9-40

Section 9-7 9-86 Expected Frequency is found by using the Poisson distribution

!

)(x

exXPxλλ−

== where 41.1100/)]4(4)11(3)31(2)30(1)24(0[ =++++=λ

Value 0 1 2 3 4 Observed Frequency 24 30 31 11 4 Expected Frequency 30.12 36.14 21.69 8.67 2.60

Since value 4 has an expected frequency less than 3, combine this category with the previous category:

Value 0 1 2 3-4 Observed Frequency 24 30 31 15 Expected Frequency 30.12 36.14 21.69 11.67

The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for X. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is

( )∑=

−=

k

i i

ii

EEO

1

220χ

6) Reject H0 if χ χo2

0 05 32 7 81> =. , .

7) ( ) ( ) ( ) ( ) 23.7

67.1167.1115

69.2169.2131

14.3614.3630

12.3012.3024 2222

20 =

−+

−+

−+

−=χ

8) Since 7.23 < 7.81 do not reject H0. We are unable to reject the null hypothesis that the distribution of

X is Poisson.

b) The P-value is between 0.05 and 0.1 using Table IV. From Minitab the P-value = 0.0649.

9-41

9-87 Expected Frequency is found by using the Poisson distribution

!

)(x

exXPxλλ−

== where 907.475/)]9(8)10(7)11(2)1(1[ =++++= …λ

Estimated mean = 4.907

Value 1 2 3 4 5 6 7 8 Observed Frequency 1 11 8 13 11 12 10 9 Expected Frequency 2.7214 6.6770 10.9213 13.3977 13.1485 10.7533 7.5381 4.6237

Since the first category has an expected frequency less than 3, combine it with the next category:

Value 1-2 3 4 5 6 7 8 Observed Frequency 12 8 13 11 12 10 9 Expected Frequency 9.3984 10.9213 13.3977 13.1485 10.7533 7.5381 4.6237

The degrees of freedom are k − p − 1 = 7 − 1 − 1 = 5 a) 1) The variable of interest is the form of the distribution for the number of flaws. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.01 5) The test statistic is

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if χ χo2

0 01 52 15 09> =. , .

7)

( ) ( )χ0

22 212 9 3984

9 39849 4 6237

4 6237=

−+ +

−=

..

..

6.955

8) Since 6.955 < 15.09 do not reject H0. We are unable to reject the null hypothesis that the distribution of the number of flaws is Poisson. b) P-value = 0.2237 (found using Minitab)

9-42

9-88 Estimated mean = 10.131

Value 5 6 8 9 10 11 12 13 14 15 Rel. Freq 0.067 0.067 0.100 0.133 0.200 0.133 0.133 0.067 0.033 0.067 Observed

(Days) 2 2 3 4 6 4 4 2 1 2

Expected (Days)

1.0626 1.7942 3.2884 3.7016 3.7501 3.4538 2.9159 2.2724 1.6444 1.1106

Since there are several cells with expected frequencies less than 3, the revised table would be:

Value 5-8 9 10 11 12-15 Observed

(Days) 7 4 6 4 9

Expected (Days)

6.1452 3.7016 3.7501 3.4538 7.9433

The degrees of freedom are k − p − 1 = 5 − 1 − 1 = 3

a) 1) The variable of interest is the form of the distribution for the number of calls arriving to a switchboard from noon to 1pm during business days. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if χ χo2

0 05 32 7 81> =. , .

7) ( ) ( ) ( ) ( ) ( ) 1.72

9433.79433.79

4538.34538.34

7501.37501.36

7016.37016.34

1452.61452.67 22222

20 =

−+

−+

−+

−+

−=χ

8) Since 1.72 < 7.81 do not reject H0. We are unable to reject the null hypothesis that the distribution for the number of calls is Poisson.

b) The P-value is between 0.9 and 0.5 using Table IV. P-value = 0.6325 (found using Minitab)

9-43

9-89 Use the binomial distribution to get the expected frequencies with the mean = np = 6(0.25) = 1.5

Value 0 1 2 3 4 Observed 4 21 10 13 2 Expected 8.8989 17.7979 14.8315 6.5918 1.6479

The expected frequency for value 4 is less than 3. Combine this cell with value 3:

Value 0 1 2 3-4 Observed 4 21 10 15 Expected 8.8989 17.7979 14.8315 8.2397

The degrees of freedom are k − p − 1 = 4 − 0 − 1 = 3 a) 1) The variable of interest is the form of the distribution for the random variable X. 2) H0: The form of the distribution is binomial with n = 6 and p = 0.25 3) H1: The form of the distribution is not binomial with n = 6 and p = 0.25 4) α = 0.05 5) The test statistic is

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if χ χo2

0 05 32 7 81> =. , .

7)

( ) ( )χ0

22 24 8 8989

8 898915 8 2397

8 2397=

−+ +

−=

..

..

10.39

8) Since 10.39 > 7.81 reject H0. We can conclude that the distribution is not binomial with n = 6 and p = 0.25 at α = 0.05. b) P-value = 0.0155 (from Minitab) 9-90 The value of p must be estimated. Let the estimate be denoted by psample

sample mean =+ + +

=0 39 1 23 2 12 3 1

750 6667( ) ( ) ( ) ( ) .

02778.0246667.0ˆ ===

nmeansamplepsample

Value 0 1 2 3

Observed 39 23 12 1 Expected 38.1426 26.1571 8.5952 1.8010

Since value 3 has an expected frequency less than 3, combine this category with that of value 2:

Value 0 1 2-3 Observed 39 23 13 Expected 38.1426 26.1571 10.3962

The degrees of freedom are k − p − 1 = 3 − 1 − 1 = 1 a) 1) The variable of interest is the form of the distribution for the number of underfilled cartons, X. 2) H0: The form of the distribution is binomial 3) H1: The form of the distribution is not binomial 4) α = 0.05 5) The test statistic is

9-44

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if χ χo2

0 05 12 384> =. , .

7) ( ) ( )χ0

22 2 239 381426

38142623 261571

26157113 10 3962

10 39=

−+

−+

−=

..

( . ).

..

1.053

8) Since 1.053 < 3.84 do not reject H0. We are unable to reject the null hypothesis that the distribution of the number of underfilled cartons is binomial at α = 0.05. b) The P-value is between 0.5 and 0.1 using Table IV. From Minitab the P-value = 0.3048. 9-91 Estimated mean = 49.6741 use Poisson distribution with λ=49.674 All expected frequencies are greater than 3. The degrees of freedom are k − p − 1 = 26 − 1 − 1 = 24 a) 1) The variable of interest is the form of the distribution for the number of cars passing through the intersection. 2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if χ χo2

0 05 242 36 42> =. , .

7) Estimated mean = 49.6741 χ0

2 769 57= . 8) Since 769.57 >>> 36.42, reject H0. We can conclude that the distribution is not Poisson at α = 0.05. b) P-value = 0 (found using Minitab) 9-92 The Expected Frequency is found by using the Poisson distribution

!

)(x

exXPxλλ−

== where 514.19105/)]1(41)1(39)1(7)1(6[ =++++= …λ

Estimated mean = 19.514

Number of Earthquakes

Observed Frequency

Expected Frequency

6 1 0.03 7 1 0.08 8 4 0.18

10 3 0.78 11 2 1.38 12 2 2.24 13 6 3.36 14 5 4.69 15 11 6.10 16 7 7.43 17 3 8.53 18 8 9.25 19 4 9.50 20 4 9.27

9-45

21 7 8.62 22 8 7.64 23 4 6.48 24 3 5.27 25 2 4.12 26 4 3.09 27 4 2.23 28 1 1.56 29 1 1.05 30 1 0.68 31 1 0.43 32 2 0.26 34 1 0.09 35 1 0.05 36 2 0.03 39 1 0.00 41 1 0.00

After combining categories with frequencies less than 3, we get the following table:

Number of Earthquakes

Observed Frequency

Expected Frequency

6-7-8-10-11-12 13.00 4.68

13 6 3.3614 5 4.6915 11 6.1016 7 7.4317 3 8.5318 8 9.2519 4 9.5020 4 9.2721 7 8.6222 8 7.6423 4 6.4824 3 5.2725 2 4.1226 4 3.09

27-28-29-30-31-32-34-35-36-39-41 16 6.38

The degrees of freedom are k − p − 1 = 16 − 1 − 1 = 14

9-46

a) 1) The variable of interest is the form of the distribution for the number of earthquakes per year of magnitude 7.0 and greater..

2) H0: The form of the distribution is Poisson 3) H1: The form of the distribution is not Poisson 4) α = 0.05 5) The test statistic is

( )χ0

22

1=

−

=∑

O EE

i i

ii

k

6) Reject H0 if 68.23214,05.0

2 => χχo

7)

( ) ( ) 48.89

38.638.616

86.486.413 22

20 =

−++

−=χ

8) Since 48.89 > 23.68 reject H0. The form of the distribution of the number of earthquakes is not Poisson.

b) P-value < 0.005

Section 9-8 9-93 1. The variable of interest is breakdowns among shift. 2. H0: Breakdowns are independent of shift. 3. H1: Breakdowns are not independent of shift.

4. α = 0.05 5. The test statistic is:

( )∑∑

= =

−=

r

i

c

j ij

ijij

EEO

1 1

220χ

6. The critical value is 592.122

6,05. =χ

7. The calculated test statistic is 65.112

0 =χ

8. χ χ02

0 05 62/> . , , do not reject H0 and conclude that the data provide insufficient evidence to claim that

machine breakdown and shift are dependent at α = 0.05. P-value = 0.070 (using Minitab)

9-94 1. The variable of interest is calls by surgical-medical patients. 2. H0:Calls by surgical-medical patients are independent of Medicare status. 3. H1:Calls by surgical-medical patients are not independent of Medicare status.

4. α = 0.01 5. The test statistic is:

( )∑∑

= =

−=

r

i

c

j ij

ijij

EEO

1 1

220χ

6. The critical value is 637.62

1,01. =χ

7. The calculated test statistic is 033.02

0 =χ

8. 21,01.0

20 χχ >/ , do not reject H0 and conclude that the evidence is not sufficient to claim that surgical-

medical patients and Medicare status are dependent. P-value = 0.85