Identify the null hypothesis and alternative hypothesis.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

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

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

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 4

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

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

Answer 5

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

Answer 6

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

Answer 7

Chapter 24, Problem 1E

a.

To determine

Identify the null hypothesis and alternative hypothesis.

Answer 8

a.

Expert Solution

Answer to Problem 1E

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained from at least one of the brand is different from the rest of the brands.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

The given information is that, a student conducted an experiment to identify whether there is any significant difference in the four popcorn brands. The number of kernels which was left unpopped were also recorded. The student measured pops of each brand four times. The F-ratio obtained from the test is 13.56.

Let μ1,μ2,μ3,μ4 be the mean number of unpopped kernels obtained each of the four brands of popcorn.

The null hypothesis is framed by assuming that all the four brands of popcorn results in the same mean number of unpopped kernels.

Null hypothesis:

H0:μ1=μ2=μ3=μ4

The mean number of unpopped kernels obtained from each of the four brands of popcorn is the same.

Alternative hypothesis:

HA:At least one of the μ's is not equal

The mean number of unpopped kernels obtained at least one of the brand is different from the rest of the brands.

Answer 13

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

Answer 14

b.

To determine

Find the degrees of freedom for the treatment sum of squares and the error sum of squares.

Answer 15

b.

Expert Solution

Answer to Problem 1E

The degrees of freedom for the treatment sum of squares is 3.

The degrees of freedom for the error sum of squares is 12.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

Degrees of freedom for treatment sum of squares:

If there are k treatments them the degrees of freedom for the treatment sum of squares is k−1.

Here, there are four brands (treatments), then the degrees of freedom is,

Substitute k as 4

k−1=4−1=3

Thus, the degrees of freedom for the treatment sum of squares is 3.

Degrees of freedom for the error sum of squares:

The error sum of squares for the experiment with N observations and having k treatments is N−k.

There are four brands of treatments and each treatment is applied four times. Hence, there will be 16(=4×4) observations.

Then the degrees of freedom is,

Substitute N as 16 k as 4.

N−k=16−4=12

Thus, the degrees of freedom for the error sum of squares is 12.

Answer 20

c.

To determine

Find the P-value and give conclusion.

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

Answer 21

c.

To determine

Find the P-value and give conclusion.

Answer 22

c.

Expert Solution

Answer to Problem 1E

The P-value is 0.00037.

There is strong evidence to conclude that the mean number of un popped kernels is different for at least one of the four brands of popcorn.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

The given information is that, assume all the required conditions for ANOVA were satisfied.

The P-value is calculated as follows:

Software procedure:

Step by step procedure to find the P-value using MINITBA is given below:

Choose Graph > Probability Distribution Plot > choose View Probability> OK.
From Distribution, choose F.
Enter Numerator df as 3 and Denominator df as 12.
Under Shaded Area tab, select X Value and choose Right Tail.
Enter the X Value as 13.56.
Click OK.

Output obtained from MINITAB is given below:

Intro Stats, Chapter 24, Problem 1E

Conclusion:

The P-value for the F-statistic is 0.00037 and the level of significance is 0.01.

The P-value is less than the level of significance.

That is, 0.00037(=P-value)<0.01(=α).

Thus, there is strong evidence to conclude that the mean number of unpopped kernels is different for at least one of the four brands of popcorn.

Answer 27

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.

Answer 28

d.

To determine

Suggest what else about the data would be useful to check the assumptions and conditions.

Answer 29

d.

Expert Solution

Explanation of Solution

The side by side boxplot of the treatments would be useful to check the similar variance conditions and also the spread of the data.

The normal probability of residuals can be used to check the normality assumption of the error term. If the points in the normal probability plot lies approximately on a straight line it assumed that the residuals are normally distributed.

Also, residual plot is used to identify pattern and spread of the residuals.