The shapes of the distribution of the given sample.

Question

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Answer 1

Question

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

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Answer 2

Question

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

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Answer 3

Question

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

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Answer 4

Question

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

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Answer 5

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

Answer 6

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Answer 7

Chapter 18, Problem 18.26E

(a)

To determine

To explain:

The shapes of the distribution of the given sample.

Answer 8

(a)

Expert Solution

Answer to Problem 18.26E

$S=17.84$

Answer 9

(a)

Expert Solution

Answer 10

(a)

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Answer 13

(b)

To determine

To explain:

The appropriate test of the given sample.

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

Answer 14

(b)

To determine

To explain:

The appropriate test of the given sample.

Answer 15

(b)

Expert Solution

Answer to Problem 18.26E

$25.0258$

Answer 16

(b)

Expert Solution

Answer 17

(b)

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

Answer 20

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

Answer 21

(c)

To determine

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.

Answer 22

(c)

Expert Solution

Answer to Problem 18.26E

$(−0.2721,−0.1023)$

Answer 23

(c)

Expert Solution

Answer 24

(c)

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Given:

$n1=25,s1=14n2=25,s2=21$

Concept used:

Formula

Standard error

$s12n1+s22n2$

Degree of freedom

$=[s12n1+s22n2]2(s12n1)2n1−1+(s12n1)2n2−1$

Significance level

$α=1−confidence$

Calculation:

To test is significance evidence of a different in elimination half-life depending

$H0:μ1=μ2H0:μ1≠μ2α=0.05$

The $95%$ confidence interval for difference

$t=x1¯−x2¯s12n1+s22n2S2=(n1−1)s12+(n2−1)s22n1+n2−2S2=(25−1)(14)2+(25−1)(21)225+25−2S2=1528848S2=318.5S=17.84$

Two sample T-test obese, healthy weight two sample T for obese or healthy weight and standard error $=25.0258$

Since p-value $=0.000$ is less than $0.05$

So, the null hypothesis are rejected

The $95%$ confidence interval for difference in mean plasma testosterone level between adolescent males

A healthy weight is

$(−0.2721,−0.1023)$

Answer 27

Ch. 18 - Prob. 18.1AYK Ch. 18 - Prob. 18.2AYK Ch. 18 - Prob. 18.3AYK Ch. 18 - Prob. 18.4AYK Ch. 18 - Prob. 18.5AYK Ch. 18 - Prob. 18.6AYK Ch. 18 - Prob. 18.7AYK Ch. 18 - Prob. 18.8AYK Ch. 18 - Prob. 18.9AYK Ch. 18 - Prob. 18.10AYK

The shapes of the distribution of the given sample.

To explain: The shapes of the distribution of the given sample.

Answer to Problem 18.26E

Explanation of Solution

To explain: The appropriate test of the given sample.

Answer to Problem 18.26E

Explanation of Solution

To explain: The difference in mean plasma testosterone level between adolescent males of the given sample.

Answer to Problem 18.26E

Explanation of Solution

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Chapter 18 Solutions

To explain:

The shapes of the distribution of the given sample.

To explain:

The appropriate test of the given sample.

To explain:

The difference in mean plasma testosterone level between adolescent males of the given sample.