ENME 392 - Homework 8 - Fa13_Solutions.pdf

8/14/2019 ENME 392 - Homework 8 - Fa13_Solutions.pdf

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ENME392 Fall 2013

Homework 8 Solutions

Total number of points: 100

1. Chapter 9: 9.14(5pts)

2. Chapter 9: 9.22 (Calculate the tolerance limit)(5pts)

9.22 n = 50, x =78.3, and s=5.6. Since 0.05t =1.677 with 49 degrees of freedom, the bound of a

lower 95% prediction interval for a single new observation is 78.3-(1.677)*(5.6)*

1 1 / 50 68.91 . So, the interval is (68.91, ) . On the other hand, with 1 95% and

0.01 , the k value for a one sided tolerance limit is 3.125 and the bound is 78.3-(3.125)*(5.6)

= 60.80. So, the tolerance interval is (60.80, ) .



9.24 This time 1 =0.99 and 0.05 with k = 2.269. So, the tolerance limit is 78.3-

(2.269)*(5.6) = 65.59. There is little cause for concern.



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Number of Candies in the Packages

Number Red Blue Yellow Green Orange Brown

0

1

23 1

4 2

5 3 1 1 1

6 4 1 3

7 4 2 3 1 1 2

8 2 5 1 4

9 2 1 2 4 2 2

10 4 4 4 3

11 3 1 2 3

12 1 2 2 3

13 2 1 1 1

14 1 3

15 1 2

16 1 1

17 1

18

19

20

21Total 17 17 17 17 17 17

To compare to the population mean, we need to find the sample average, or

x

E( x ) xf ( x ) . This is obtained by taking each cell value, multiplying by the

corresponding value for Number, and dividing by 17.

x

E( x ) xf ( x ) Expected Value

Number Red Blue Yellow Green Orange Brown

0 0.000 0.000 0.000 0.000 0.000 0.000

1 0.000 0.000 0.000 0.000 0.000 0.000

2 0.000 0.000 0.000 0.000 0.000 0.000


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3 0.000 0.000 0.176 0.000 0.000 0.000

4 0.471 0.000 0.000 0.000 0.000 0.000

5 0.882 0.294 0.294 0.000 0.000 0.294

6 1.412 0.000 0.000 0.353 0.000 1.059

7 1.647 0.824 1.235 0.412 0.412 0.824

8 0.941 0.000 2.353 0.471 0.000 1.882

9 1.059 0.529 1.059 2.118 1.059 1.059

10 0.000 2.353 2.353 2.353 1.765 0.000

11 0.000 1.941 0.647 1.294 1.941 0.000

12 0.000 0.706 0.000 1.412 1.412 2.118

13 0.000 1.529 0.000 0.765 0.765 0.765

14 0.000 0.824 0.000 0.000 2.471 0.000

15 0.000 0.882 0.000 0.000 1.765 0.000

16 0.000 0.000 0.000 0.941 0.000 0.941

17 0.000 1.000 0.000 0.000 0.000 0.000

18 0.000 0.000 0.000 0.000 0.000 0.000

19 0.000 0.000 0.000 0.000 0.000 0.000

20 0.000 0.000 0.000 0.000 0.000 0.000

21 0.000 0.000 0.000 0.000 0.000 0.000 sum

Total =

Expected Value 6.412 10.882 8.118 10.118 11.588 8.941 56.06

Percent 11.4 19.4 14.5 18.0 20.7 15.9

% in Population 13 24 14 16 20 13

# in Population 7.28 13.44 7.84 8.96 11.2 7.28 56

We have added the known population values at the bottom of the table (in blue) from the

M&M web site.

We also need to find the variances.


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12. The following data on the use of the CRC were collected. Sample refers to variousdata collection times, and the counts at that time are given to the right. Is there a difference in

the use of the gym by males and females? (I.e., could the samples come from the samepopulation?) Address this question by making appropriate plots and by applying the techniques

in Chapter 9. Discuss your approach. If it is possible to test assumptions, then do so, and dont

use those assumptions if they are not valid.(15 pts)

Sample Male Female Sample Male Female Sample Male Female

1 54 73 13 236 177 25 194 146

2 33 31 14 28 22 26 82 35

3 86 39 15 185 62 27 192 824 107 46 16 90 48 28 75 45

5 19 31 17 174 95 29 57 35

6 233 89 18 203 109 30 178 132

7 91 27 19 157 85 31 164 101

8 104 54 20 65 29 32 52 55

9 57 36 21 49 55 33 85 24

10 92 11 22 37 67 34 123 63

11 309 259 23 233 61 35 195 96

12 78 21 24 181 62 - - -

This problem should be approached using pairing, since the numbers vary widely (comparesamples 10 and 11, for instance) for different observation times. Start by plotting the data.


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The data for number of males in the CRC is bimodal, indicating two underlying populations, onethat has essentially the same distribution as for the females, centered at 60-90, and a second that

is higher, centered at 210. Interesting. There are some observation periods when there are large

numbers of men in the CRC, way higher than usual, and no corresponding times for women,

except for one occasion.

Next, calculate and plot the differences.

Sample Male Female Difference

1 54 73 -19

2 33 31 2

3 86 39 47

4 107 46 61

5 19 31 -12

6 233 89 144

7 91 27 64

8 104 54 50

9 57 36 2110 92 11 81

11 309 259 50

12 78 21 57

13 236 177 59

14 28 22 6

15 185 62 123

0

2

4

6

8

10

12

14

30 60 90 120 150 180 210 240 270 300 330

Number

Frequency

Male

Female


11/12

16 90 48 42

17 174 95 79

18 203 109 94

19 157 85 72

20 65 29 36

21 49 55 -6

22 37 67 -30

23 233 61 172

24 181 62 119

25 194 146 48

26 82 35 47

27 192 82 110

28 75 45 30

29 57 35 22

30 178 132 4631 164 101 63

32 52 55 -3

33 85 24 61

34 123 63 60

35 195 96 99

average 54.142857

std dev 45.948272

0

2

4

6

8

10

12

-60 -30 0 30 60 90 120 150 180 210 240

Number

Frequency

Difference


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There are, on average, 54 more men than women in the gym, but the standard deviation is large,

and there are times (negative numbers) when there are more women.

Lets find a 95% confidence interval on this difference. We dont know , so we cant use thenormal distribution.

Now, the t-test presumes that the population is normally distributed. The histogram ofdifferences suggests that the differencesmay, in fact, be normally distributed (even though the

population of male attendance is not).

The equation to use is

2 2d d

/ D /

s sd t d t

n n

and the values of the variables are:

n = 35, d = 54.1, sd= 44.9, and twith v = n1 = 34 degrees of freedom

Looking up v = 34, we find that the table only has values at 30 and 40. To get an approximate

value for 34, interpolate.

t0.025,30= 2.042 and t0.025,40= 2.021, sot0.025,34= 2.042 - (2.042 - 2.021)*0.4 = 2.042 - 0.021*0.4 = 2.0420.0084 = 2.034

This far down the table, we are closely approaching the normal distribution, so we could have

used that instead.

2

44.9 44.954 1 2.034 54 1 2.034 54 1 2.034*7.775 54 1 15.8

5.91635

d/

sd t . . . .

n

So we are 95% sure that the mean difference is between 38.35 and 69.9. This difference does

not span zero, so we can be confident that it is real.

Note: This problem illustrates the value of plotting the raw data. It helps us to understand and

interpret the difference. It would be interesting to know which times of day or days of the week

have the large imbalances. The data histogram also shows that we should definitely not haveused an unpaired test, for while the differences may be normally distributed, the population for

the males was not, making a t-test invalid.

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ENME 392 - Homework 8 - Fa13_Solutions.pdf