To construct: Each specified confidence interval and interpret the interval.

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Answer 1

Question

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

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Answer 2

Question

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

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Answer 3

Question

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

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Answer 4

Question

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

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Answer 5

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

Answer 6

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

Answer 7

Chapter 9.5, Problem 20E

To determine

To construct:

Each specified confidence interval and interpret the interval.

Answer 8

Expert Solution & Answer

Answer to Problem 20E

Solution:

The confidence interval is $(0.4126, 2.4637)$ .

Answer 9

Expert Solution & Answer

Answer 10

Expert Solution & Answer

Answer 11

Explanation of Solution

Confidence interval is used to compare two population variances, estimate the value of the ratio or fraction, $σ12σ22$ .

1. If the confidence interval contains 1 that means the two population variances are unequal at a particular level of confidence.

2. If both endpoints of the interval are greater than 1 it concludes that $σ12$ is larger than $σ22$ .

3. If both endpoints of the interval are less than 1 it concludes that $σ12$ is smaller than $σ22$ .

Given Information:

From the given information, Amy’s practice run is $n1=16$ and Veronica’s practice run is $n2=15$ . Their population variances are $s12=2.560$ and $s22=2.519$ at 90% level of confidence.

Formula used:

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

The formula to calculate the point estimate is,

$s12s22$

Where, $s1$ and $s2$ are the sample variances for the respective samples.

For the given level of confidence $c$ , the level of significance is $α=1−c$ .

The right and left endpoints for $F$ -distribution are $α2$ and $1−α2$ respectively. The critical values are $Fα2$ and $F1−α2$ that can be determined in the $F$ -distribution table using the column for $df1=n1−1$ and row for $df2=n2−1$ .

Calculation:

Substitute 2.560 for $s12$ and 2.519 for $s22$ in the point estimate formula as,

$s12s22=2.5602.519=1.0163$

At 90% level of confidence or $c=0.90$ , the level of significance is,

$α=1−0.90=0.10$

The right endpoint for $F$ -distribution is $α2$ . Substitute $0.10$ for $α$ in $α2$ .

$α2=0.102=0.05$

Similarly the left endpoint for $F$ -distribution is $1−α2$ . Substitute $0.10$ for $α$ in $α2$ .

$1−α2=1−0.05=0.95$

The degrees of freedom are,

$df1=n1−1$ …… $(1)$

And,

$df2=n2−1$ …… $(2)$

The sample sizes are given as $n1=16$ and $n2=15$ .

Substitute 16 for $n1$ in equation $(1)$ .

$df1=n1−1=16−1=15$

Substitute 15 for $n2$ in equation $(2)$ .

$df2=n2−1=15−1=14$

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $α2$ , the value of $Fα2$ is $2.4630$ .

For the column of $df1=15$ and the row of $df2=14$ in the $F$ -distribution table under the area of $1−α2$ , the value of $F1−α2$ is $0.4125$ .

The formula to calculate the confidence interval is,

$(s12s22.1Fα2)<σ12σ22<(s12s22.1F(1−α2))$ .

Substitute $1.0163$ for $s12s22$ , 2.4630 for $Fα/2$ and 0.4125 for $F(1−α/2)$ in the above formula of confidence interval,

$Confidence interval=(1.0163×12.4630)<σ12σ22<(1.0163×10.4125)=(0.4126)<σ12σ22<(2.4637)$

Thus, the confidence interval is $(0.4126, 2.4637)$ .

Interpretation:

Since the value 1 is in the interval $(0.4126, 2.4637)$ , the data do not provide evidence at 90% level of confidence that, the population variances of Amy’s run time and Veronika’s run time are unequal.

Answer 12

Ch. 9.1 - Prob. 1E Ch. 9.1 - Prob. 2E Ch. 9.1 - Prob. 3E Ch. 9.1 - Prob. 4E Ch. 9.1 - Prob. 5E Ch. 9.1 - Prob. 6E Ch. 9.1 - Prob. 7E Ch. 9.1 - Prob. 8E Ch. 9.1 - Prob. 9E Ch. 9.1 - Prob. 10E

To construct: Each specified confidence interval and interpret the interval.

To construct: Each specified confidence interval and interpret the interval.

Answer to Problem 20E

Explanation of Solution

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To construct:

Each specified confidence interval and interpret the interval.