Chapter 8: Interval Estimation

In a random sample of 400 registered voters, 120 indicated they plan to vote for Trump for President. Determine a 95% confidence interval for the proportion of all the registered voters who will vote for Trump.

(.26, .34) .3 +- sqrt(.3(1-.3)/ 400)

What is the symbol for the population mean?

μ

As the sample size increases, the margin of error:

decreases

A sample of 100 footballs showed an average air pressure of 13 psi. The standard deviation of the population is known to be .25 psi. What is the standard error of the mean?

.025 .25/sqrt(100)

The CEO of a company wants to estimate the percent of employees who use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. What is the estimate of the standard error of the proportion ?

.039 sigma sub p bar = sqrt(.353(1-.353)/ 150)

Suppose a 95% confidence interval, based upon a sample of size 25, for the mean number of hours of sleep that college students get per night was 5 to 8 hours. How many students should be surveyed in order to cut the width of the interval down from 3 hours to 1.5 hours?

100

A professor would like to estimate the average number of hours his students spend doing work and studying throughout the semester for the course he teaches. He wants to estimate μ with 99% confidence and with a margin of error of at most 2 hours. From past experience, he believes that the standard deviation of the number of hours students spent is 8 hours. How many students need to be surveyed to meet these requirements?

107 (2.576)^2(8)^2/2^2

An elementary school teacher asked a random sample of 12 of her students what their favorite number was. Assume the population of responses would follow a normal distribution. The students stated that their favorite numbers are: 2 10 7 4 0 5 6 4 4 6 1 100 What is the appropriate degrees of freedom to use when calculating a 95% confidence interval for μ ?

11 n-1

The sample size needed to provide a margin of error of 2 or less with 95% confidence when the population standard deviation is equal to 11 is:

117 (1.96)^2(11)^2/2^2

In order to determine an interval for the mean of a population with unknown standard deviation, a sample of 24 items is selected. The mean of the sample is determined to be 23. The number of degrees of freedom for reading the t value is:

23

A researcher is interested to determine the average age at which people obtain their first credit card. If past information shows a mean of 22 years and a standard deviation of 2 years, what size sample should be taken so that at 95% confidence the margin of error will be 3 months or less?

246 (1.96)^2(2)^2/.25^2

The t value for a 99% confidence interval estimation based upon a sample of size 10 is:

3.250

A researcher is interested to determine the average number of years new teachers remain in the classroom. If past information shows a standard deviation of 5 years, what size sample should be taken so that at 95% confidence the margin of error will be 6 months or less?

385 (1.96)^2(5)^2/.5^2

margin of error

The ± value added to and subtracted from a point estimate in order to develop an interval estimate of a population parameter.

An approximate value of a population parameter that provides limits and believed to contain the value of the parameter is known as the

interval estimate

For a fixed sample size, n, in order to have a higher degree of confidence, the margin of error and the width of the interval:

must be larger

A poll of 600 voters showed 210 that were in favor of stricter gun control measures. Develop a 90% confidence interval estimate for the proportion of all the voters who are in favor of stricter gun control measures.

n 600 210 point estimate 0.3500 tα/2 1.64 margin of error 0.0319 Lower 0.32 Upper 0.38

The sampling distribution of can be approximated by a normal distribution as long as:

np>5 and n(1-p)>5

When "s" is used to estimate "σ," the margin of error is computed by using the:

t distribution

To compute the necessary sample size for an interval estimate of a population proportion, all of the following procedures are recommended when p* is unknown except:

using .95 as an estimate.

To compute the necessary sample size for an interval estimate of a population mean, all of the following procedures are recommended when σ is unknown except:

using σ = 1

What is the symbol for the sample mean?

x̄

A sample of 100 footballs showed an average air pressure of 13 psi. The standard deviation of the population is known to be .25 psi. With a .90 probability, the margin of error is approximately equal to:

.041 MOE= Z*(sigma/sqrt(n))1.645*(.25/sqrt(100))

The CEO of a company wants to estimate the percent of employees who use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. What is the margin of error?

.076 1.96 sqrt(.353(1-.353)/150)

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to check out was 3.0 minutes. It is known that the standard deviation of the population of checkout times is one minute. The standard error of the mean is equal to:

.100 1/sqrt(100)

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to checkout was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a .95 probability, the sample mean will provide a margin of error of:

.196 1.96(1/sqrt(100))

The CEO of a company wants to estimate the percent of employees who use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged on to Facebook that day. What is the point estimate of the proportion of the population who logged on to Facebook that day?

.35 53/150

The CEO of a company wants to estimate the percent of employees who use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. Compute the 95% confidence interval for the population proportion.

.35+- 1.96*sqrt(.35(1-.35)/150)

When computing the sample size needed to estimate a proportion within a given margin of error for a specific confidence level, what planning value of p* should be used when no estimate of p* is available?

.50

Sales data from a family-owned store shows that 78% of a random sample of 300 customers pay using a credit card. Compute the 99% confidence interval for the population proportion.

.72 to .84 .78 +- 2.576 sqrt(.78(1-.78)/300)

Male college basketball players have to weigh-in during season, and this information is published. We can, therefore, know the standard deviation of the entire population. Suppose we do not know the population mean and wanted to estimate it. Suppose we took a random sample of 25 male college basketball players and recorded their weights. The sample mean was found to be 220 lbs. The population standard deviation was 5 lbs. What is the standard error of the mean?

1 5/sqrt(25)

An elementary school teacher asked a random sample of 12 of her students what their favorite number was. Assume the population of responses would follow a normal distribution. The students stated that their favorite numbers are: 2 10 7 4 0 5 6 4 4 6 1 100 Create a 95% confidence interval for μ.

12.42 +- 2.201(27.7/sqrt(12))

A sample of 100 footballs showed an average air pressure of 13 psi. The standard deviation of the population is known to be .25 psi. The 99% confidence interval for the true mean air pressure for all footballs is:

12.936 to 13.06413 +- 2.576(.25/sqrt(100))

A random sample of 144 observations has a mean of 20, a median of 21, and a mode of 22. The population standard deviation is known and is equal to 4.8. The 95.44% confidence interval for the population mean is:

19.200 to 20.800.

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. If we would like to capture the population mean with 95% confidence, the margin of error would be:

2.262(9/sqrt(10))

In an interval estimation for a proportion of a population, the z value for 99% confidence is:

2.576

Male college basketball players have to weigh-in during season, and this information is published. We can, therefore, know the standard deviation of the entire population. Suppose we do not know the population mean and wanted to estimate it. Suppose we took a random sample of 25 male college basketball players and recorded their weights. The sample mean was found to be 220 lbs. The population standard deviation was 5 lbs. With a .99 probability, the margin of error is approximately equal to:

2.576 2.576(5/sqrt(25))

The z value for a 99% confidence interval estimation is:

2.58

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. What is the standard error of the mean?

2.846 s/sqrt(n)9/sqrt(10)

Male college basketball players have to weigh-in during season, and this information is published. We can, therefore, know the standard deviation of the entire population. Suppose we do not know the population mean and wanted to estimate it. Suppose we took a random sample of 25 male college basketball players and recorded their weights. The sample mean was found to be 220 lbs. The population standard deviation was 5 lbs. The 99% confidence interval is:

220 +- 2.576(5/sqrt(25))

A local health center noted that in a sample of 400 patients 80 were referred to them by the local hospital. What size sample would be required to estimate the proportion of hospital referrals with a margin of error of .04 or less at 95% confidence?

385 n=(z a/2)p(1-p)/E^2(1.96)^2*(.2)(1-.2)/.04^2

An elementary school teacher asked a random sample of 12 of her students what their favorite number was. Assume the population of responses would follow a normal distribution. The students stated that their favorite numbers are: 2 10 7 4 0 5 6 4 4 6 1 100 Suppose the student who said "100" was left out of the data set. Create a revised 95% confidence interval for μ.

4.45 +- 2.228 (2.84/sqrt(11))

The manager of a grocery store has taken a random sample of 100 customers. The average length of time it took these 100 customers to checkout was 3 minutes. It is known that the standard deviation of the population of checkout times is 1 minute. With a .95 probability, the sample mean will provide a margin of error of .196. If we wanted to cut this margin of error in half, how many customers would we need to include in our sample?

400 (1.96)^2(1)^2/.098^2

An analyst for a cell phone company would like to know the average age at which people obtain their first cell phone. Past records show a mean of 14 years and a standard deviation of 8 years. What sample size should be taken so that at 99% confidence the margin of error will be 1 year or less?

425 n= z^2*sigma^2/E^2=(2.576)^2*(8)^2/1^2

The CEO of a company wants to estimate the percent of employees who use company computers to go on Facebook during work hours with 95% confidence. He selects a random sample of 150 of the employees and finds that 53 of them logged onto Facebook that day. The CEO would like to cut the margin of error in half without changing the confidence level. How large of a sample size is needed?

600 To cut the margin of error in half, the sample size must quadruple

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. The 95% confidence interval for the true mean number of pushups that can be done is:

8.56 to 21.4415 +- 2.262(9/sqrt(10))

t distribution

A family of probability distributions that can be used to develop an interval estimate of a population mean whenever the population standard deviation σ is unknown and is estimated by the sample standard deviation s.

degrees of freedom

A parameter of the t distribution. When the t distribution is used in the computation of an interval estimate of a population mean, the appropriate t distribution has n − 1 degrees of freedom, where n is the size of the sample.

A statistics teacher started class one day by drawing the names of 10 students out of a hat and asked them to do as many pushups as they could. The 10 randomly selected students averaged 15 pushups per person with a standard deviation of 9 pushups. Suppose the distribution of the population of number of pushups that can be done is approximately normal. Which of the following statements is true?

A t distribution should be used because σ is unknown.

interval estimate

An estimate of a population parameter that provides an interval believed to contain the value of the parameter. For the interval estimates in this chapter, it has the form: point estimate ± margin of error.

An elementary school teacher asked a random sample of 12 of her students what their favorite number was. Assume the population of responses would follow a normal distribution. The students stated that their favorite numbers are: 2 10 7 4 0 5 6 4 4 6 1 100 Suppose we were to create a 95% confidence interval for μ. What effect does the value 100 have on the width of the confidence interval?

It makes the interval wider.

σ known

The case when historical data or other information provide a good value for the population standard deviation prior to taking a sample. The interval estimation procedure uses this known value of σ in computing the margin of error.

confidence level

The confidence associated with an interval estimate. For example, if an interval estimation procedure provides intervals such that 95% of the intervals formed using the procedure will include the population parameter, the interval estimate is said to be constructed at the 95% confidence level. for an interval estimate.

confidence coefficient

The confidence level expressed as a decimal value. For example, .95 is the confidence coefficient for a 95% confidence level.

σ unknown

The more common case when no good basis exists for estimating the population standard deviation prior to taking the sample. The interval estimation procedure uses the sample standard deviation s in computing the margin of error.

Level of significance

The probability that the interval estimation procedure will generate an interval that does not contain μ.

Practical Signficance

The real-world impact the result of statistical inference will have on business decisions.

confidence interval

another name for an interval estimate

In interval estimation, as the sample size becomes larger, the interval estimate:

becomes narrower.

As the number of degrees of freedom for a t distribution increases, the difference between the t distribution and the standard normal distribution:

becomes smaller

When the level of confidence decreases, the margin of error:

becomes smaller

Using an α = .04, a confidence interval for a population proportion is determined to be .65 to .75. If the level of significance is decreased, the interval for the population proportion:

becomes wider

For a fixed confidence level and population standard deviation, if we would like to cut our margin of error in half, we should take a sample size that is:

four times as large as the original sample size.

In general, higher confidence levels provide larger confidence intervals. One way to have high confidence and a small margin of error is to:

increase the sample size.

We can reduce the margin of error in an interval estimate of p by doing any of the following except:

increasing the planning value p* to .5.

The probability that the interval estimation procedure will generate an interval that does not contain µ is known as the:

level of significance.

The value added and subtracted from a point estimate in order to develop an interval estimate of the population parameter is known as the:

margin of error.

For a fixed confidence level and population standard deviation, if we would like to cut our margin of error to 1/3 of the original size, we should take a sample size that is:

nine times as large as the original sample size.

The margin of error in an interval estimate of the population mean is a function of all of the following except the:

sample mean.

From a population that is normally distributed, a sample of 30 elements is selected and the standard deviation of the sample is computed. For the interval estimation of μ, the proper distribution to use is the:

t distribution with 29 degrees of freedom.

For the interval estimation of μ when σ is known and the sample is large, the proper distribution to use is:

the normal distribution.