Interpret computer output: The following MINITAB output presents a 98% confidence interval. a. Fill in the blanks: We are__ confident that the population mean is between and _ b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval. c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0. d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Question

Want to see more full solutions like this?

Answer 1

Textbook Question

Chapter 7.1, Problem 66E

Interpret computer output: The following MINITAB output presents a 98% confidence interval.

Chapter 7.1, Problem 66E, Interpret computer output: The following MINITAB output presents a 98% confidence interval. a. Fill

a. Fill in the blanks: We are______ confident that the population mean is between____ and _____
b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval.
c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.
d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

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.

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

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

1 No. 2 3 4 Binomial Prob. X n P Answer 5 6 4 7 8 9 10 12345678 8 3 4 2 2552 10 0.7 0.233 0.3 0.132 7 0.6 0.290 20 0.02 0.053 150 1000 0.15 0.035 8 7 10 0.7 0.383 11 9 3 5 0.3 0.132 12 10 4 7 0.6 0.290 13 Poisson Probability 14 X lambda Answer 18 4 19 20 21 22 23 9 15 16 17 3 1234567829 3 2 0.180 2 1.5 0.251 12 10 0.095 5 3 0.101 7 4 0.060 3 2 0.180 2 1.5 0.251 24 10 12 10 0.095

step by step on Microssoft on how to put this in excel and the answers please Find binomial probability if: x = 8, n = 10, p = 0.7 x= 3, n=5, p = 0.3 x = 4, n=7, p = 0.6 Quality Control: A factory produces light bulbs with a 2% defect rate. If a random sample of 20 bulbs is tested, what is the probability that exactly 2 bulbs are defective? (hint: p=2% or 0.02; x =2, n=20; use the same logic for the following problems) Marketing Campaign: A marketing company sends out 1,000 promotional emails. The probability of any email being opened is 0.15. What is the probability that exactly 150 emails will be opened? (hint: total emails or n=1000, x =150) Customer Satisfaction: A survey shows that 70% of customers are satisfied with a new product. Out of 10 randomly selected customers, what is the probability that at least 8 are satisfied? (hint: One of the keyword in this question is “at least 8”, it is not “exactly 8”, the correct formula for this should be = 1- (binom.dist(7, 10, 0.7,…

Kate, Luke, Mary and Nancy are sharing a cake. The cake had previously been divided into four slices (s1, s2, s3 and s4). What is an example of fair division of the cake S1 S2 S3 S4 Kate $4.00 $6.00 $6.00 $4.00 Luke $5.30 $5.00 $5.25 $5.45 Mary $4.25 $4.50 $3.50 $3.75 Nancy $6.00 $4.00 $4.00 $6.00

Answer 2

Textbook Question

Chapter 7.1, Problem 66E

Interpret computer output: The following MINITAB output presents a 98% confidence interval.

Chapter 7.1, Problem 66E, Interpret computer output: The following MINITAB output presents a 98% confidence interval. a. Fill

a. Fill in the blanks: We are______ confident that the population mean is between____ and _____
b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval.
c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.
d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

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.

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

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

Textbook Question

Chapter 7.1, Problem 66E

Interpret computer output: The following MINITAB output presents a 98% confidence interval.

Chapter 7.1, Problem 66E, Interpret computer output: The following MINITAB output presents a 98% confidence interval. a. Fill

a. Fill in the blanks: We are______ confident that the population mean is between____ and _____
b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval.
c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.
d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

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.

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

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

Textbook Question

Chapter 7.1, Problem 66E

Interpret computer output: The following MINITAB output presents a 98% confidence interval.

Chapter 7.1, Problem 66E, Interpret computer output: The following MINITAB output presents a 98% confidence interval. a. Fill

a. Fill in the blanks: We are______ confident that the population mean is between____ and _____
b. Use the appropriate critical value along with the information in the computer output to construct a 95% confidence interval.
c. Find the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.
d. Find the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

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.

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

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

Textbook Question

Answer 6

Textbook Question

Answer 7

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

Answer 8

a.

Expert Solution

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

Answer 9

a.

Expert Solution

Answer 10

a.

Expert Solution

Answer 11

Expert Solution

Answer 12

To determine

Identify for what percent it is confident that the population mean will lie between.

Answer 13

Answer to Problem 66E

We are 98% confident that the population mean is between 0.2133 and 5.1007.

Answer 14

Explanation of Solution

From the display of MINITAB output, the 98% confidence interval for the population mean is $(0.2133,5.1007)$ .

Hence, there is 98% confidence that the population mean will lie between 0.2133 and 5.1007.

Answer 15

b.

Expert Solution

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

Answer 16

b.

Expert Solution

Answer 17

b.

Expert Solution

Answer 18

Expert Solution

Answer 19

To determine

Construct a 95% confidence interval using the appropriate critical value and the information in the output.

Answer 20

Answer to Problem 66E

The 95% confidence interval using the appropriate critical value and the information in the output is $(0.598,4.716)$

Answer 21

Explanation of Solution

Calculation:

Critical value:

If $α$ is a number between 0 and 1 or $0<α<1$ , then

The notation $zα$ represents the z-score with an area $α$ to the right.
The notation $zα2$ represents the z-score with an area $α2$ to the right.

Here the confidence level is provided as 95% that is, the significance level is 0.05.

$zα2=z0.052=z0.025$

For critical value $z0.025$ :

Software procedure:

Step-by-step software procedure to obtain the critical value using MINITAB software is as follows,

Choose Stat > Graphs > Probability Distribution Plot.
Choose View Probability.
From Distribution, choose ‘Normal’ distribution.
Enter the Mean as 0 and Standard deviation as 1.
Click Shaded Area tab.
Choose Probability and right tail for the region of the curve to shade.
Enter the Probability as 0.025.
Click OK in all the dialogue boxes.

Output using MINITAB software is as follows,

Essential Statistics, Chapter 7.1, Problem 66E

Thus, the critical value is 1.96.

Confidence interval:

The interval which is used to estimate the value of a parameter is termed as confidence interval.

$Confidence interval=Point estimate±Margin of error=x¯±zα2σn$

Substitute, $x¯$ as 2.657, $zα2$ as 1.96, $σ$ as 8 and n as 58.

$Confidence interval=2.657±1.96(858)=2.657±1.96(1.0505)=2.657±2.0589=(0.598,4.716)$

Thus, the 95% confidence interval is $(0.598,4.716)$ .

Answer 22

c.

Expert Solution

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

Answer 23

c.

Expert Solution

Answer 24

c.

Expert Solution

Answer 25

Expert Solution

Answer 26

To determine

Obtain the sample size needed so that the 98% confidence interval will have a margin of error of 1.0.

Answer 27

Answer to Problem 66E

The sample size needed so that the 98% confidence interval will have a margin of error of 1.0 is 345.

Answer 28

Explanation of Solution

Calculation:

Sample size:

Assume m as the desired margin of error, $σ$ is the population standard deviation and $zα2$ as the critical value of a confidence interval. The sample size n needed for the confidence interval will have margin of error approximately equal to m is,

$n=(zα2⋅σm)2$

If the value of n obtained from the formula is not a whole number, round it up to the nearest whole number. By rounding up, it can be sure that the margin of error is no greater than the desired value m.

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 98% confidence level is observed as 2.326.

Substitute $zα2$ as 2.326, $σ$ as 8 and m as 1.0 in the formula,

$n=(2.326⋅81.0)2=(18.608)2=346.2577≈347$

Thus, the sample size is 347.

Answer 29

d.

Expert Solution

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,

$n=(1.96⋅81.0)2=(15.68)2=245.8624≈246$

Thus, the sample size is 246.

Answer 30

d.

Expert Solution

Answer 31

d.

Expert Solution

Answer 32

Expert Solution

Answer 33

To determine

Obtain the sample size needed so that the 95% confidence interval will have a margin of error of 1.0.

Answer 34

Answer to Problem 66E

The sample size needed so that the 95% confidence interval will have a margin of error of 1.0 is 246.

Answer 35

Explanation of Solution

Calculation:

From the bottom row of Table A.3: Critical Values for the Student’s t Distribution, the critical value $zα2$ for 95% confidence level is observed as 1.96.

Substitute $zα2$ as 1.96, $σ$ as 8 and m as 1.0 in the formula,