# The population mean annual salary

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

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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