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Category: Calculus and Above
Satisfied Customers: 8801
Experience:  B.S. in Civil Engineering
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# Find the margin of error

find the margin of error for c s and n
c+.90 s=3.1, n=100

Hi,

Welcome! Thank you for using JustAnswer.

I assume that the notation "c + .90" was supposed to be "c = 0.90" and is meant to indicate a 90% confidence interval.

If so, the margin of error is:

E = z * s / √n

For a 90% confidence interval, use z = 1.645.

Substituting the value for z, s, and n gives:

E = (1.645)(3.1) / √100

E = 0.5100
(rounded to four decimal places)

Thanks

Customer: replied 3 years ago.

Construct the confidence interval for the population

c=0.90, x=8.1 s=0.2 and n=53

Hi,

Here is the solution for this one:

For a 90% confidence interval, use z = 1.645.

The margin of error is then:

E = z * s / √n = (1.645)(0.2)/√53 = 0.0452

The lower limit of the confidence interval is then:

x-bar - E = 8.1 - 0.0452 = 8.0548

The upper limit of the confidence interval is then:

x-bar + E = 8.1 + 0.0452 = 8.1452

The confidence interval is then (8.0548, 8.1452)

(If all you want is the final answer, please let me know. It will speed things up if I don't have to type the solution out. Of course, if you're interested in the steps, I'm fine with including it.)

Thanks

Customer: replied 3 years ago.

construct the confidence interval for population. Round to one decimal place as needed.

c=0.95, x (x has line over it)=15.9 , s=4.0, and n=70

Hi,

The confidence interval is (14.9629, 16.8371).

Thanks

Customer: replied 3 years ago.

please adjust for a 95% confidence interval for u is? round to one decimal place

Hi,

I'm sorry, I overlooked the part about rounding off.

The interval is (15.0, 16.8).

Thanks

Customer: replied 3 years ago.

Question 6

Use the confidence interval to find the estimated margin of error. Then find the sample mean.

A biologist reports a confidence interval of (4.7,5.1) when estimating the mean height (in centimeters) of a sample of seedlings

The estimated margin of error is?

Hi,

The estimated margin of error is 0.2

The sample mean is 4.9

Thanks

Customer: replied 3 years ago.

Q7 of 20

Find the minimum sample size n needed to estimate u for the given values of c,s, and E.

C=0.98,s=7.4,E=1

Assume that a preliminary sample of has at least 30 members

n=? (round up to the nearest whole number)

Hi,

The minimum sample size would be 297.

Thanks

Ryan, Engineer
Satisfied Customers: 8801
Experience: B.S. in Civil Engineering
Customer: replied 3 years ago.
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. Interpret the result and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.A random sample of 55 home theather systems has a mean price of \$143.00 and a standard deviation is \$16.80.Construct 90% confidence interval for the poplulation mean.The 90% confidence interval is? (round to two decimal places as needed)

Hi,

The 90% confidence interval is (139.27, 146.73)

The interpretation is that we are 90% certain that the true population mean lies within the limits of the 90% confidence interval.

The 95% confidence interval is (138.56, 147.44)

The interpretation is that we are 95% certain that the true population mean lies within the limits of the 95% confidence interval.

The 95% confidence interval is wider than the 90% confidence interval.

Thanks

Customer: replied 3 years ago.

Q9 of 20

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 convenient, use technology to construct the confidence intervals.

A random sample of 44 gas grills has a mean price of \$646.30 and a standard deviation of \$57.20.

The 90% confidence interval is? (round to two decimal places as needed)

Hi,

The 90% confidence interval is (632.11, 660.49)

The 95% confidence interval is (624.09, 668.51)

The 95% confidence interval is wider.

Thanks

Customer: replied 3 years ago.

FYI 95% interval was not correct...

Q10 of 20

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

A random sample pf 32 eight ounce servings of different juice drinks has a mean of 76.6 calories and a standard deviation of 45.2 calories.

The 90% confidence interval is ? (round to one decimal place as needed)

Hi,

I'm sorry. I looked at the wrong line in the spreadsheet I set up. The one I typed was the 99% interval.

It may be too late, but the 95% interval should have been (629.40, 662.20)

For #10:

The 90% confidence interval is (63.5, 89.7)

The 95% confidence interval is (60.9, 92.3)

The 95% confidence interval is wider.

Thanks

Customer: replied 3 years ago.

Q11

People were polled on how many books they read the previous year. How many subjects are needed to estimate the number of books read the previous year within one book with 90% confidence? The intial survey results indicate that O (looks kind of like an o symbol) o=13.9 books.

A 90% confidence level requires blank subjects

Hi,

The answer here depends on which value your course uses for a 90% confidence level.

If it uses z = 1.645, then the answer is 523

If, instead, they round it off to z = 1.65, then the answer is 527

We have used 1.645 in previous problems in this set, so hopefully that is the value that they want you to use. This issue didn't come up earlier because it didn't change the value of the answer like it does here.

Thanks

Customer: replied 3 years ago.

Q12 of 20

A doctor wants to estimate the HDLcholestrol 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 o=13.1? Suppose the doctor would be content with 90% confidence. How does the degrease in confidence affect the sample size required?

A 99% confidence level requires Blank subjects

Hi,

The 99% confidence level requires 285 subjects.

Decreasing the confidence level would reduce the required sample size.

Thanks

Customer: replied 3 years ago.

pls add the 90% confidence level for q12

Hi,

The required sample size for the 90% confidence level would be 117. (Using the 1.645 value for z.)

Thanks

Customer: replied 3 years ago.

Q13 of 20

Construct the indicated confidence interval for the population mean u using (a) a t-distribution. (b) if you had a incorrectly used a normal distribution, which interval would be wider?

c=0.95, x (with line over it) x=13.3, s=3.0 n=10

(a) the 95% confidence interval using a t distribution is? ( round to one decimal place as needed)

Hi,

The 95% confidence interval using the t-distribution is (11.2, 15.4).

If you had used the normal distribution the 95% confidence interval would be (11.4, 15.2). The interval using the normal distrubition is narrower than the interval using the t-distribution.

Thanks

Customer: replied 3 years ago.

Q14 of 20

In the following 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.

(a) in a random sample of 10 adults from a near by county, the mean waste generated per person per days was 5.26 pounds and the standard deviation was 1.79 pounds.

(b) Repeat part 9a), assuming the same statistics came from a sample size of 600. Compare the results.

(a) for the sample of 10 adults, the 90% confidence interval is ? 9 round to two decimal places as needed)

Hi,

a) The 90% confidence interval is (4.22, 6.30).

b) The 90% confidence interval is (4.89, 5.63).

The confidence interval in part b is narrower due to the larger sample size.

Thanks

Customer: replied 3 years ago.

b) was wrong not sure if you based if of 600.

Q15 of 20

Use the given confidence interval to find the margin of error and the sample proportion

(0.786,0.810)

E=blank (type an integer or a decimal)

Hi,

Yes, I did. I don't know what happened there. It should have been (5.14, 5.38). I had the numbers in the right places, but something happened when pushing it all through the calculator.

For #15:

E = 0.012

The sample proportion is 0.798

Customer: replied 3 years ago.

Q16 of 20

In a survey of 626 males ages 18-64, 391 say they have gone to the dentist in the past year.

Construct 90% and 95% confidence intervals for the population proportion. Interpret the results and compare the widths of the confidence intervals. If convenient, use technology to construct the confidence intervals.

The 90% confidence interval for the population proportion p is? (round the final answers to the nearest thousandth as needed. Round all intermediate values to the nearest thousandth as needed)

Hi,

The 90% confidence interval for the population proportion is (0.593, 0.657).

The interpretation of this interval is that we can be 90% certain that the true population proportion is between the limits of the interval.

The 95% confidence interval for the population proportion is (0.587, 0.663).

The interpretation of this interval is that we can be 95% certain that the true population proportion is between the limits of the interval.

The 95% confidence interaval is wider.

Thanks

Customer: replied 3 years ago.

Q17 of 20

In a survey of 6000 women, 3431 say they change their nail polish once a week. Construct a 99% confidence interval for the population proportion of women who change their nail polish once a week.

A 99% confidence interval for the population proportion is? Round to the nearsest thousandth

Hi,

The 99% confidence interval for the population proportion is (0.555, 0.588).

Thanks

Customer: replied 3 years ago.

Q18 of 20

A researcher wishes to estimate, with 90% confidence the proportion of adults who have high speed internet access. her estimate must be accurate within 1% of the true proportion.

(a) find the minimum sample size needed, using a prior student that found that 52% of the respondents said they have high speed internet access.

(b) No preliminary estimate is available. Find the minimum sample size needed.

a) what is the minimum sample size needed using a prior study that found that %52 of the respondents said they have highspeed internet access?

n=? round up to the nearest whole number as needed

Hi,

For part (a), the required minimum sample size is 6,755.

For part (b), the required minimum sample size is 6,766.

Thanks

Ryan, Engineer
Satisfied Customers: 8801
Experience: B.S. in Civil Engineering

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