median education of heads of household living in mobile homes in a certain area was 11.6 years. Suppose that a similar survey conducted in another revealed the educational level of heads of household shown in Table 2.11. Find the point estimate, and construct 95% confidence interval for population median.

Educational level (years of school completed) of heads

of house residing in mobile homes

13 6 6 12 12 10 9 11

14 8 7 16 15 8 7

Answer: One can be 95% confident that the population median is between 7 and 13 Answer: One can be 95% confident that the population median is between 7 and 13

Please refer the solution for the exercise by using this link

http://appliednonparametricstatistic.blogspot.com/

LINK TO THE VIDEO

http://youtu.be/C4ZvkSWE25M