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Stevewh, Bachelor's Degree
Category: Calculus and Above
Satisfied Customers: 7610
Experience:  I teach Calculus and Probability in an University since 1994
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# Find the margin of error for the given values of c, s, and

### Customer Question

Find the margin of error for the given values of c, s, and n.
C=0.95, s=2.3, n=49
E=_________
(Round to three decimal places as needed)

Construct the confidence interval for the population mean µ.
C= 0.98, x=7.7, s=0.9, and n=60
A 98% confidence interval for µ is _____, ______ (Round to two decimal places as needed.)

Construct the confidence interval for the population mean µ.
C= 0.95, x=16.9, s=10.0, and n=80
A 95% confidence interval for µ is _____, ______ (Round to two decimal places as needed.)
Use the confidence interval to find the estimated margin of error. Then find the sample mean.
A biologist reports a confidence interval of (3.2,4.4) when estimating the mean height in (in centimeters) of a sample of seedlings. The estimated margin of error is__________

Find the minimum sample size n needed to estimate µ for the given values of c, s, and E.
C=0.95, s=6.7, and E=1
Assume that a preliminary sample has at least 30 members.
N=__________ (Round up to the nearest whole number)

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals or the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 40 home theater systems has a mean price of \$150.00 and a standard deviation of \$17.90.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is (____,____).
(Round to two decimal places as needed.)

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If the convenient, use technology to construct the confidence intervals.
A random sample of 43 gas grills has a mean price of \$633.90 and a standard deviation of \$56.90.
The 90% confidence interval is ( _____,_____). (Round to two decimal places as needed.)

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If the convenient, use technology to construct the confidence intervals.

A random sample of 43 eight-ounce servings of different juice drinks has a mean of 79.6 calories and a standard deviation of 43.9 calories.

The 90% confidence interval is ( _____,_____). (Round to one decimal places as needed.)

People were polled on how many book they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 95% confidence? Initial survey results indicate that σ=11.2 books.

A 95% confidence level requires ______ subjects.

(Round up to the nearest whole number as needed.)

A doctor wants to estimate the HDL cholesterol of all 20-to 29- year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99% confidence assuming σ = 12.8? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires ____ subjects.

(Round up to the nearest whole number as needed.)

Construct the indicated confidence interval for the population mean µ using (a) a t-distribution. (b) If you had incorrectly used a normal distribution, which interval would be wider?
C= 0.95, x=13.2, s=4.0, n=5

The 95% confidence interval using a t-distribution is (____,_____).
(Round to one decimal place as needed.)

In the situation, assume the random variable is normally distributed and use a normal distribution or a t-distribution to construct a 90% confidence interval for the population mean. If convenient, use technology to construct the confidence interval.
In a random sample of 10 adults from a nearby county, the mean waste generated
Per person per day was 4.17 pounds and the standard deviation was 1.87 pounds.
Repeat part (a), assuming the same statistics came from a sample size of 450. Compare the results.

For the sample of 10 adults, the 90% confidence intervals is (____,____).
(Round to two decimal places as needed.)

Use the given confidence interval to find the margin of error and the sample proportion. (0.669,0.697)
E=_____ (Type an integer or a decimal)
P=_____(Type an integer or a decimal)
Submitted: 4 years ago.
Category: Calculus and Above
Expert:  Stevewh replied 4 years ago.
Hi,

You posted about 20 questions here.....

When do you need the answers ?

Steve
Customer: replied 4 years ago.
before 3:00 today
Expert:  Stevewh replied 4 years ago.
Wht is your time now ?

Steve
Customer: replied 4 years ago.
1:52
Expert:  Stevewh replied 4 years ago.
Only the answers i think ?

Steve
Customer: replied 4 years ago.
yes
Expert:  Stevewh replied 4 years ago.
Ok,

I will start now

Steve
Expert:  Stevewh replied 4 years ago.
But stay online, because maybe i have some doubt about the symbols used
Expert:  Stevewh replied 4 years ago.
I have a doubt about the symbol used as "s"

All the "s" are "s" indeed or some s is "σ", that´s important

Steve
Expert:  Stevewh replied 4 years ago.
Are you there ?

I am working on this set but please check the symbols "s"

Steve
Customer: replied 4 years ago.
Thats what they have on the questions s
Customer: replied 4 years ago.
For what question so i can go back to it and look at it
Expert:  Stevewh replied 4 years ago.
For all of them where you find a "s"

Steve
Expert:  Stevewh replied 4 years ago.
I did them but since there are a lot of questions can i post them in 3 sets ?

(3 answers must be accepted )

Thanks

Steve
Customer: replied 4 years ago.
Yes thank you
Expert:  Stevewh replied 4 years ago.
Ok,

First set:

Find the margin of error for the given values of c, s, and n.
C=0.95, s=2.3, n=49
E= 0.644
(Round to three decimal places as needed)

Construct the confidence interval for the population mean µ.
C= 0.98, x=7.7, s=0.9, and n=60
A 98% confidence interval for µ is (7.43, 7.97)(Round to two decimal places as needed.)

Construct the confidence interval for the population mean µ.
C= 0.95, x=16.9, s=10.0, and n=80
A 95% confidence interval for µ is (14.71, 19.09) (Round to two decimal places as needed.)
Use the confidence interval to find the estimated margin of error. Then find the sample mean.
A biologist reports a confidence interval of (3.2,4.4) when estimating the mean height in (in centimeters) of a sample of seedlings. The estimated margin of error is__0.6________

Find the minimum sample size n needed to estimate µ for the given values of c, s, and E.
C=0.95, s=6.7, and E=1
Assume that a preliminary sample has at least 30 members.
N=173 (Round up to the nearest whole number)

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals or the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

Thanks

Steve
Expert:  Stevewh replied 4 years ago.
You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals or the population mean. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

A random sample of 40 home theater systems has a mean price of \$150.00 and a standard deviation of \$17.90.

Construct a 90% confidence interval for the population mean.

The 90% confidence interval is (145.34,154.66).
(Round to two decimal places as needed.)

95% is (144.45,155.55)

A random sample of 43 eight-ounce servings of different juice drinks has a mean of 79.6 calories and a standard deviation of 43.9 calories.

The 90% confidence interval is (68.6 , 90.6). (Round to one decimal places as needed.)

People were polled on how many book they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 95% confidence? Initial survey results indicate that σ=11.2 books.

A 95% confidence level requires 482 subjects.

(Round up to the nearest whole number as needed.)

A doctor wants to estimate the HDL cholesterol of all 20-to 29- year old females. How many subjects are needed to estimate the HDL cholesterol within 2 points with 99% confidence assuming σ = 12.8? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires 272 subjects.

(Round up to the nearest whole number as needed.)

Decreases
Expert:  Stevewh replied 4 years ago.
3rd set

Construct the indicated confidence interval for the population mean µ using (a) a t-distribution. (b) If you had incorrectly used a normal distribution, which interval would be wider?
C= 0.95, x=13.2, s=4.0, n=5

The 95% confidence interval using a t-distribution is (8.2,18.2).
(Round to one decimal place as needed.)

Using z-distribution: (9.7,16.7)

In the situation, assume the random variable is normally distributed and use a normal distribution or a t-distribution to construct a 90% confidence interval for the population mean. If convenient, use technology to construct the confidence interval.
In a random sample of 10 adults from a nearby county, the mean waste generated
Per person per day was 4.17 pounds and the standard deviation was 1.87 pounds.
Repeat part (a), assuming the same statistics came from a sample size of 450. Compare the results.

For the sample of 10 adults, the 90% confidence intervals is (3.09,5.25).
(Round to two decimal places as needed.)

for n = 450: (4.03,4.31)

Use the given confidence interval to find the margin of error and the sample proportion. (0.669,0.697)
E=0.014 (Type an integer or a decimal)
P= 0.683(Type an integer or a decimal)

Sorry if something is missing or wrong, did my best

Thanks

Steve
Customer: replied 4 years ago.
Number 3. Was to be round to one decimal place as needed , i said two my mistake.

Number 7. Was to be round to two decimals place as needed, i said one my mistake.

Number 8. Can you tell the 95% confidence interval is_____,____(Round to one decimal place as needed)

This problem the numbers was a mistake.

A doctor wants to estimate the HDL cholesterol of all 20-to 29- year old females. How many subjects are needed to estimate the HDL cholesterol within 3 points with 99% confidence assuming σ = 19.1? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires _____subjects. (Round up to the nearest whole number as needed.)
A 95% confidence level requires _____subjects. (Round up to the nearest whole number as needed.)
Expert:  Stevewh replied 4 years ago.
Hi,

Can i still help with these questions ?

I appreciate if you accept the previous ones

Steve
Customer: replied 4 years ago.
Yes i still need them.

Thank You
Expert:  Stevewh replied 4 years ago.

Number 3)

Construct the confidence interval for the population mean µ.
C= 0.95, x=16.9, s=10.0, and n=80
A 95% confidence interval for µ is (14.7, 19.1) (Round to two decimal places as needed.)

Which question is number 7) ??

Number 8)

You are given the sample mean and the sample standard deviation. Use this information to construct the 90% and 95% confidence intervals for the population mean. Which interval is wider? If the convenient, use technology to construct the confidence intervals.
A random sample of 43 gas grills has a mean price of \$633.90 and a standard deviation of \$56.90.
The 90% confidence interval is ( 619.63,648.17). (Round to two decimal places as needed.)
The 95% confidence interval is ( 616.89,650.91).

A doctor wants to estimate the HDL cholesterol of all 20-to 29- year old females. How many subjects are needed to estimate the HDL cholesterol within 3 points with 99% confidence assuming σ = 19.1? Suppose the doctor would be content with 95% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires 269 subjects. (Round up to the nearest whole number as needed.)
A 95% confidence level requires 156 subjects. (Round up to the nearest whole number as needed.)

The sample size required to 95% confidence decreases

I appreciate if you accept my previous answers

Thanks

Steve

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