3) H1: The form of the distribution is not Poisson.

9.7.3 The estimated mean = 49.6741. Based on a Poisson distribution with

= 49.674, the expected frequencies are shown in

the following table. All expected frequencies are greater than 3.

The degrees of freedom are k − p − 1 = 26 − 1 − 1 = 24.

Vechicles per minute

Frequency

Expected

frequency

40 or less

14

277.6847033

41

24

82.66977895

49

185

168.5430665

50

171

167.4445028

51

150

163.091274

52

110

155.7963895

53

102

146.0197251

54

96

134.3221931

a)

1) Interest is the form of the distribution for the number of cars passing through the intersection.

9-22

4) The test statistic is

( )

=

−

=2

2

0

1

.

kii

ii

OE

E

9.7.4 Estimated mean = 10.131

Value

5 or

less

6

7

8

9

10

11

12

13

14

15 or

more

Rel. Freq

0.067

0.067

0

0.100

0.133

0.200

0.133

0.133

0.067

0.033

0.067

Because there are several cells with expected frequencies less than 3, a revised table follows. We note that there are

other reasonable alternatives to combine cells. For example, cell 7 could also be combined with cell 8.

Value

7 or less

8

9

10

11

12 or more

a)

1) Interest is on the form of the distribution for the number of calls arriving to a switchboard from noon to 1pm during

business days.

6)

Applied Statistics and Probability for Engineers, 7th edition 2017

9-23

Section 9-8

9.8.1 1) Interest is on the species distribution.

9.8.2 1) Interest is on the survival distrbution.

9.8.3 1) Interest is on the distribution of breakdowns among shift.

9.8.4 1) Interest is on the distribution of calls by surgical–medical patients.

Applied Statistics and Probability for Engineers, 7th edition 2017

9-24

0

9.8.5 1) Interest is on the distribution of statistics and OR grades.

9.8.6 1) Interest is on the distribution of deflections.

9.8.7 1) Interest is on the distribution of failures of an electronic component.

9.8.8 1) Interest is on the distribution of opinion on core curriculum change.

Applied Statistics and Probability for Engineers, 7th edition 2017

9-25

9.8.9 a)

1) Interest if on the distribution of successes.

7)

22

0 0.05,1

, reject H0 and conclude that the number of successes and the stone size are not independent.

1 2 All

1 55 25 80

Section 9-9

9.9.1 a)

1) The parameter of interest is median titanium content.

Applied Statistics and Probability for Engineers, 7th edition 2017

9-26

b)

1) Parameter of interest is the median titanium content

9.9.2 a)

1) The parameter of interest is the median of pH.

9.9.3 a)

1) The parameter of interest is the mean ball diameter

Applied Statistics and Probability for Engineers, 7th edition 2017

6) Usually zeros are dropped from the ranking and the sample size is reduced. The sum of the positive ranks is

observation

Difference xi − 0.265

Signed Rank

1

0

–

6

0

–

9

0

–

7) Conclusion: because w− = 9 is not less than or equal to the critical value

*

w

, we fail to reject the null

9.9.4 a)

1) Parameter of interest is the median impurity level.

9-28

9.9.5 1) The parameter of interest is the mean dying time of the primer

Observation

Difference xi − 1.5

Sign Rank

1.5

0

–

1.5

0

–

1.5

0

–

9.9.6 1) The parameter of interest is mean hardness

observation

Difference xi − 60

Sign Rank

4

0

−

8

−1

−1

Applied Statistics and Probability for Engineers, 7th edition 2017

Section 9-10

9.10.1 a)

1) The parameter of interest is the mean molecular weight of a raw material from a new supplier,

.

2, 3) The null and alternative hypotheses that must be tested are as follows (δ = 50):

b)

4) The test statistic is

9.10.2 a)

1) The parameter of interest is the mean absorption rate of the new product,

.

2, 3) The null and alternative hypotheses that must be tested are as follows (δ = 0.50):

b)

Test the first hypothesis (H0:

= 18.50 vs. H1:

≤ 18.50):

9-30

4) The test statistic is

4) The test statistic is

9.10.3 a)

1) The parameter of interest is the mean bond strength,

.

9.10.4 a)

Applied Statistics and Probability for Engineers, 7th edition 2017

9-31

b)

4) The test statistic is

Section 9-11

9.11.1

= − + + + =

2

02[ln(0.15) ln(0.83) ... ln(0.13)] 37.40

with 2m = 2(8) = 16 degrees of freedom.

9.11.3 H0:

μ

= 22

9.11.4 H0: σ = 0.2

Supplemental Exercises

9.S5 a) Degrees of freedom = n – 1 = 16 – 1 = 15.

Applied Statistics and Probability for Engineers, 7th edition 2017

9-32

9.S6 a)

===

1.5

SE Mean 0.401

NN

, so n = 14

9.S7 a) Degree of freedom = n – 1 = 25 – 1 = 24.

9.S8 a) The null hypothesis is

= 300 versus

< 300

Applied Statistics and Probability for Engineers, 7th edition 2017

9-33

9.S9

n

Test Statistic

P-value

Conclusion

a)

50

−

= = −

−

0

0.095 0.10 0.12

0.10(1 0.10) / 50

z

0.4522 × 2 = 0.9044

Fail to reject H0

9.S10

=Sample Mean ˆ,p

=−

Sample Variance ˆˆ

(1 )pp

n

Sample Size, n

Sampling Distribution

Sample Mean

Sample Variance

a)

50

Normal

p

−(1 )

50

pp

9.S11 a)

= 0.05

Applied Statistics and Probability for Engineers, 7th edition 2017

9-34

b)

= 0.01

n = 100,

−

= + = − = =

0.01

0.5 0.6 (2.33 2.0) (0.33) 0.6293

0.5(0.5) / 100

z

c)

= 0.05

d)

− − −

=

−

2

/2 0 0

0

(1 ) (1 )

z p p z p p

npp

9.S12 a) Rejecting a null hypothesis provides a stronger conclusion than failing to reject a null hypothesis. Therefore, place

b)

1) The parameter of interest is the mean weld strength,

.

5) Because no value of

is given, we calculate the P-value.

Applied Statistics and Probability for Engineers, 7th edition 2017

7) There is some modest evidence to support the claim that the weld strength exceeds 150 psi. If we used

= 0.01 or

9.S13 Assume the data follow a normal distribution.

1) The parameter of interest is the standard deviation of the concentration,

.

9.S14 Assume the data follow a normal distribution.

a)

1) The parameter of interest is the standard deviation,

.

9.S15 a) In order to use t statistics in hypothesis testing, we need to assume that the underlying distribution is normal.

1) The parameter of interest is the true mean concentration of suspended solids,

.

9-36

6)

=59.87,x

s = 12.50, n = 60

7) Because 6.12 > –1.65, we fail to reject the null hypothesis. There is insufficient evidence to conclude that the true

mean concentration of suspended solids is less than 50 ppm at

= 0.05.

Interval Obs. Frequency. Exp. Frequency.

x 45.50 9 7.5

45.50< x 51.43 5 7.5

The test statistic is

9.S16 Divide the real line under a standard normal distribution into eight intervals with equal probability. These intervals are

Interval Obs. Frequency. Exp. Frequency.

x 5383.307 12 12.5

5383.307 < x 5399.218 14 12.5

The test statistic is

Applied Statistics and Probability for Engineers, 7th edition 2017

9-37

9.S17 a)

1) The parameter of interest is the true mean sugar concentration,

.

7) Because 6.10 > 2.093, reject the null hypothesis. There is sufficient evidence that the true mean sugar

From Table V, the t0 value in absolute value is greater than the value corresponding to 0.0005 with 19 degrees of

freedom. Therefore, 2(0.0005) = 0.001 > P-value.

d) 95% two-sided confidence interval

− +

0.025,19 0.025,19

ss

x t x t

nn

Applied Statistics and Probability for Engineers, 7th edition 2017

9-38

e) The normality plot below indicates that the normality assumption is reasonable.

9.S18

a)

( )

−−

= = = −

−

0

0

00

53 225(0.25) 0.5004

225(0.25)(0.75)

1

x np

z

np p

9.S19 a) In order to use the

2 statistic in hypothesis testing and confidence interval construction, we need to assume that the

underlying distribution is normal.

Applied Statistics and Probability for Engineers, 7th edition 2017

9-39

9.S20 a) In order to use the

2 statistic in hypothesis testing and confidence interval construction, we need to assume that the

underlying distribution is normal.

1) The parameter of interest is the true variance of tissue assay,

2.

9.S21 a)

1) The parameter of interest is the standard deviation,

.

Applied Statistics and Probability for Engineers, 7th edition 2017

## Trusted by Thousands of

Students

Here are what students say about us.

###### Resources

###### Company

Copyright ©2022 All rights reserved. | CoursePaper is not sponsored or endorsed by any college or university.