Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Question

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Answer 1

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

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Answer 2

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

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

Answer 3

Question

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

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

Features Features Normal distribution is characterized by two parameters, mean (µ) and standard deviation (σ). When graphed, the mean represents the center of the bell curve and the graph is perfectly symmetric about the center. The mean, median, and mode are all equal for a normal distribution. The standard deviation measures the data's spread from the center. The higher the standard deviation, the more the data is spread out and the flatter the bell curve looks. Variance is another commonly used measure of the spread of the distribution and is equal to the square of the standard deviation.

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

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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 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

Answer 6

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Answer 7

Chapter 9, Problem 66CE

a.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Answer 8

a.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 520 (=0.52×1,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 520 and in Number of trials, enter 1,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 1

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.489, 0.551).

Answer 13

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

Answer 14

b.

To determine

Find the probability that the Democratic candidate is actually leading with.

Answer 15

b.

Expert Solution

Answer to Problem 66CE

The probability that the Democratic candidate is actually leading with is 0.102.

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

Explanation of Solution

Calculation:

In this case, the sample size (=1,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0158 (=(0.52)×(1−0.52)1,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0158.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0158)=P(z<−1.27)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –1.27.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 2

Thus, the probability that the Democratic candidate is actually leading is 0.102.

Answer 20

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

Answer 21

c.

To determine

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

Answer 22

c.

Expert Solution

Answer to Problem 66CE

The 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

The probability that the Democratic candidate is actually leading is 0.0132.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

Explanation of Solution

Calculation:

In this case, the number of voters favoring the Republican candidate is 1,560 (=0.52×3,000).

Step-by-step procedure to find the 95% confidence interval for the proportion favoring the Republican candidate using MINITAB software:

Choose Stat > Basic Statistics > 1 Proportion.
Choose Summarized data.
In Number of events, enter 1,560 and in Number of trials, enter 3,000.
Check Options, enter Confidence level as 95.0.
Choose not equal in alternative.
Click OK in each dialog box.

Output is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 3

From the output, the 95% confidence interval for the proportion favoring the Republican candidate is (0.502, 0.538).

In this case, the sample size (=3,000) is larger. Therefore, the mean and standard deviation of the sampling distribution using central limit theorem is (p,p(1−p)n).

The mean (p) is 0.52 and the standard deviation is 0.0091 (=(0.52)×(1−0.52)3,000).

Hence, the sampling distribution follows normal with mean of 0.52 and standard deviation of 0.0091.

The probability that the Democratic candidate is actually leading is obtained as follows:

P(x<0.5)=(x−pp(1−p)n<0.5−0.520.0091)=P(z<−2.20)

Step-by-step procedure to find the probability value using MINITAB software:

Choose Graph > Probability Distribution Plot >View Probability > OK.
From Distribution, choose ‘Normal’ distribution.
Enter Mean as 0 and Standard deviation as 1.
Click the Shaded Area tab.
Choose X Value and Left Tail for the region of the curve to shade.
Enter the data value as –2.20.
Click OK.

Output using MINITAB software is obtained as follows:

STATISTICAL TECHNIQUES-ACCESS ONLY, Chapter 9, Problem 66CE , additional homework tip 4

Thus, the probability that the Democratic candidate is actually leading is 0.0139.

Answer 27

Ch. 9 - Prob. 1SR Ch. 9 - Prob. 1E Ch. 9 - A sample of 81 observations is taken from a normal...Ch. 9 - Prob. 3E Ch. 9 - Suppose you know and you want an 85% confldence...Ch. 9 - Prob. 5E Ch. 9 - Prob. 6E Ch. 9 - Prob. 7E Ch. 9 - Prob. 8E Ch. 9 - Prob. 2SR

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Videos

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Answer to Problem 66CE

Explanation of Solution

Find the probability that the Democratic candidate is actually leading with.

Answer to Problem 66CE

Explanation of Solution

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.

Answer to Problem 66CE

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Videos

Construct a 95% confidence interval for the proportion favoring the Republican candidate.

Answer to Problem 66CE

Explanation of Solution

Find the probability that the Democratic candidate is actually leading with.

Answer to Problem 66CE

Explanation of Solution

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000. Find the probability that the Democratic candidate is actually leading.

Answer to Problem 66CE

Explanation of Solution

Want to see more full solutions like this?

Chapter 9 Solutions

Construct a 95% confidence interval for the proportion favoring the Republican candidate when the number of voters is 3,000.

Find the probability that the Democratic candidate is actually leading.