The two sample standard deviations.

Question

Want to see more full solutions like this?

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 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

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!

Students have asked these similar questions

A well-known company predominantly makes flat pack furniture for students. Variability with the automated machinery means the wood components are cut with a standard deviation in length of 0.45 mm. After they are cut the components are measured. If their length is more than 1.2 mm from the required length, the components are rejected. a) Calculate the percentage of components that get rejected. b) In a manufacturing run of 1000 units, how many are expected to be rejected? c) The company wishes to install more accurate equipment in order to reduce the rejection rate by one-half, using the same ±1.2mm rejection criterion. Calculate the maximum acceptable standard deviation of the new process.

5. Let X and Y be independent random variables and let the superscripts denote symmetrization (recall Sect. 3.6). Show that (X + Y) X+ys.

8. Suppose that the moments of the random variable X are constant, that is, suppose that EX" =c for all n ≥ 1, for some constant c. Find the distribution of X.

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 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

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 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 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Want to see more full solutions like this?

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

Answer 4

Question

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 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

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 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Answer 6

Chapter 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

Answer 7

Chapter 21, Problem 21.26E

(a)

To determine

To find: The two sample standard deviations.

Answer 8

(a)

Expert Solution

Answer to Problem 21.26E

The two sample standard deviations are 0.25 and 0.15.

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given info:

In the given information, nAlcohol=25 , nPlacebo=25 , x¯Alcohol=0.25 , x¯Placebo=0.12 , SEMAlcohol=0.05 , and SEMPlacebo=0.03 .

Calculation:

The sample standard deviation for alcohol is,

SEMAlcohol=sAlcoholnAlcoholsAlcohol=SEMAlcoholnAlcohol=0.05(25)=0.25

The sample standard deviation for placebo is,

SEMPlacebo=sPlacebonPlacebosPlacebo=SEMPlacebonPlacebo=0.03(25)=0.15

Thus, the two sample standard deviations are 0.25 and 0.15.

Answer 13

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

Answer 14

(b)

To determine

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

Answer 15

(b)

Expert Solution

Answer to Problem 21.26E

The degrees of freedom are 24.

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

The degrees of freedom for the two-sample t procedures with the conservative Option 2 is,

df=Smaller of nAlcohol−1 and nPlacebo−1=25−1=24

Thus, the degrees of freedom are 24.

Answer 20

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Answer 21

(c)

To determine

To find: The 90% confidence interval for the mean difference between the two groups.

Answer 22

(c)

Expert Solution

Answer to Problem 21.26E

The 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

Critical value:

From the table of “t Distribution critical values”, the critical value for 24 degrees of freedom with 90% confidence level is 1.711.

Confidence interval:

The 90% confidence interval for the mean difference between the two groups is,

CI=(x¯Alcohol−x¯Placebo)±t*sAlcohol2nAlcohol+sPlacebo2nPlacebo=(x¯Alcohol−x¯Placebo)±t*SEMAlcohol2+SEMPlacebo2=0.25−0.12±1.7110.052+0.032=0.13±0.0998

=(0.13−0.0998,0.13+0.0998)=(0.0302,0.2298)

Thus, the 90% confidence interval for the mean difference between the two groups is (0.0302,0.2298) .

Answer 27

Ch. 21.1 - Prob. 1AYK Ch. 21.1 - Prob. 2AYK Ch. 21.1 - Prob. 3AYK Ch. 21.1 - Prob. 4AYK Ch. 21.3 - Prob. 5AYK Ch. 21.3 - Prob. 6AYK Ch. 21.3 - Prob. 7AYK Ch. 21.4 - Prob. 8AYK Ch. 21.5 - Prob. 9AYK Ch. 21.5 - Prob. 10AYK

The two sample standard deviations.

Concept explainers

Videos

To find: The two sample standard deviations.

Answer to Problem 21.26E

Explanation of Solution

To find: The degrees of freedom by using conservative Option 2 for two-sample t procedures.

Answer to Problem 21.26E

Explanation of Solution

To find: The 90% confidence interval for the mean difference between the two groups.

Answer to Problem 21.26E

Explanation of Solution

Want to see more full solutions like this?

Chapter 21 Solutions