To find out what population contrast L expresses this comparison.

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 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

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Students have asked these similar questions

In this problem, we consider a Brownian motion (W+) t≥0. We consider a stock model (St)t>0 given (under the measure P) by d.St 0.03 St dt + 0.2 St dwt, with So 2. We assume that the interest rate is r = 0.06. The purpose of this problem is to price an option on this stock (which we name cubic put). This option is European-type, with maturity 3 months (i.e. T = 0.25 years), and payoff given by F = (8-5)+ (a) Write the Stochastic Differential Equation satisfied by (St) under the risk-neutral measure Q. (You don't need to prove it, simply give the answer.) (b) Give the price of a regular European put on (St) with maturity 3 months and strike K = 2. (c) Let X = S. Find the Stochastic Differential Equation satisfied by the process (Xt) under the measure Q. (d) Find an explicit expression for X₁ = S3 under measure Q. (e) Using the results above, find the price of the cubic put option mentioned above. (f) Is the price in (e) the same as in question (b)? (Explain why.)

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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 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

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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 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

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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 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

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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 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Answer 6

Chapter 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Answer 7

Chapter 26, Problem 26.6AYK

(a)

To determine

To find out what population contrast L expresses this comparison.

Answer 8

(a)

Expert Solution

Answer to Problem 26.6AYK

The population contrast Lis (1)(μC)+(−12)(μR)+(−12)(μE) .

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast Lthat expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Answer 13

(b)

To determine

To give the sample contrast that estimates L and its standard error.

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

Answer 14

(b)

To determine

To give the sample contrast that estimates L and its standard error.

Answer 15

(b)

Expert Solution

Answer to Problem 26.6AYK

The sample contrast that estimates Lis −68.59 and its standard erroris 15.859 .

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. Thus, the population contrast L that expresses this comparison can be as:

L=(c1)(μC)+(c2)(μR)+(c3)(μE)=(1)(μC)+(−12)(μR)+(−12)(μE)

Then the sample contrast that estimates L is calculated as:

⇒L^=(1)(x¯C)+(−12)(x¯R)+(−12)(x¯E)=(1)(605.63)+(−12)(657.31)+(−12)(691.13)=−68.59

And the standard error is calculated as:

SE=(54.42)×1219+(−12)216+(−12)215=15.859

Answer 20

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

Answer 21

(c)

To determine

To explain is there good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food and state the hypotheses in terms of the population contrast L and carry out a test.

Answer 22

(c)

Expert Solution

Answer to Problem 26.6AYK

Yes, there is good evidence that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food.

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Now, we will define the hypotheses as:

H0:L=0Ha:L<0

The value of test statistics is as:

t=L^SE=−68.5915.859=−4.32

And P-value is calculated as:

P<0.005

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

P<0.05⇒Reject H0

Thus, we have sufficient evidence to conclude that the mean weight of rats given chow only is lower than the average for the two groups of rats allowed access to cafeteria food

Answer 27

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Answer 28

(d)

To determine

To explain how much smaller is the mean weight of rats give chow only than the average for the two groups of rats allowed access to cafeteria food and give a 95% confidence interval.

Answer 29

(d)

Expert Solution

Answer to Problem 26.6AYK

We are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.

Answer 30

(d)

Expert Solution

Answer 31

(d)

Expert Solution

Answer 32

Expert Solution

Answer 33

Explanation of Solution

In the question, it is given that figure 26.1 gives basic ANOVA output for the study of the effects of diet type on rat body weights described in example. And from part (b) we have,

The sample contrast that estimates L is −68.59 and its standard error is 15.859 .

Thus, the 95% confidence interval can be calculated as:

L^±t*SE=(−68.59)±2.021(15.859)=(−100.64,−36.54)

Thus, we are 95% confident that the mean weight of rats give chow only is smaller than the average for the two groups of rats allowed access to cafeteria food by about −100.64 and−36.54 grams.