Chapter 9

A student found that out of 100 cars that passed an intersection 20 were made by Ford. Create a 95% confidence interval for the population proportion of cars made by Ford.

.12 to .28 .2 +- 1.96 sqrt((.2)(.8)/100)

if there is no estimate of the population proportion, what value should be used?

0.5

Which of the following statements is the best definition of a point estimate?

a sample statistic that estimates a population parameter. (I.E. a single item (point) of information)

all of the following are true about the z and t distributions except which one of the following?

both the distributions have a standard deviation of 1

which of the factors listed below determine the width of a confidence interval? select all that apply

the chosen level of confidence the size of the standard error

For a given confidence level, how does a confidence interval calculated using the t-value compare to one calculated using the z-value?

the confidence interval from the t-value is wider

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) What is the best estimate of this value?

18

A large jar contains a mixture of white and black beans. The population proportion of beans was found to be .25. what sample size would be needed to estimate the proportion of black with an error of no more than 0.05 and a confidence level of 95%?

289 n=0.25(.75)*(1.96/.05)^2

when we speak of an '90% confidence level', what are we referring to?

a 90% probability that a parameter will be within an interval to be constructed i.e. p(a<x<b) = .9

when the population standard deviation is known, the confidence interval for the population mean is based on the:

z-stat

Bob Nale is the owner of Nale's Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 1.80 gallons. From his records, he selects a random sample of 80 sales and finds the mean number of gallons sold is 7.40. Develop a 99% confidence interval for the population mean. (Use z Distribution Table.) (Round your answers to 2 decimal places.)

(6.88, 7.92) z-value for 99% CI is 2.58 CI = (xbar - z*sigma/sqrt(n), xbar + z*sigma/sqrt(n))= (7.4 - 2.58*1.8/sqrt(80), 7.4 + 2.58*1.8/sqrt(80))= (6.8808, 7.9192)= (6.88, 7.92) A confidence interval is centered on the sample mean. Its width is determined by the confidence level and the size of the standard error of the mean. In this case, the standard error is 1.8/√80 and the z value at a 99% level is 2.576.The end points of the confidence interval are 6.882 and 7.918, found by 7.40±2.576(1.8/80‾‾‾√)7.40±2.576(1.8/80) .

calculate a 95% confidence interval for a sample mean of 20 with a sample standard deviation of 10 and a sample size of 9.

12.31 to 27.69 20 +- 2.306 (10/sqrt(9)) use t-distribution

ou wish to estimate the mean number of travel days per year for salespeople. The mean of a small pilot study was 150 days, with a standard deviation of 34 days. If you want to estimate the population mean within 7 days, how many salespeople should you sample? Use the 98% confidence level. (Use z Distribution Table.) (Round up your answer to the next whole number.)

128 You are estimating a population mean. Substitute the maximum allowable error (7), the value of z for the desired level of confidence (2.326) and the standard deviation (34) into formula [9-5]. The required sample size is 128, found by rounding up n = {(2.326 × 34)/7}2 = 127.63835.

Suppose you want to estimate the population mean with 99% confidence with a margin of error of 2.5. if the estimate of the population standard deviation is 11, what sample size is required?

129 n=(2.58*11 /2.5)^2

Calculate a 90% confidence interval for a sample mean of 15 with a sample standard deviation of 5 and a sample size of 25.

13.29 to 16.71 15+- 1.711 (5/sqrt(25))

The mean number of travel days per year for salespeople employed by three hardware distributors needs to be estimated using a 90% level of confidence. For a small pilot study, the mean was 150 days and the standard deviation was 14 days. If the population mean is estimated within two days, how many salespeople should be sampled? (Please use Student's t distribution (Appendix B.5) at infinite degrees of freedom for three decimal accuracy of the required z value.)

133

The mean weight of trucks traveling on a particular section of I-475 is not known. A state highway inspector needs to estimate the population mean. He selects and weighs a random sample of 49 trucks. The mean weight of the 49 trucks was 15.8 tons. Assume for purposes of this problem that the population standard deviation is known, and it is 3.8 tons. What is the 95% confidence interval for the population mean?

14.7 and 16.9 Here, confidence level is 0.95. \begin{array}{c}\\{\rm{For}}\,\left( {1 - \alpha } \right) = 0.95\\\\\alpha = 0.05\\\end{array}For(1−α)=0.95α=0.05​ From the Standard Normal Table, the required {Z_{0.05}}Z0.05​ The value {Z_{0.05}}Z0.05​ is obtained below: That is, P\left( {Z \le z} \right) = 0.05P(Z≤z)=0.05 Procedure for finding the z-value is listed below: 1.From the table of standard normal distribution, locate the probability value of 0.05. 2.Move left until the first column is reached. Note the value as 1.9 3.Move upward until the top row is reached. Note the value as 0.06. 4.The intersection of the row and column values gives the area to the two tail of z. That is, P\left( {Z \le 1.96} \right) = 0.05P(Z≤1.96)=0.05 . From standard normal table, the required {Z_{0.05}}Z0.05​ value for 95% confidence level is 1.96. Explanation|Common mistakes|Hint for next step Locate the probability of 0.05 in the standard normal table and identify the corresponding row and column values to obtain the value of {Z_{0.05}}Z0.05​ . The 95% confidence interval for the population mean is obtained below: From the information given, a random sample of 49 trucks is considered. That is, n = 49n=49 , the mean weight is 15.8 tons. That is \bar x = 15.8xˉ=15.8 , and the population standard deviation is 3.8 tons. That is, \sigma = 3.8σ=3.8 . The confidence interval is, \begin{array}{c}\\{\rm{Confidence interval}} = 15.8 \pm \left( {1.96} \right)\left( {\frac{{3.8}}{{\sqrt {49} }}} \right)\\\\ = 15.8 \pm \left( {1.96 \times 0.5429} \right)\\\\ = 15.8 \pm 1.064\\\\ = \left( {15.8 - 1.064,15.8 + 1.064} \right)\\\end{array}Confidenceinterval=15.8±(1.96)(49​3.8​)=15.8±(1.96×0.5429)=15.8±1.064=(15.8−1.064,15.8+1.064)​ = \left( {14.736,\,\,16.864} \right)=(14.736,16.864) Thus, the 95% confidence interval for the population mean is \left( {14.736,\,\,16.864} \right)(14.736,16.864) .

a sample size 16 is taken from a normally distributed population with standard deviation 8. The sample mean is 22. Which of the following is a 95% confidence interval for the population mean?

18.08 to 25.92 22+- 1.96(8/sqrt(16))

A random sample of 20 items is selected (from a population with an unknown population standard deviation). When computing a confidence interval for the population mean, what number of degrees of freedom should be used to determine the appropriate t-value?

19 Here degree of freedom for the t-test is df=n-1=20-1=19 where n shows the number of observations.

the formula for estimating the sample size for a study is n=(zsd/E)^2 . match the variables to their description

n= size of the sample z=standard normal value for the chosen confidence level sd= the population standard deviation E=the max allowable error

Which of the following must be true so that the standard normal distribution can be used to construct a confidence interval for the population proportion?

npi >=5 and n(1-pi)>=5

Knowing the population standard deviation allows us to choose an appropriate sample size for a study. Which two of the following could be used ot estimate the population standard deviation?

one sixth of the population range the sample standard deviation from a pilot study

a confidence interval for the population proportion is calculated using the formula p+-z sqrt(p(1-p)/n) . match the variables to their description

p- the sample proportion z- the value associated with the confidence level n-the sample size pi- the population proportion that is being estimated

A sample mean is the best point estimate of the ______.

population mean

The central limit theorem tells us which of the following?

the standard deviation of the sample means is the standard error, sd/sqrt(n) the sampling distribution follows the normal probability distribution

The binomial conditions must be met before we can develop a confidence interval for a population proportion. Which two of the following are binomial conditions?

we can define two outcomes, success and failure the probability of success is the same for all trials

Choose the correct formula for calculating the confidence interval for the mean when the population standard deviation is NOT known

xbar +- t* s/sqrt(n)

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) For a 98% confidence interval, what is the value of t? (Round your answer to 3 decimal places.)

2.552 The degrees of freedom are the sample size less one. There are 18 degrees of freedom. Use Appendix B.5 with a 98% confidence level. You find the t value is 2.552.

Mileage tests were conducted on a randomly selected sample of 100 newly developed automobile tires. The results showed that the mean tread life was 50,000 miles, with a standard deviation of 3,500 miles. What is the best estimate of the mean tread life in miles for the entire population of these tires?

50,000 We know that the sample mean is the best point- estimate of the population mean . Given : In sample of 100 newly developed automobile tires, the average tread wear was found to be 50,000 miles and with a standard deviation of 3,500 miles. i.e. s=3,500 miles. Then, the best estimate of the average tread life in miles for the population of these tires is 50,000 miles

A random sample of 85 supervisors revealed that they worked an average of 6.5 years before being promoted. The population standard deviation was 1.7 years. Using the 95% level of confidence, what is the confidence interval for the population mean?

6.14 and 6.86

A sample size 25 is taken from a normally distributed population with standard deviation 6. The sample mean is 10. Which of the following is a 99%

6.9 to 13.2 10+- 2.57(6/ sqrt(25))

Suppose you want to estimate the population mean with 95% confidence with a margin of error of 2. If the estimate of the population standard deviation is 8, what sample size is required?

62 n=((1.96*8)/2)^2 =61.47 round UP

the central limit theorem tells us that the sample means follow a normal distribution with mean u and standard deviation (i.e. standard error) of sd/sqrt(n) . this lets us use z-values to set confidence intervals. match the confidence levels to the z-values

68% -> z=1, interval u+-sd / sqrt(n) 95% -> z=1.96, interval u+-1.96sd / sqrt(n) 99% -> z=2.58, interval u+- 2.58sd / sqrt(n)

Bob Nale is the owner of Nale's Quick Fill. Bob would like to estimate the mean number of gallons of gasoline sold to his customers. Assume the number of gallons sold follows the normal distribution with a population standard deviation of 1.80 gallons. From his records, he selects a random sample of 80 sales and finds the mean number of gallons sold is 7.40. What is the point estimate of the population mean? (Round your answer to 2 decimal places.)

7.40 xbar = 7.4 A point estimate is a single statistic that estimates the value of a population parameter. The sample mean of 7.40 gallons is our best estimate of the population mean.

Suppose 1,600 of 2,000 registered voters sampled said they planned to vote for the Republican candidate for president. Using the 95% level of confidence, what is the interval estimate for the population proportion (to the nearest 10th of a percent)?

78.2 and 81.8%

Suppose that, after using the sample size formula, you find n=83.25. What sample size should you use?

84

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) What is the value of the population mean?

It's unknown The population mean is unknown, but the best estimate is 18, the sample mean.

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) What about 25 eggs?

No 25 is greater than 19.76. That is not reasonable because it is not inside the confidence interval.

What should you do if your calculation for appropriate sample gives a fractional number?

Round up to the next whole number

Which of the following items are valid considerations in the choice of sample size? Select all that apply

The population dispersion the desired level of confidence the margin of error the researcher will tolerate

in choosing a sample size for a study, we need to know the population standard deviation. Which of the following could be used to estimate the population standard deviation? select all that apply

The sample standard deviation from a pilot study the population standard deviation from a comparable study

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) Develop the 98% confidence interval for the population mean. (Round your answers to 2 decimal places.)

[16.24, 19.76] A confidence interval is centered on the sample mean. Its width is determined by the confidence level and the size of the standard error of the mean. In this case, the standard error is 3/√18 and the t value at a 98% level is 2.552. The end points of the confidence interval are 16.24 and 19.76, found by 18±2.552(319√).18±2.552(319).

Which of the following considerations require a larger sample size? select all

a higher level of confidence a smaller margin on error

constructing a confidence interval to estimate the population mean from a sample, which of the following steps are necessary only when sd is not known? select all that apply

find the sample sd find the t-value for the proper distribution, based on the sample size

Which of the following statements correctly describe the role of the population standard deviation, sd, in creating a confidence interval for the population mean? select all that apply

if sd is known, we use it to calculate the confidence interval if sd is not known, we estimate it using the sample standard deviation

which of the following is true about the range of the sample proportion?

it can never be less than 0 or greater than 1

identify which of the following are traits that apply to the meaning of "proportion". select all that apply

it can refer to either a sample or a population it refers to a fraction, ratio, or percent

The owner of Britten's Egg Farm wants to estimate the mean number of eggs produced per chicken. A sample of 19 chickens shows they produced an average of 18 eggs per month with a standard deviation of 3 eggs per month. (Use t Distribution Table.) Would it be reasonable to conclude that the population mean is 18 eggs?

yes 18 is between 16.24 and 19.76. That is reasonable because it is inside the confidence interval.