To outline the design of the experiment.

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 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

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3. Explain why the following statements are not correct. a. "With my methodological approach, I can reduce the Type I error with the given sample information without changing the Type II error." b. "I have already decided how much of the Type I error I am going to allow. A bigger sample will not change either the Type I or Type II error." C. "I can reduce the Type II error by making it difficult to reject the null hypothesis." d. "By making it easy to reject the null hypothesis, I am reducing the Type I error."

Given the following sample data values: 7, 12, 15, 9, 15, 13, 12, 10, 18,12 Find the following: a) Σ x= b) x² = c) x = n d) Median = e) Midrange x = (Enter a whole number) (Enter a whole number) (use one decimal place accuracy) (use one decimal place accuracy) (use one decimal place accuracy) f) the range= g) the variance, s² (Enter a whole number) f) Standard Deviation, s = (use one decimal place accuracy) Use the formula s² ·Σx² -(x)² n(n-1) nΣ x²-(x)² 2 Use the formula s = n(n-1) (use one decimal place accuracy)

Table of hours of television watched per week: 11 15 24 34 36 22 20 30 12 32 24 36 42 36 42 26 37 39 48 35 26 29 27 81276 40 54 47 KARKE 31 35 42 75 35 46 36 42 65 28 54 65 28 23 28 23669 34 43 35 36 16 19 19 28212 Using the data above, construct a frequency table according the following classes: Number of Hours Frequency Relative Frequency 10-19 20-29 |30-39 40-49 50-59 60-69 70-79 80-89 From the frequency table above, find a) the lower class limits b) the upper class limits c) the class width d) the class boundaries Statistics 300 Frequency Tables and Pictures of Data, page 2 Using your frequency table, construct a frequency and a relative frequency histogram labeling both axes.

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 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

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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 24, Problem 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

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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 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

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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 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

Answer 6

Chapter 24, Problem 24.1AYK

a.

To determine

To outline the design of the experiment.

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

Answer 7

Chapter 24, Problem 24.1AYK

a.

To determine

To outline the design of the experiment.

Answer 8

a.

Expert Solution

Explanation of Solution

Given:

The given values are-

n=36P<0.001

Calculation:

Randomly select four groups of fruit flies, then gave each group a different dosage of caffeine (control, 1 mg/ml, 2.5 mg/ml, and 5 mg/ml), then measure the rest time of each fruit fly during a 12-hour dark period.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.

Answer 13

b.

To determine

To find out which means does this test compare, and also state the null and alternative hypothesis for the test in the setting.

Answer 14

b.

Expert Solution

Explanation of Solution

Given:

The following is the given graph-

Practice of Statistics in the Life Sciences, Chapter 24, Problem 24.1AYK

Calculation:

The ANOVA F test compares the means of the four groups. Let be the mean rest time for the control group, be the mean rest time for group of a dosage of 1 mg/ml, be the mean rest time for group of a dosage of 2.5 mg/ml, and be the mean rest time for group of a dosage of 5 mg/ml. The hypotheses are:

H 0: μ1=μ2=μ3=μ4 .

H a: not all μ1,μ2,μ3 and μ4 are equal

The graph shows that as the dosage of caffeine increases, the rest time decreases. The P-value of the ANOVA F test of P < 0.0001 leads a rejection of the null hypothesis. So at least one of the average rest times of the four groups is different.