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
