To construct: The stemplot for Diabetic Mice and Normal Mice. To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Question

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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 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

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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 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

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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

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 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

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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 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

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 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

Answer 6

Chapter 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

Answer 7

Chapter 24, Problem 24.58SE

(a)

To determine

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Answer 8

(a)

Expert Solution

Answer to Problem 24.58SE

Stemplot for Diabetic Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 1

Stemplot for Normal Mice:

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 2

Yes, it appears that potentials in the two groups differ in a systematic way.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info:

The data shows the measured the difference in electrical potential between the Diabetic Mice and Normal Mice.

Calculation:

Stemplot for Diabetic Mice:

Software procedure:

Step by step procedure to construct Stemplot for Diabetic Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Diabetic Mice.
Click OK.

Observation:

From Stemplot for Diabetic Mice, it is clear that the left side of the stemplot extended larger than the right side of the stemplot. Hence, the distribution of the Diabetic Mice is left-skewed with one low outlier.

Stemplot for Normal Mice:

Software procedure:

Step by step procedure to construct Stemplot for Normal Mice by using the MINITAB software:

Choose Graph > Stem-and-Leaf.
In Graph variables, enter the column of Normal Mice.
Click OK.

Observation:

From Stemplot for Normal Mice, it is clear that the right side of the stemplot extended same as in the left side of the stemplot. Hence, the distribution of the Normal Mice is symmetric.

Justification:

From the Stemplot for Diabetic Mice and Normal Mice, it can be observed that the distribution of the two groups is symmetric without considering the outliers. Hence, it appears that potentials in the two groups differ in a systematic way.

Answer 13

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Answer 14

(b)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

Answer 15

(b)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 3

From the MINITAB output, the value of the t-statistic is 3.08 and the P-value is 0.004.

CONCLUDE:

The P-value is 0.004 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.004(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Answer 20

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

Answer 21

(c)

To determine

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

Answer 22

(c)

Expert Solution

Answer to Problem 24.58SE

The conclusion is that, there is a significant difference in mean potentials between the two groups.

No, the outlier does not affect the conclusion.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

PLAN:

Check whether or not there is a significant difference in mean potentials between the two groups.

State the test hypotheses.

Null hypothesis:

H0:μ1=μ2

Alternative hypothesis:

Ha:μ1≠μ2

SOLVE:

Test statistic and P-value:

Software procedure:

Step by step procedure to obtain test statistic and P-value without outlier using the MINITAB software:

Choose Stat > Basic Statistics > 2-Sample t.
Choose Sample in different columns.
In First, enter the column of Diabetic Mice.
In Second, enter the column of Normal Mice.
Choose Options.
In Confidence level, enter 95.
In Alternative, select not equal.
Click OK in all the dialogue boxes.

Output using the MINITAB software is given below:

BASIC PRACTICE OF STATISTICS >C<, Chapter 24, Problem 24.58SE , additional homework tip 4

From the MINITAB output, the value of the t-statistic is 3.84 and the P-value is 0.000.

CONCLUDE:

The P-value is 0.000 and the significance level is 0.05.

Here, the P-value is less than the significance level.

That is, 0.000(=P-value)<0.05(=α) .

Therefore, by the rejection rule, it can be concluded that there is evidence to reject H0 at α=0.05 .

Thus, there is a significant difference in mean potentials between the two groups.

Justification:

From the results, it can be observed that the conclusions for inference with outlier and without outlier are same. Hence, the outlier does not affect the conclusion.

Answer 27

Ch. 24 - Prob. 24.1TY Ch. 24 - Prob. 24.2TY Ch. 24 - Prob. 24.3TY Ch. 24 - Prob. 24.4TY Ch. 24 - Prob. 24.5TY Ch. 24 - Prob. 24.6TY Ch. 24 - Prob. 24.7TY Ch. 24 - Prob. 24.8TY Ch. 24 - Prob. 24.9TY Ch. 24 - Prob. 24.10TY

To construct: The stemplot for Diabetic Mice and Normal Mice. To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Videos

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Answer to Problem 24.58SE

Explanation of Solution

To test: Whether there is a significant difference in mean potentials between the two groups or not.

Answer to Problem 24.58SE

Explanation of Solution

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.

Answer to Problem 24.58SE

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions

To construct: The stemplot for Diabetic Mice and Normal Mice. To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Videos

To construct: The stemplot for Diabetic Mice and Normal Mice. To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

Answer to Problem 24.58SE

Explanation of Solution

To test: Whether there is a significant difference in mean potentials between the two groups or not.

Answer to Problem 24.58SE

Explanation of Solution

To test: Whether there is a significant difference in mean potentials between the two groups or not. To check: Whether the outlier affects the conclusion or not.

Answer to Problem 24.58SE

Explanation of Solution

Want to see more full solutions like this?

Chapter 24 Solutions

To construct: The stemplot for Diabetic Mice and Normal Mice.

To check: Whether it appears that potentials in the two groups differ in a systematic way or not.

To test: Whether there is a significant difference in mean potentials between the two groups or not.

To check: Whether the outlier affects the conclusion or not.