Solution for Exercise Confident Interval for median based on sign test

8/3/2019 Solution for Exercise Confident Interval for median based on sign test

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SOLUTION FOR 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. Constructthe possible confidence interval for 95% confidence interval for the population

median.

DEATH DUE SEVERE WEATHER

53 39 39 67 69 4025 33 30 130 94

Step 1: Arrange the data in ordered array

25 30 33 39 39 40 53 67 69 94 130

Step 2: Determine the amount of sample

n = 11

n


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Case 1 : The wider interval and the higher confidence.

Let = 1

When = 1, cumulative probability is 0.0059

Thus, 100[1-2(0.0059)] = 98.82%

Case 2 : Narrow interval and lower confidence

Let = 2


Thus, 100[1-2(0.0328)] = 93.44%

Step 5 Determine the ( )th value in the ordered array as ML and ( )th

value in the ordered array as MU for Case 1 and Case 2

Case 1 :

Let

ML ; ML = 30

MU ; MU = 94

Case 2 :

Let = 2

ML ; ML = 33

MU ; MU = 69

Rank

1 2 3 4 5 6 7 8 9 10 11

Value

25 30 33 39 39 40 53 67 69 94 130

Conclusion;

Case 1: Hence, one can be 98.82% confidence that death due severe weather

per years is between 30 and 94 person, based on sample 11 years.

Case 2: Hence, one can be 93.44% confidence that death due severe weather

per years is between 30 and 94 person, based on sample 11 years.


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QUESTION 2

A game commissioner studies the number of hunting in counties in Western

Pennsylvania. A sample of counties is selected and the numbers 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 1721 11 17 8 15


8 9 10 11 11 13 15 17 17 21

Step 2: Determine the amount of sample

n = 10

n


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Let = 1


Thus, 100[1-2(0.011)] = 97.84%

Case 2 : Narrow interval and lower confidence

Let = 2


Thus, 100[1-2(0.0547)] = 89.06%

Step 5 Determine the ( )th value in the ordered array as ML and ( )th

value in the ordered array as MU for Case 1 and Case 2

Case 1 :

Let

ML ; ML = 9

MU ; MU = 17

Case 2 :

Let = 2

ML ; ML = 10

MU ; MU = 17

Rank

1 2 3 4 5 6 7 8 9 10

Value

8 9 10 11 11 13 15 17 17 21

Conclusion;

Case 1: Hence, one can be 97.84%confidence that hunting accidents in counties

in Western Pennsylvania is between 9 and 17.

Case 2: Hence, one can be 89.06% confidence that hunting accidents in

counties in Western Pennsylvania is between 10 and 17.


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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 area revealed the educational level of heads of household

shown in table below. Find the point estimate, and construct the approximate

95% confidence interval for population median

Educational level(years of school complete) of heads of household residing in mobile homes

13 6 12 12 10 9 11 14 8 7 16 15 8 7 6


Educational level(years of school complete) of heads of household residing in mobile homes

6 6 7 7 8 8 8 10 11 12 12 13 14 15 16

Step 2: Use large sample approximation

Find K+1

Use the large-sample approximation; let us apply the normal approximation

procedure of the data.

For a confidence coefficient of 0.95

z

Ran

k

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15

6 6 7 7 8 8 8 10 11 12 12 13 14 15 16

Step 3: determine value for ML and MU

ML = K+1 = rank 4

The fourth value in the ordered array is ML=7

Let,

MU =


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The fourth value from the right is MU= 13

Step 4: Make the conclusion

Hence, one can be 95% confident that the population median education level of

household living in mobile homes in a certain area is between 7 and 13.

CONFIDENCE INTERVAL FOR MEDIAN BASED ON SIGN TEST

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Solution for Exercise Confident Interval for median based on sign test