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The population mean annual salary for environmental compliance specialists is about

$66 comma 00066,000.

A random sample of 4040 specialists is drawn from this population. What is the probability that the mean salary of the sample is less than $62 comma 50062,500?

Assume sigmaσequals=$6 comma 4006,400.

Calculate the z value:

z = (x-bar – mu) / (sigma / √n) = (62,500 – 66,000) / (6400 / √40) = -3.4587

Then, since the question asks for the probability that the average is “less than 62,500”, we want the area to the left of z = -3.4587:

Find the probability and interpret the results. If convenient, use technology to find the probability. During a certain week the mean price of gasoline was

$2.7062.706 per gallon. A random sample of 3333 gas stations is drawn from this population. What is the probability that the mean price for the sample was between

$2.6992.699 and $2.7272.727 that week? Assume sigmaσequals=$0.0450.045.

The probability that the sample mean was between $2.6992.699 and $2.7272.727

is nothing m.

The interpretation of the result from the previous question is the first choice:

The answer the gasoline price problem is:

The height of women ages 20-29 is normally distributed, with a mean of

64.464.4 inches. Assume sigmaσequals=2.72.7 inches. Are you more likely to randomly select 1 woman with a height less than 65.165.1 inches or are you more likely to select a sample of 2727 women with a mean height less than 65.165.1

inches? Explain.

What is the probability of randomly selecting 1 woman with a height less than

65.165.1 inches?

A news reporter reports the results of a survey and states that 45% of those surveyed responded “yes” with a margin of error of “plus or minus 5%.” Explain what this means. Choose the correct answer below. A. The percent of the population who fall under the “no” category is 5% less than the “yes” category. B.It means that 5% of the surveys are invalid. C. The true percent of the population who fall under the “yes” category is either 40% or 50%. D.The percent of the population who fall under the “yes” category probably lies between 40% and 50%.

Find the critical value z Subscript czc necessary to form a confidence interval at the level of confidence shown below.cequals=0.830.83z Subscript czcequals=nothingm

zc = 1.372 round off as directed

The site doesn’t charge on a “per day” basis like that. Also, please be forewarned that if you are offered some kind of a “subscription” for the site, that their policies do not allow subscriptions to be used in the homework categories.The best approach for your situation is to post a set of questions, and adjust the price to a level that you are comfortable with for that amount of work involved. If one of the experts is willing to take on the assignment for the offered price, then they will contact you.How many more questions do you need help with tonight?The interpretation of the result from the previous question is the first choicewhat do you mean

Find the margin of error for the given values of c, s, and n.cequals=0.900.90,sequals=3.63.6,nequals=100100Eequals=nothingm

In a random sample of 5050 refrigerators, the mean repair cost was $134.00134.00

and the population standard deviation is $16.5016.50.

A 9090% confidence interval for the population mean repair cost is left parenthesis 130.16 comma 137.84 right parenthesis(130.16,137.84).Change the sample size to

nequals=100100. Construct a 9090% confidence interval for the population mean repair cost. Which confidence interval is wider? Explain. Construct a 9090% confidence interval for the population mean repair cost. The 9090% confidence interval is

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

The margin of error is then:

The nequals=5050 confidence interval is wider because a smaller sample is taken, giving less information about the population. B. The nequals=100100

confidence interval is wider because a larger sample is taken, giving more information about the population. C. The two intervals are the same size because the confidence interval is based on the level of confidence and sample standard deviation.

A random sample of fifty dash seven fifty seven 200-meter swims has a mean time of

3.5043.504 minutes and a standard deviation of 0.0700.070 minutes. A 9090%

confidence interval for the population mean time is left parenthesis 3.489 comma 3.519 right parenthesis(3.489,3.519).Construct a 9090% confidence interval for the population mean time using a standard deviation of 0.040.04 minutes. Which confidence interval is wider? Explain. The 9090% confidence interval

A. The s equals=0.0700.070 confidence interval is wider because of the increased variability within the sample. B. The sequals=0.040.04 confidence interval is wider because of the decreased variability within the sample. C.The two intervals are the same size because the confidence interval is based on the level of confidence and sample size.

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 44 points with 99%

confidence assuming sigma equals 11.7 question mark σ=11.7?

Suppose the doctor would be content with 90 % 90% confidence. How does the decrease in confidence affect the sample size required? A 99% confidence level requires

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 44 points with 99% confidence assuming

σ=11.7?

Suppose the doctor would be content with 90% confidence. How does the decrease in confidence affect the sample size required?

A beverage company uses a machine to fill one-liter bottles with water (see figure). Assume that the population of volumes is normally distributed.

(a) The company wants to estimate the mean volume of water the machine is putting in the bottles within 1 milliliter. Determine the minimum sample size required to construct a

90% confidence interval for the population mean. Assume the population standard deviation is

55 milliliters.

(b) Repeat part (a) using an error tolerance of 22 milliliters. Which error tolerance requires a larger sample size? Explain.

A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).

(a) The company wants to estimate the mean length the machine is cutting the plastic within

0.125 inch. Determine the minimum sample size required to construct a 90%

confidence interval for the population mean. Assume the population standard deviation is

A machine cuts plastic into sheets that are 30 feet (360 inches) long. Assume that the population of lengths is normally distributed. Complete parts (a) and (b).

(a) The company wants to estimate the mean length the machine is cutting the plastic within 0.125 inch. Determine the minimum sample size required to construct a 90% confidence interval for the population mean. Assume the population standard deviation is 0.25 inch.

n = 11 (Round up to the nearest whole number as needed.)

(b) Repeat part (a) using an error tolerance of 0.0625 inch.

A soccer ball manufacturer wants to estimate the mean circumference of soccer balls within 0.050.05 in.

(a) Determine the minimum sample size required to construct a 9090% confidence interval for the population mean. Assume the population standard deviation is 0.250.25 in.

(b) Repeat part (a) using a standard deviation of 0.400.40 in. Which standard deviation requires a larger sample size? Explain.

A population standard deviation of 0.40 in. requires a larger sample size. Due to the increased variability in the population, a larger sample size is needed to ensure the desired accuracy.

With 9090% confidence, it can be said that the confidence interval contains the sample mean closing stock price.

B.

The confidence interval contains 9090% of the closing stock prices.

C.

With 9090% confidence, it can be said that the confidence interval contains the true mean closing stock price.

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

cequals=0.950.95, x overbarxequals=13.813.8, sequals=4.04.0, nequals=66

(a) The 9595% confidence interval using a t-distribution is left parenthesis nothing comma nothing right parenthesis .

(a) (9.6, 18.0)

(b) (10.6, 17.00)

The interval using the t-distribution is wider.

Construct the indicated confidence interval for the population mean muμ using a t-distribution.

cequals=0.9999, x overbarxequals=113113, sequals=1010, nequals=2222

The confidence interval is left parenthesis nothing comma nothing right parenthesis .

(107.0, 119.0)

In a random sample of six six microwave ovens, the mean repair cost was $65.0065.00 and the standard deviation was $13.5013.50. Assume the variable is normally distributed and use a t-distribution to construct a 9595% confidence interval for the population mean muμ. What is the margin of error of muμ?

The 9595% confidence interval for the population mean muμ is left parenthesis nothing comma nothing right parenthesis .

In a random sample of 5454 bolts, the mean length was 1.491.49 inches and the standard deviation was 0.080.08 inch. use a normal distribution or a t-distribution to construct a 8080% confidence interval for the mean.

Which distribution should be used to construct the 8080% confidence interval?

A.

use a t-distribution because the lengths are normally distributed and sigmaσ is known.

B.

Use a t-distribution because ngreater than or equals≥30.

C.

use a normal distribution because the lengths are normally distributed and sigmaσ is known.

D.

use a normal distribution because ngreater than or equals≥30.

The 8080% confidence interval is left parenthesis nothing comma nothing right parenthesis .

You take a random survey of 25 sports cars and record the miles per gallon for each. The data are listed to the right. Assume the miles per gallon are normally distributed. Use a normal distribution or a t-distribution to construct a 9090% confidence interval for the mean.

2222

2424

1313

1616

2323

2424

2020

1818

1818

2424

1616

2424

2323

1616

2323

2323

2626

2121

2727

1717

1919

2121

1313

2424

2727

Which distribution should be used to construct the 9090% confidence interval?

A.

use a t-distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

B.

use a normal distribution because nless than<30, the miles per gallon are normally distributed and sigmaσ is unknown.

C.

use a t-distribution because nless than<30 and sigmaσ is known.

D.

use a normal distribution because nless than<30 and sigmaσ is known.

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 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

46924692 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 44 points with 99 %99% confidence assuming sigma equals 11.7 question mark σ=11.7? Suppose the doctor would be content with 90 %90% confidence. How does the decrease in confidence affect the sample size required?

A 99% confidence level requires

46924692 subjects.

(Round up to the nearest whole number as needed

For the 90% confidence interval, the required sample size is:

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

(0.7530.753,0.7790.779)

Eequals=

In a survey of 90009000 women, 64316431 say they change their nail polish once a week. Construct a 99 %99% confidence interval for the population proportion of women who change their nail polish once a week.

A 9999% confidence interval for the population proportion is (

A researcher wishes to estimate, with 9090% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 11% of the true proportion.

a) Find the minimum sample size needed, using a prior study that found that 3434% 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 3434% of the respondents said they have high-speed Internet access?

a) 51

b) 56

A researcher wishes to estimate, with 9090% confidence, the proportion of adults who have high-speed Internet access. Her estimate must be accurate within 11% of the true proportion.

a) Find the minimum sample size needed, using a prior study that found that 3434% 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 3434% of the respondents said they have high-speed Internet access?

nequals=

5151 (Round up to the nearest whole number as needed.)

a) 6073

b) 6766