The ANOVA table for the data in the Exercise 12.5.

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Answer 1

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

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Answer 2

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

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Answer 3

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

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Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 4

Question

Definition Definition Measure of central tendency that is the average of a given data set. The mean value is evaluated as the quotient of the sum of all observations by the sample size. The mean, in contrast to a median, is affected by extreme values. Very large or very small values can distract the mean from the center of the data. Arithmetic mean: The most common type of mean is the arithmetic mean. It is evaluated using the formula: μ = 1 N ∑ i = 1 N x i Other types of means are the geometric mean, logarithmic mean, and harmonic mean. Geometric mean: The nth root of the product of n observations from a data set is defined as the geometric mean of the set: G = x 1 x 2 ... x n n Logarithmic mean: The difference of the natural logarithms of the two numbers, divided by the difference between the numbers is the logarithmic mean of the two numbers. The logarithmic mean is used particularly in heat transfer and mass transfer. ln x 2 − ln x 1 x 2 − x 1 Harmonic mean: The inverse of the arithmetic mean of the inverses of all the numbers in a data set is the harmonic mean of the data. 1 1 x 1 + 1 x 2 + ...

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

Want to see more full solutions like this?

Subscribe now to access step-by-step solutions to millions of textbook problems written by subject matter experts!

Answer 5

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

Answer 6

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.

Answer 7

Chapter 12.1, Problem 26E

To determine

To find: The ANOVA table for the data in the Exercise 12.5.

Answer 8

Expert Solution & Answer

Answer to Problem 26E

Solution: The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	10323
Error	57	325052	5702.67
Total	59	336825.33

The P-value is 0.362749.

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Given: The data are provided in the referred exercise 12.5 of three joysticks to compare the time taken to complete a navigation mission by each joystick, which is also mentioned in the table given below:

Joystick	Mean	s	n
1	279	78	20
2	245	68	20
3	258	80	20

Calculation: The ANOVA table requires the following entries, which are mentioned below in the table along with their formula of calculation:

Sources	Degree of freedom	Sum of squares	Mean sum of square	F
Groups	I−1	∑groupsni(x¯i−x¯)2	SSGDFG	MSG/MSE
Error	N−I	∑groups(ni−1)si2	SSEDFE
Total	N−1	∑obs(xij−x¯)2

Sum of square of group can be calculated by the formula given below:

∑groupsni(x¯i−x¯)2

x¯=279+245+2583=260.67

All the required calculation is provided in the table given below:

n	x¯i	(x¯i−x¯)	(x¯i−x¯)2	ni×(x¯i−x¯)2
20	279	279−260.67=18.33	335.99	6719.78
20	245	245−260.67=−15.67	245.55	4910.98
20	258	258−260.67=−2.67	7.13	142.58

SSG=6719.78+4910.98+142.58=11773.3

Sum of square of error can be calculated by the formula given below:

∑groups(ni−1)si2

All the required calculation is provided in the table given below:

ni−1	si	si2	(ni−1)si2
20−1=19	78	78×78=6084	6084×19=115596
20−1=19	68	68×68=4624	4624×19=87856
20−1=19	80	80×80=6400	6400×19=121600

SSE=115596+87856+121600=325052

Also, the sum of squares of all the observations is calculated by adding the sum of squares of groups and sum of squares of error, which is shown below:

∑obs(xij−x¯)2=∑groupsni(x¯i−x¯)2+∑groups(ni−1)si2=11773.3+325052=336825.33

There are total three groups, so I is equal to 3 and the total number of observations are 60 as there are 20 observations in each group. Therefore, the degree of freedom for the groups, error and total are provided in the table below:

Groups=I−1=3−1=2

Error=N−I=60−3=57

Total=N−1=60−1=59

Mean sum of square of groups can be calculated by the following formula:

MSG=SSGDFG

Substitute the values of SSG and its respective degree of freedom in the above formula as follows:

MSG=SSGDFG=11773.332=5886.67

Mean sum of square of error can be calculated by the formula mentioned below:

MSE=SSEDFE

Substitute the values of SSE and its respective degree of freedom in the above formula as follows:

MSE=SSEDFE=32505257=5702.67

The F-statistic can be calculated by the formula mentioned below:

F=MSGMSE

Substitute the values of MSG and MSE in the formula as follows:

F=MSGMSE=5886.675702.67=1.0323

The required ANOVA table is provided below:

Sources	Degree of freedom	Sum of square	Mean sum of square	F
Groups	2	11773.33	5886.67	1.0323
Error	57	325052	5702.67
Total	59	336825.33

The P-value on the basis of F statistic and degree of freedom can be calculated by using the software Microsoft excel as shown in the snapshot below:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 1

Insert the values of x, degree of freedom1, and degree of freeom2, and press enter. The output is shown below in the snapshot:

EBK INTRODUCTION TO THE PRACTICE OF STA, Chapter 12.1, Problem 26E , additional homework tip 2

The null hypothesis (H0) and alternative hypothesis (Hα) are shown below:

H0: μ1=μ2=μ3

Ha: Means are not all equal.

The P-value 0.362749 is greater than the 0.05 level of significance, therefore the null hypothesis is not rejected.

Interpretation: Since the Null hypothesis is not rejected, so it can be concluded that the mean time of all three joysticks to complete the navigation mission is same.