Alta - Chapter 8 - Confidence Intervals - Part 1

The developmental department for a National Lawn Care Services Provider is interested in improving the products and services. The germination periods, in days, for grass seed are normally distributed with a population standard deviation of 5 days and an unknown population mean. If a random sample of 17 types of grass seed is taken and results in a sample mean of 52 days, find a 80% confidence interval for the population mean.

(50.45,53.55) Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=52. We can use the formula to find the error bound: EBM=(zα2)(σn−−√) =(1.282)(517−−√) ≈(1.282)(1.213) ≈1.55 So, the error bound (EBM) is 1.55. So we can write this confidence interval as: (52−1.55,52+1.55) or (50.45,53.55). Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: Z Interval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 5 for σ (standard deviation), 52 for x¯ (sample mean), 17 for n (sample size), and 0.8 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 80% confidence that the true population mean is between 50.45 and 53.55 days.

Calculating Confidence Intervals Using Student's t-Distribution

A confidence interval is a range of values, written with a point estimate and a margin or error, and has the form: (point estimate - margin of error, point estimate + margin of error) ...with the value of (point estimate - margin of error) as the lower value, and the (point estimate + margin of error) as the upper value. The point estimate is the sample statistic that is used to estimate the population parameter. So, when we are trying to estimate the population mean, the point estimate is the sample mean, x¯. The margin of error is the error bound. When the population standard deviation is unknown, the error bound (EBM) is calculated with a t-score, instead of z-score (which is used when the standard deviation is known). The error bound is found with the following formula: EBM=(tα2)(sn−−√) ...where tα2 is the t-score that corresponds with the confidence level α, s is the sample standard deviation, and n is the sample size. So, using the sample mean as the point estimate, and the EBM as the margin of error, the confidence interval at a confidence level, α, can be written as: (x¯−EBM,x¯+EBM) *We can also use a calculator to find these confidence intervals. The steps for using a calculator are below in the example.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children.

A philanthropist is deciding where to locate her foundation to combat poverty. According to the Small Area Income and Poverty Estimates (SAIPE) data, the state with the largest population of children in poverty is California. Using the poverty estimate 90% confidence interval lower and upper bounds for ages 0−17, what is the sample mean? Enter a whole number.

The following data set provides borough populations in New York City for 2000 and 2010. Which pattern puts the boroughs in ascending order of population change from 2000 to 2010?

Queens, Staten Island, Brooklyn, Manhattan, Bronx Be sure to order the boroughs by their population change, not their total population.

Suppose the number of sales per ad in a magazine is normally distributed. If the population standard deviation is 4 sales, what minimum sample size is needed to be 95% confident that the sample mean is within 2 sales of the true population mean?

The formula for sample size is n=z2σ2EBM2. In this formula, z=zα2=z0.025=1.96, because the confidence level is 95%. From the problem, we know that σ=4 and EBM=2. Therefore, n=z2σ2EBM2=(1.96)2(4)222≈15.37. Use n=16 to ensure that the sample size is large enough.

The t-score has the same interpretation as the z-score. It measures how far the sample mean (x¯) is from the population mean (μ).

Yes The t-score has the same interpretation as the z-score. It measures how far the sample mean (x¯) is from the population mean (μ).

Suppose that the cash flow, in dollars , for a pharmacy are normally distributed with an unknown mean and standard deviation. The cash flow of 40 randomly sampled pharmacies are used to estimate the mean of the population. What t-score should be used to find the 80% confidence interval for the population mean?

$1.304$1.304​ The sample used was 40 pharmacies, so n=40. To find the degrees of freedom: df=n−1=40−1=39 The confidence level is given in the scenario: 80%. So,α=1−CL=1−0.8=0.2But we want to use the value for α2, which is 0.22=0.1. Using the table, we need to find the row for 39 degrees of freedom, and the column for t0.1. So, the t-score we would use to find the 80% confidence interval is 1.304.We could also use a calculator and input 1−α2=0.9 and the degrees of freedom into invT, in which case we would enter invT(0.9,39).

Suppose the times of delivery from an area distributor are normally distributed. If the population standard deviation is 2 hours, what minimum sample size is needed to be 90% confident that the sample mean is within 1 hour of the true population mean?

$11\text{ deliveries}$11 deliveries​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.05=1.645because the confidence level is 90%. From the problem, we know that σ=2 and EBM=1. Therefore,n=z2σ2EBM2=(1.645)2(2)212≈10.82Use n=11 to ensure that the sample size is large enough.

The liabilities of departments are normally distributed with an unknown population mean and standard deviation. If a random sample of 36 departments is taken to estimate the mean liabilities, what t-score should be used to find a 95% confidence interval estimate for the population mean?

$2.030$2.030​ The sample used was 36 departments, so n=36. To find the degrees of freedom: df=n−1=36−1=35 The confidence level is given in the scenario: 95%. So,α=1−CL=1−0.95=0.05But we want to use the value for α2, which is 0.052=0.025. Using the table, we need to find the row for 35 degrees of freedom, and the column for t0.025. So, the t-score we would use to find the 95% confidence interval is 2.030.We could also use a calculator and input 1−α2=0.975 and the degrees of freedom into invT, in which case we would enter invT(0.975,35).

The heights of lighthouse replica models are normally distributed with an unknown population mean and standard deviation. If the heights are obtained for a random sample of 42 models to estimate the mean height, what t-score should be used to find a 99% confidence interval estimate for the population mean? Use the portion of the table below or a calculator. If you use a calculator, round your answer to three decimal places.

$2.701$2.701​ Degrees of freedom are n−1=42−1=41. Confidence level: 99% . α=1−CL=1−0.99=0.01 We want the value for α2, which is 0.012=0.005. Using the table, find the row for 41 degrees of freedom, and the column for t0.005. So, the t -score for the 99% confidence interval is 2.701.

Precipitation inches in the summer in a remote area are normally distributed with an unknown population mean and standard deviation. If precipitation is measured on a random sample of 39 days to estimate the mean annual inches, what t-score should be used to find a 99% confidence interval estimate for the population mean? Use the portion of the table below or a calculator. If you use a calculator, round your answer to three decimal places.

$2.712$2.712​ Degrees of freedom are n−1=39−1=38. Confidence level: 99% . α=1−CL=1−0.99=0.01 We want the value for α2, which is 0.012=0.005 . Using the table, find the row for 38 degrees of freedom, and the column for t0.005 . So, the t -score for the 99% confidence interval is 2.712.

Suppose the shipping periods, in days, for medical equipment are normally distributed. If the population standard deviation is 3 days, what minimum sample size is needed to be 90% confident that the sample mean is within 1 day of the true population mean?

$25\text{ medical equipment}$25 medical equipment​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.05=1.645because the confidence level is 90%. From the problem, we know that σ=3 and EBM=1. Therefore,n=z2σ2EBM2=(1.645)2(3)212≈24.35Use n=25 to ensure that the sample size is large enough.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. For ages 0−17 in the capital of the United States, the District of Columbia, what is the error bound using the poverty estimate 90% confidence interval lower and upper bounds?

$3,408$3,408​ The full range of the confidence level is calculated by subtracting the smaller number from the larger. The EBM will be half that value. 36,355−29,539=6,816 6,8162=3,408=EBM

The number of profits per store is normally distributed. If a random sample of stores is taken and the confidence interval is (104.25, 111.75), what is the error bound?Give just a number for your answer. For example, if you found that the EBM was 2, you would enter 2.

$3.75$3.75​ To find the error bound, subtract the lower value of the confidence interval from the upper value and divide the difference by 2: 111.75−104.252=3.75.

Suppose the yearly bonuses, in thousands, of the technicians in a company are normally distributed. If the population standard deviation is 3 thousand, what minimum sample size is needed to be 95% confident that the sample mean is within 1 thousand of the true population mean? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score. Round only at the final step.

$35\text{ technicians}$35 technicians​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.025=1.96because the confidence level is 95%. From the problem, we know that σ=3 and EBM=1. Therefore,n=z2σ2EBM2=(1.96)2(3)212≈34.57Use n=35 to ensure that the sample size is large enough.

Suppose the total employee turnovers per year in a company are normally distributed. For a random sample of years, the confidence interval (84.50, 93.50) is generated. Find the error bound.Give just a number for your answer. For example, if you found that the EBM was 2, you would enter 2.

$4.5$4.5​ To find the error bound, subtract the lower value of the confidence interval from the upper value and divide the difference by 2: 93.5−84.52=4.5.

The population standard deviation for the complaints of the employees of a company per year is 10 complaints. If we want to be 95% confident that the sample mean is within 3 complaints of the true population mean, what is the minimum sample size that should be taken? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table above for the z-score. Round only at the final step.

$43\text{ employees}$43 employees​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.025=1.96because the confidence level is 95%. From the problem, we know that σ=10 and EBM=3. Therefore,n=z2σ2EBM2=(1.96)2(10)232≈42.68Use n=43 to ensure that the sample size is large enough.

Suppose the prices of new snacks are normally distributed. If the population standard deviation is 14 dollars, what minimum sample size is needed to be 95% confident that the sample mean is within 4 dollars of the true population mean?

$48\text{ new snacks}$48 new snacks​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.025=1.96because the confidence level is 95%. From the problem, we know that σ=14 and EBM=4. Therefore,n=z2σ2EBM2=(1.96)2(14)242≈47.06Use n=48 to ensure that the sample size is large enough.

Suppose we know that the hourly earnings of a company has a confidence interval of (72,86), with a sample mean of 79. Find the error bound (EBM). Give just a number for your answer. For example, if you found that the EBM was 2, you would enter 2.

$7$7​ Since we know the sample mean is 79, we can subtract 79 from the upper value of the confidence interval, 86. 86−79=7 So the error bound is 7.

The following data set provides a summary statistic of 5,742 interviews of commuters.

$\text{EBM=}14.23$EBM=14.23​ The first step is to find the z-value. Instead of using zα, we need to use: zα2=0.052=z0.025=1.960 EBM=zα2⋅σn−−√ =1.960⋅423.63407−−−−√ =1.960⋅7.26 EBM=14.23

The following data set provides a summary statistic of 5,742 interviews of commuters. Public transportation planners sampled US workers who did not own vehicles. Their seven commuting choices are normally distributed with a population standard deviation of 719.4 people and an unknown population mean. If a random sample of 5742 workers is taken and results in a sample mean of 820.4 square feet, find the error bound (EBM) of the confidence interval with a 90% confidence level. Round your answer to TWO decimal places.

$\text{EBM=}15.62$EBM=15.62​ The first step is to find the z-value. Instead of using zα, we need to use: zα2=0.102=z0.05=1.645 EBM=zα2⋅σn−−√ =1.645⋅719.45742−−−−√ =1.645⋅9.49 EBM=15.62

Suppose liabilities of each site from Company A are normally distributed and have a known population standard deviation of 4 liabilities and an unknown population mean. A random sample of 16 sites is taken and gives a sample mean of 68 liabilities. Find the error bound (EBM) of the confidence interval with a 98% confidence level. Round your answer to THREE decimal places.

$\text{EBM=}2.326$EBM=2.326​ We can use the formula to find the error bound: EBM=(zα2)(σn−−√) We know that σ=4 and n=16. We are also given that the confidence level (CL) is 98%, or 0.98. So, we can calculate alpha (α).α=1−CL=1−0.98=0.02Since α=0.02, we know thatα2=0.022=0.01The value of z0.01 is 2.326. Now we can substitute the values into the formula to find the error bound.EBM=(zα2)(σn−−√)=(2.326)(416−−√)≈(2.326)(1.000)≈2.326So, the error bound (EBM) is 2.326.

The following data set provides a summary statistic of 5,742 interviews of commuters. US workers earning $75,000 or more who did not own vehicles had seven commuting choices normally distributed with a population standard deviation of 61.4 people and an unknown population mean. If a random sample of 410 workers is taken and results in a sample mean of 58.6 square feet, find the error bound (EBM) of the confidence interval with a 90% confidence level. Round your answer to TWO decimal places.

$\text{EBM=}4.99$EBM=4.99​ The first step is to find the z-value. Instead of using zα, we need to use: zα2=0.102=z0.05=1.645 EBM=zα2⋅σn−−√ =1.645⋅61.4410−−−√ =1.645⋅3.03 EBM=4.99

The cost of spare parts per machine are normally distributed with a population standard deviation of 158 dollars and an unknown population mean. If a random sample of 23 machines is taken and results in a sample mean of 1790 dollars, find the error bound (EBM) of the confidence interval with a 98% confidence level. Round your answer to THREE decimal places.

$\text{EBM=}76.631$EBM=76.631​ We can use the formula to find the error bound: EBM=(zα2)(σn−−√) We know that σ=158 and n=23. We are also given that the confidence level (CL) is 98%, or 0.98. So, we can calculate alpha (α).α=1−CL=1−0.98=0.02Since α=0.02, we know thatα2=0.022=0.01The value of z0.01 is 2.326. Now we can substitute the values into the formula to find the error bound.EBM=(zα2)(σn−−√)=(2.326)(15823−−√)≈(2.326)(32.945)≈76.631So, the error bound (EBM) is 76.631.

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant looks again at the data set about total cheese sales, in which the sample mean is 164.5 and the sample standard deviation is 103. Find the 95% confidence interval estimate. Round your answer to ONE decimal place.

$\text{The 95% confidence interval estimate is }\ -41.5\ \text{ , }370.5$The 95% confidence interval estimate is −41.5 , 370.5​ 95% of the area under the curve is two standard deviations away from the mean. EBM=2⋅σ EBM=2⋅103=206 x¯−EBM=164.5−206=−41.5 x¯+EBM=164.5+206=370.5 We can say that the 95% confidence interval is (−41.5,370.5).

Suppose that the commissions of salesmen, in thousands of dollars, at a dealership are normally distributed with an unknown mean and standard deviation. A random sample of 24 salesmen is taken and gives a sample mean of 27 thousand and a sample standard deviation of 6 thousand.

$\text{ebm=}2.53$ebm=2.53​ We can calculate the EBM with the formula: EBM=(tα2)(sn−−√) The question tells us s=6 and n=24. We need to find the t-value before we can find EBM. The degrees of freedom aredf=n−1=24−1=23The confidence level is 95%, so α=1−0.95=0.05, and α2=0.025. Now, we can use the table above to find the t-value for t0.025 with 23 degrees of freedom, which is 2.069. Now we can calculate the EBM.EBM=(tα2)(sn−−√)=(2.069)(624−−√)≈(2.069)(1.225)≈2.53

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant is looking to expand her business. She looks at the data set about cheese establishments in six categories, in which the sample mean is 3,324.3 and the sample standard deviation is 2,463.8. Find the rounded upper level of the 68% confidence interval estimate. Round your answer to ONE decimal place.

$\text{the rounded upper level of the 68% confidence interval estimate=}5788.1$the rounded upper level of the 68% confidence interval estimate=5788.1​ The upper bound for 68% is one standard deviation away from the mean. EBM=σ x¯+EBM=3324.3+2463.8=5788.1

A Company CEO is trying to estimate the population mean for a certain set of data representing each company sites' equity. The sample mean is 32, and the error bound for the mean is 4, at a 95% confidence level. (So, x¯=32 and EBM = 4.) Find and interpret the confidence interval estimate.

$\text{we can estimate with 95% confidence that the true value of the population mean is between }\ 28\ \text{ and }36$we can estimate with 95% confidence that the true value of the population mean is between 28 and 36​ We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. The EBM is dependent on the confidence level, which is 95%. This corresponds to the range of data that is two standard deviations away from the mean. Here σ=2 and μ=32. So the confidence interval estimate, at the 95% confidence level, is (32−4,32+4), and calculating these values gives the interval (28,36).

A market researcher is trying to estimate the population mean for a certain set of data representing ad sales. The sample mean is 18 sales, and the error bound for the mean is 3 sales, at a 99.7% confidence level. (So, x¯=18 and EBM = 3.) Find and interpret the confidence interval estimate.

$\text{we can estimate with 99.7% confidence that the true value of the population mean is between }\ 15\ \text{ and }21$we can estimate with 99.7% confidence that the true value of the population mean is between 15 and 21​ We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. The EBM is dependent on the confidence level, which is 99.7%. This corresponds to the range of data that is three standard deviations away from the mean. Here σ=1 and μ=18. So the confidence interval estimate, at the 99.7% confidence level, is (18−3,18+3), and calculating these values gives the interval (15,21).

Suppose days of maintenance per machine per month are normally distributed and have a known population standard deviation of 1 day and an unknown population mean. A random sample of 20 machines is taken and gives a sample mean of 13 days. Find the confidence interval for the population mean with a 80% confidence level. z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576 You may use a calculator or the common z values above. Round your answer to TWO decimal places.

1$12.71$12.71​ 2$13.29$13.29​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=13. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(1.282)(120−−√)≈(1.282)(0.224)≈0.29 So, the error bound (EBM) is 0.29. So we can write this confidence interval as: (13−0.29,13+0.29) or (12.71,13.29).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 13 for x¯ (sample mean), 20 for n (sample size), and 0.8 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 80% confidence that the true population mean is between 12.71 and 13.29 days.

Suppose that the commissions of salesmen, in thousands of dollars, at a dealership are normally distributed with an unknown mean and standard deviation. A random sample of 24 salesmen is taken and gives a sample mean of 27 thousand and a sample standard deviation of 6 thousand. (a) The EBM, margin of error, for a 95% confidence interval estimate for the population mean using the Student's t-distribution is 2.53. (b) Find a 95% confidence interval estimate for the population mean using the Student's t-distribution. Round the final answers to two decimal places.

1$24.47$24.47​ 2$29.53$29.53​ As calculated above, the EBM is EBM=(tα2)(sn−−√)=(2.069)(624−−√)≈(2.069)(1.225)≈2.53 So we can write this confidence interval as: (27−2.53,27+2.53) or (24.47,29.53).Using a TI-83, 83+, or 84+ calculator, press STAT and arrow over to TESTS. Arrow down to 8:TInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 6 for Sx, 27 for xb, 24 for n, and 0.95 for C-level. Arrow down to Calculate and press ENTER. So, we estimate with 95% confidence that the true population mean is between 24.47 and 29.53 inches.

The operating expenses per process in a factory yearly are normally distributed with a population standard deviation of 24 thousand dollars and an unknown population mean. If a random sample of 20 processes is taken and results in a sample mean of 274 thousand dollars, find a 95% confidence interval for the population mean. Round your answer to two decimal places and in the same form as the values in the question. (Do not add extra zeros for thousands or dollar signs.)

1$263.48$263.48​ 2$284.52$284.52​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=274. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(1.96)(2420−−√)≈(1.96)(5.367)≈10.52 So, the error bound (EBM) is 10.52. So we can write this confidence interval as: (274−10.52,274+10.52) or (263.48,284.52).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 274 for x¯ (sample mean), 20 for n (sample size), and 0.95 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 95% confidence that the true population mean is between 263.48 and 284.52 thousand dollars.

Suppose math SAT scores for female students are normally distributed and have a known population standard deviation of 114 and an unknown population mean. A random sample of 60 female students is tested and yields a sample mean of a 499 math SAT score. Find the confidence interval for the population mean with a 95% confidence level. Round your answer to TWO decimal places.

1$470.15$470.15​ 2$527.85$527.85​ EBM=(zα2)(σn−−√) zα=0.052=0.025=zα2=1.960 EBM=(1.960)(11460−−√) =(1.960)(14.72)=28.85=EBM The confidence interval (CI) is:x¯±EBM=499±28.85=(470.15,527.85)

Suppose critical reading SAT scores for male students are normally distributed and have a known population standard deviation of 117 and an unknown population mean. A random sample of 50 male students is tested and yields a sample mean of a 499 critical reading SAT score. Find the confidence interval for the population mean with an 80% confidence level. Round your answer to TWO decimal places.

1$477.79$477.79​ 2$520.21$520.21​ EBM=(zα2)(σn−−√) zα=0.202=0.10=zα2=1.282 EBM=(1.282)(11750−−√) =(1.282)(16.55)=21.21=EBM The confidence interval (CI) is:x¯±EBM=499±21.21=(477.79,520.21)

Suppose math SAT scores for students are normally distributed and have a known population standard deviation of 118 and an unknown population mean. A random sample of 40 students is tested and yields a sample mean of a 514 math SAT score. Find the confidence interval for the population mean with a 90% confidence level. Round your answer to TWO decimal places.

1$483.31$483.31​ 2$544.69$544.69​ EBM=(zα2)(σn−−√) zα=0.102=0.05=zα2=1.645 EBM=(1.645)(11840−−√) =(1.645)(18.66)=30.69=EBM The confidence interval (CI) is:x¯±EBM=514±30.69=(483.31,544.69)

Suppose that the advertising costs of a new product are normally distributed with an unknown mean and standard deviation. A random sample of 21 new products is taken and gives a sample mean of 95 dollars and a sample standard deviation of 8 dollars.

1$91.36$91.36​ 2$98.64$98.64​ As calculated above, the EBM is EBM=(tα2)(sn−−√)=(2.086)(821−−√)≈(2.086)(1.746)≈3.64 So we can write this confidence interval as: (95−3.64,95+3.64) or (91.36,98.64).Using a TI-83, 83+, or 84+ calculator, press STAT and arrow over to TESTS. Arrow down to 8:TInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 8 for Sx, 95 for xb, 21 for n, and 0.95 for C-level. Arrow down to Calculate and press ENTER. So, we estimate with 95% confidence that the true population mean is between 91.36 and 98.64 points.

What value of z should Duane use in his second sample size formula if he needed to be 95% confident?

1.960 $1.960$1.960​ To get a 95% confidence level, instead of using zα, we need to use: zα2=0.052=z0.025=1.960

For quality purposes, a regional produce packaging company is studying the corn production of local farmers. The population standard deviation for the number of corn kernels on an ear of corn is 94 kernels. If we want to be 90% confident that the sample mean is within 17 kernels of the true population mean, what is the minimum sample size that should be taken?

83 The formula for sample size is n=(z2α2σ2)(EBM2) In this formula, z=zzα2 =z0.05 =1.645 because the confidence level is 90%. From the problem, we know that σ=94 and EBM=17. Therefore, n=(z2α2σ2)(EBM2) =(1.6452)(942)(172) ≈82.74 Use n=83 to ensure that the sample size is large enough.

The number of employees per department are normally distributed with a population standard deviation of 198 employees and an unknown population mean. If a random sample of 22 departments is taken and results in a sample mean of 1460 employees, find the error bound (EBM) of the confidence interval with a 80% confidence level. Round your answer to THREE decimal places.

Answer Explanation Correct answers:$\text{EBM=}54.118$EBM=54.118​ We can use the formula to find the error bound: EBM=(zα2)(σn−−√) We know that σ=198 and n=22. We are also given that the confidence level (CL) is 80%, or 0.8. So, we can calculate alpha (α).α=1−CL=1−0.8=0.2Since α=0.2, we know thatα2=0.22=0.1The value of z0.1 is 1.282. Now we can substitute the values into the formula to find the error bound.EBM=(zα2)(σn−−√)=(1.282)(19822−−√)≈(1.282)(42.214)≈54.118So, the error bound (EBM) is 54.118.

Assume the distribution of commute times to a major city follows the normal probability distribution and the standard deviation is 4.5. minutes. A random sample of 38 commute times is given below in minutes. Use Excel to find the 98% confidence interval for the mean travel times in minutes. Round your answers to one decimal place and use ascending order. 21 15 35 40 40 34 32 11 18 35 9 31 15 29 41 37 40 10 31 14 39 14 11 33 38 21 35 33 34 27 31 26 35 27 32 18 40 17

Answer: The 98% confidence interval would be given by (25.9;29.3) Step-by-step explanation: 1) Previous concepts A confidence interval is "a range of values that's likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval". The margin of error is the range of values below and above the sample statistic in a confidence interval. Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean". The data is: 21, 15, 35, 40 ,40 ,34, 32, 11, 18, 35, 9, 31, 15, 29, 41, 37, 40, 10, 31, 14, 39, 14, 11, 33, 38, 21, 35, 33, 34, 27, 31, 26, 35, 27, 32, 18, 40, 17 2) Compute the sample mean and sample standard deviation. In order to calculate the mean and the sample deviation we need to have on mind the following formulas: =AVERAGE(21, 15, 35, 40 ,40 ,34, 32, 11, 18, 35, 9, 31, 15, 29, 41, 37, 40, 10, 31, 14, 39, 14, 11, 33, 38, 21, 35, 33, 34, 27, 31, 26, 35, 27, 32, 18, 40, 17) On this case the average is =STDEV.S(21, 15, 35, 40 ,40 ,34, 32, 11, 18, 35, 9, 31, 15, 29, 41, 37, 40, 10, 31, 14, 39, 14, 11, 33, 38, 21, 35, 33, 34, 27, 31, 26, 35, 27, 32, 18, 40, 17) The sample standard deviation obtained was s=10.184 represent the population standard deviation 3) Find the critical value t* Use the formula for a CI to find upper and lower endpoints In order to find the critical value we need to take in count that our sample size n =38 >30 and on this case we know about the population standard deviation, so on this case we need to use the z distribution. Since our interval is at 98% of confidence, our significance level would be given by and . We can find the critical values in excel using the following formulas: "=NORM.INV(0.01,0,1)" for "=T.INV(1-0.01,0,1)" for The confidence interval for the mean is given by the following formula: And we can use Excel to calculate the limits for the interval Lower interval : "=27.605 -2.33*(4.5/SQRT(38))" =25.904 Upper interval : "=27.605 +2.33*(4.5/SQRT(38))" =29.306 So the 98% confidence interval would be given by (25.904;29.306)

Find the Sample Size Required to Estimate a Population Mean With a Given Confidence Level in Business Applications Calculating Sample Size to Estimate a Population Mean In business statistics research, it is important to establish a balance between the sample size and the allocated budget. A larger sample provides accurate results, but can be expensive. The sample size is critical in consumer research and quality control due to population extent.

Calculating Sample Size for a Given Confidence Level If researchers desire a specific margin of error, then they can use the error bound formula to calculate the required sample size in order to get that error. The error bound formula for a population mean when the population standard deviation is known is: EBM=(zα2)(σn√), where σ is the standard deviation and n is the sample size. So, the formula to find a given sample size is found by solving this error bound formula for n: n=(z2α2σ2)(EBM2) In this formula, the z-value is zα2, which is dependent on the desired confidence level (CL) of the study. The value for α is calculated by subtracting the CL from 1: α=1−CL. We divide α by 2 to find the desired z-value. For example, if we want a 95% confidence level, then α=1−0.95=0.5, and α2=0.025. So, we would use the z-value for z0.025. These z-values, corresponding to confidence levels, can be found in a Standard Normal Table. The most commonly used z-values are below with their corresponding areas under the curve.

A global technology firm offers an entrance exam for all the app developer candidates as part of their job application process. A sample estimated the standard deviation of 2015−2016 scores to be 194 points. You are researching the average exam score. You want to know how many people you should survey if you want to know, at a 98% confidence level, that the sample mean entrance exam score is within 50 points of the true mean entrance exam score. What value for z should you use in the sample size formula?

Correct answers:$2.326$2.326​ zα2 is dependent on the desired confidence level (CL) of the study. The value for α is calculated by subtracting the CL from 1: α=1−CL. We divide α by 2 to find the desired z-value. In this example, we want a 98% confidence level, so α=1−0.98=0.02, and α2=0.022=0.01. So, we would use the z-value for z0.01, which is 2.326.

The birth weights of premature infants in a rural area are normally distributed with an unknown population mean and standard deviation. If a random sample of 41 premature infants is taken to estimate the mean birth weight, what t-score should be used to find a 98% confidence interval estimate for the population mean? Use the portion of the table below or a calculator. If you use a calculator, round your answer to three decimal places.

Correct answers:$2.423$2.423​ Degrees of freedom are n−1=41−1=40. Confidence level: 98% . α=1−CL=1−0.98=0.02 We want the value for α2, which is 0.022=0.01. Using the table, find the row for 40 degrees of freedom, and the column for t0.01. So, the t -score for the 98% confidence interval is 2.423.

Suppose price of seasonal outfits are normally distributed and have a known population standard deviation of 17 dollars and an unknown population mean. A random sample of 15 outfits is taken and gives a sample mean of 308 dollars. Find the confidence interval for the population mean with a 90% confidence level.

Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=308. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(1.645)(1715−−√)≈(1.645)(4.389)≈7.22 So, the error bound (EBM) is 7.22. So we can write this confidence interval as: (308−7.22,308+7.22) or (300.78,315.22).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 308 for x¯ (sample mean), 15 for n (sample size), and 0.9 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 90% confidence that the true population mean is between 300.78 and 315.22 dollars.

Question The commute times for the workers in a city are normally distributed with an unknown population mean and standard deviation. If a random sample of 27 workers is taken and results in a sample mean of 22 minutes and sample deviation of 3 minutes, find a 95% confidence interval estimate for the population mean using the Student's t-distribution.

Correct answer: (20.81,23.19), with EBM=(2.056)(327√) We can calculate the EBM with the formula: EBM=(tα2)(sn−−√) The question tells us s=3 and n=27. We need to find the t-value before we can find EBM. The degrees of freedom are df=n−1=27−1=26. The confidence level is 95%, α=1−0.95=0.05, and α2=0.025. So, we can use the table above to find the t-value for t0.025 with 26 degrees of freedom, which is 2.056. Now we can calculate the EBM. EBM=(tα2)(sn−−√) =(2.056)(3sqrt27) ≈(2.056)(0.577) ≈1.19 So we can write this confidence interval as: (22−1.19,22+1.19) or (20.81,23.19). Using a TI−83, 83+, or 84+ calculator, press STAT and arrow over to TESTS. Arrow down to 8:TInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 3 for Sx, 22 for x, 27 for n, and 0.95 for C-level. Arrow down to Calculate and press ENTER. So, we estimate with 95% confidence that the true population mean wait time for all commuters is between 20.81 and 23.19 minutes.

You are researching the average entrance exam score, and you want to know how many people you should survey if you want to know, at a 98% confidence level, that the sample mean score is within 50 points. From above, we know that the population standard deviation is 194, and z0.01=2.326. What is the minimum sample size that should be surveyed? ​

Correct answer: 82 people From the problem, we know that σ=194 and EBM=50. Because we are looking at a 98% confidence level, we will use the z-value, z0.001=2.326. In order to find the number of people who should be surveyed, we will use the formula and substitute values into the equation. n=(z2α2σ2)(EBM2) =(2.3262)(1942)(502) ≈(5.410)(37636)2500 ≈203621/2500 ≈81.448 We will use n=82. Therefore, 82 people in your town should be surveyed in order to be 98% confident that we are within 50 points of the true population mean score.

The following data set provides a summary statistic of 5,742 interviews of commuters. The percentage of people selecting public transportation in this sample went from 39.1% of those making under $25,000, to 45.5% of those making between $25,000 and $75,000, to 46.8% of those making $75,000 and over. What statement about statistics would be true?

Correct answer: As the income increased, use of public transportation increased. As the income increased from under $25,000 to $75,000 or more, use of public transportation increased from 39.1% to 46.8%.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. Arrange the following states in ascending order of their poverty estimate 90% CI lower bound of the mean for all ages: Florida,Georgia,Illinois,NewYork,Texas

Correct answer: Illinois, Georgia, New York, Florida, Texas Sort the data by the poverty estimate 90% CI lower bound of the population mean for all ages (Column D), and you will find these states just under the leader, California. Listed in ascending order (smallest to largest), you find the order as Illinois, Georgia, New York, Florida, Texas.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. Arrange the following states in ascending order of their sample mean using the poverty estimate 90% CI upper and lower bounds for all ages: Colorado,Maryland,Minnesota,Nevada,Oregon

Correct answer: Nevada, Minnesota, Maryland, Oregon, Colorado Sorting the data by the either of the poverty estimate 90% CI upper or lower bounds of the population mean for all ages, which you have just done in the previous problems, will give you the correct ascending (or descending) order of the sample mean, which is midway between those values. This order is Nevada, Minnesota, Maryland, Oregon, Colorado.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. Arrange the following states in descending order of their poverty estimate 90% CI upper bound of the mean for all ages: Alaska, New Hampshire, North Dakota, Vermont, Wyoming

Correct answer: New Hampshire, North Dakota, Alaska, Vermont, Wyoming Sort the data by the poverty estimate 90% CI upper bound of the population mean for all ages (Column E), and you will find these states. Listed in descending order (largest to smallest), you find the order as New Hampshire, North Dakota, Alaska, Vermont, Wyoming. Or, since the data is symetrical, the data sorted from the previous problem by the poverty estimate 90% CI lower bound of the population mean for all ages will be in the right order as the last five states.

The number of degrees of freedom is the same number as the sample size.

Correct answer: No The degrees of freedom are one unit less than the sample size.

A national sports store, specializing in equipment and clothing, wants to determine the parameters for daily sales per associate. There are stores in 40 states with over 400 locations. The population mean of a set of data is unknown. The sample mean is 30, and the error bound for the mean is 6, at a 68% confidence level. (So, x¯=30 and EBM=6.) Find and interpret the confidence interval estimate.

Correct answer: We can estimate, with 68% confidence that the true value of the population mean is between 24 and 36. We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. The EBM is dependent on the confidence level, which is 68%. This corresponds to the range of data that is one standard deviation away from the mean. Here σ=6 and μ=30. So the confidence interval estimate, at the 68% confidence level, is (30-6,30+6), and calculating these values gives the interval (24,36).

An online store wants to estimate the intervals on the time customers spend on the website before purchasing an item. The population mean of a set of data is unknown. The sample mean is 18 minutes, and the error bound for the mean is 10, at a 95% confidence level. (So, x¯=18 and EBM=10.) Find and interpret the confidence interval estimate.

Correct answer: We can estimate, with 95% confidence that the true value of the population mean is between 8 and 28. We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. The EBM is dependent on the confidence level, which is 95%. This corresponds to the range of data that is two standard deviations away from the mean. Here σ=5 and μ=18. So the confidence interval estimate, at the 95% confidence level, is (18−10,18+10), and calculating these values gives the interval (8,28).

Suppose critical reading SAT scores for female students are normally distributed and have a known population standard deviation of 112 and an unknown population mean. A random sample of 50 female students is tested and yields a sample mean of a 494 critical reading SAT score. Find the confidence interval for the population mean with a 95% confidence level. Round your answer to TWO decimal places.

Correct answers: 462.96,525.04​ EBM=(zα2)(σn−−√) zα=0.052=0.025=zα2=1.960 EBM=(1.960)(11250−−√) =(1.960)(15.84)=31.04=EBM The confidence interval (CI) is:x¯±EBM=494±31.04=(462.96,525.04)

Question The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. A philanthropist is deciding where to locate her foundation to combat poverty. According to the Small Area Income and Poverty Estimates (SAIPE) data, the state with the largest population of children in poverty is California. Using the poverty estimate 90% confidence interval lower and upper bounds for ages 0−17, what is the sample mean? Enter a whole number.

Correct answers: 1,901,985​ The sample mean sits halfway between the lower and upper bounds of the confidence level. The full range is calculated by subtracting the smaller number from the larger. The midpoint will be half that value. 1,928,686−1,875,284=53,402 53,4022=26,701=EBM The mean:1,875,284+26,701=1,901,985

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. According to the SAIPE data, the state with the smallest population of children in poverty is Vermont. Using the poverty estimate 90% confidence interval lower and upper bounds for ages 0−17, what is the sample mean?

Correct answers: 16,009​ The sample mean sits halfway between the lower and upper bounds of the confidence level. The full range is calculated by subtracting the smaller number from the larger. The midpoint will be half that value. 17,370−14,648=2,722 2,7222=1,361=EBM The mean:14,648+1,361=16,009

The following data set provides information on energy usage in the City of Chicago during 2010. Your start-up business is looking into the energy usage in the community of Austin. Your sample size is 12 months. Use the sample mean and standard deviation in killowatts (KWH) for the community Austin. You will first calculate the 90% confidence level for the amount of electricity used. What is the margin of error? Round your final answer to two decimal places.

Correct answers: 3574.20​ df=12−1=11 Margin of error =(tα2)⋅sn√. To find the t-value, we need to use:tα2=0.102=t0.05=1.796Margin of error =1.796⋅6893.8712√≈1.796⋅1990.09≈3,574.20

What value of z should Duane use in his first sample size formula if he needed to be 90% confident?

Correct answers:$1.645$1.645​ To get a 90% confidence level, instead of using zα, we need to use: zα2=0.102=z0.05=1.645

Suppose that the weights, in pounds, of dogs in a city are normally distributed with an unknown mean and standard deviation. The dog weights of 17 randomly sampled dogs are used to estimate the mean of the population. What t-score should be used to find the 90% confidence interval for the population mean?

Correct answers:$1.746$1.746​ The sample used was 17 dogs, so n=17. To find the degrees of freedom: df=n−1=17−1=16 The confidence level is given in the scenario: 90%. So,α=1−CL=1−0.9=0.1But we want to use the value for α2, which is 0.12=0.05. Using the table, we need to find the row for 16 degrees of freedom, and the column for t0.05. So, the t-score we would use to find the 90% confidence interval is 1.746.We could also use a calculator and input 1−α2=0.95 and the degrees of freedom into invT, in which case we would enter invT(0.95,16).

Suppose that the length, in words, of the essays written for a contest are normally distributed with an unknown mean and standard deviation. The essay lengths of 29 randomly sampled essays are used to estimate the mean of the population. What t-score should be used to find the 95% confidence interval for the population mean?

Correct answers:$2.048$2.048​ The sample used was 29 essays, so n=29. To find the degrees of freedom: df=n−1=29−1=28 The confidence level is given in the scenario: 95%. So,α=1−CL=1−0.95=0.05But we want to use the value for α2, which is 0.052=0.025. Using the table, we need to find the row for 28 degrees of freedom, and the column for t0.025. So, the t-score we would use to find the 95% confidence interval is 2.048.We could also use a calculator and input 1−α2=0.975 and the degrees of freedom into invT, in which case we would enter invT(0.975,28).

Suppose that the finishing time for cyclists in a race are normally distributed with an unknown mean and standard deviation. The finishing times of 42 randomly sampled cyclists are used to estimate the mean of the population. What t-score should be used to find the 99% confidence interval for the population mean?

Correct answers:$2.701$2.701​ The sample used was 42 cyclists, so n=42. To find the degrees of freedom: df=n−1=42−1=41 The confidence level is given in the scenario: 99%. So,α=1−CL=1−0.99=0.01But we want to use the value for α2, which is 0.012=0.005. Using the table, we need to find the row for 41 degrees of freedom, and the column for t0.005. So, the t-score we would use to find the 99% confidence interval is 2.701.We could also use a calculator and input 1−α2=0.995 and the degrees of freedom into invT, in which case we would enter invT(0.995,41).

The ratings of customer satisfaction surveys are normally distributed with an unknown population mean and standard deviation. If a random sample of 18 surveys is taken to estimate the mean ratings, what t-score should be used to find a 99% confidence interval estimate for the population mean?

Correct answers:$2.898$2.898​ The sample used was 18 surveys, so n=18. To find the degrees of freedom: df=n−1=18−1=17 The confidence level is given in the scenario: 99%. So,α=1−CL=1−0.99=0.01But we want to use the value for α2, which is 0.012=0.005. Using the table, we need to find the row for 17 degrees of freedom, and the column for t0.005. So, the t-score we would use to find the 99% confidence interval is 2.898.We could also use a calculator and input 1−α2=0.995 and the degrees of freedom into invT, in which case we would enter invT(0.995,17).

The lengths of text messages are normally distributed. If a random sample of text messages is taken and the confidence interval is (17.75, 22.25), what is the sample mean x¯¯¯?

Correct answers:$20.0$20.0​ To find the sample mean x¯¯¯, average the upper and lower endpoints of the confidence interval: 17.75+22.25/2=20.0.

The following data set provides information of the 20 largest cities by population to compare 2000 to 2010. Duane is interested in investing in real estate in a major city. It is important for him to know about the growth of the population, so he does not invest in a stagnant area. He is looking at assumed normally distributed growth since 2000 within the 20 largest cities. This data set has a population standard deviation of 18,603, what is the minimum sample size to be 99% confident that the sample mean is within 100 of the true population mean?

Correct answers:$229,646$229,646​ The first step is to find the z-value. Instead of using zα, we need to use: zα2=z0.005=2.576 The formula for deciding the sample size n is the z-value squared times sigma squared divided by EBM squared:(2.5762)⋅(18,6032)1002n≈229,645.37Rounding up to ensure the minimum sample required is achieved, the minimum sample size is 229,646.

Suppose we know that a confidence interval is (36,42), with a sample mean of 39. Find the error bound (EBM). Give just a number for your answer. For example, if you found that the EBM was 2, you would enter 2.

Correct answers:$3$3​ Since we know the sample mean is 39, we can subtract 39 from the upper value of the confidence interval, 42. 42−39=3 So the error bound is 3.

The largest healthcare system in the state is analyzing the data on geriatric patients to improve quality of services. In this case, the age of patients with coronary conditions. Suppose we know that a confidence interval is (76,82), with a sample mean of 79. Find the error bound (EBM). Give just a number for your answer. For example, if you found that the EBM was 2, you would enter 2.

Correct answers:$3$3​ Since we know the sample mean is 79, we can subtract 79 from the upper value of the confidence interval, 82. 82−79=3 So the error bound is 3.

The following data set provides confidence levels of 90% for poverty levels of 50 states, the whole US, and Washington, D.C. for adults and children. For ages 0−17 in the capital of the United States, the District of Columbia, what is the error bound using the poverty estimate 90% confidence interval lower and upper bounds?

Correct answers:$3,408$3,408​ The full range of the confidence level is calculated by subtracting the smaller number from the larger. The EBM will be half that value. 36,355−29,539=6,816 6,8162=3,408=EBM

Suppose the prices of new snacks are normally distributed. If the population standard deviation is 14 dollars, what minimum sample size is needed to be 95% confident that the sample mean is within 4 dollars of the true population mean? z0.10 - 1.282 z0.05 - 1.645 z0.025 - 1.960 z0.01 - 2.326 z0.005 - 2.576 Use the table above for the z-score. Round only at the final step.

Correct answers:$48\text{ new snacks}$48 new snacks​ The formula for sample size is n=z2σ2EBM2 In this formula,z=zα2=z0.025=1.96because the confidence level is 95%. From the problem, we know that σ=14 and EBM=4. Therefore,n=z2σ2EBM2=(1.96)2(14)242≈47.06Use n=48 to ensure that the sample size is large enough.

The following data set provides information on energy usage in the City of Chicago during 2010. Your start-up business is looking into the energy usage in the community of Austin. Your sample size is 12 months. Use the sample mean and standard deviation in killowatts (KWH) for the community Austin. You will first calculate the 99% confidence level for the amount of electricity used. What is the margin of error?

Correct answers:$6181.22$6181.22​ df=12−1=11 Margin of error =(tα2)⋅sn√. To find the t-value, we need to use:tα2=0.012=t0.005=3.106Margin of error =3.106⋅6893.8712√≈3.106⋅1990.09.≈6,181.22

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant examines the data set about the product sales of cheese as a % of total sales, in which the sample mean is 15.8 and the sample standard deviation is 8.9. Find the 68% confidence interval. Round your answer to ONE decimal place.

Correct answers:$\text{68% confidence interval estimate }\ 6.9\ \text{ , }24.7$68% confidence interval estimate 6.9 , 24.7​ 68% of the area under the curve is one standard deviation away from the mean. EBM=σ x¯−EBM=15.8−8.9=6.9 x¯+EBM=15.8+8.9=24.7 We can say that the 68% confidence interval is (6.9,24.7).

US workers earning between $25,000 and $75,000 who did not own vehicles had seven commuting choices normally distributed with a population standard deviation of 276.3 people and an unknown population mean. If a random sample of 1925 workers is taken and results in a sample mean of 275.1 square feet, find the error bound (EBM) of the confidence interval with a 99% confidence level. Round your answer to TWO decimal places. z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576

Correct answers:$\text{EBM=}16.22$EBM=16.22​ The first step is to find the z-value. Instead of using zα, we need to use: zα2=0.012=z0.005=2.576 EBM=zα2⋅σn−−√ =2.576⋅276.31925−−−−√ =2.576⋅6.30 EBM=16.22

The overtime, in hours, per month of managers is normally distributed with an unknown population mean and standard deviation. A random sample of 27 managers is taken and results in a sample mean of 20 hours per month and sample standard deviation of 6 hours per month. (a) Find the EBM, margin of error, for a 98% confidence interval estimate for the population mean using the Student's t-distribution. Use the portion of the table above or a calculator. Round the final answer to two decimal places.

Correct answers:$\text{ebm=}2.86$ebm=2.86​ We can calculate the EBM with the formula: EBM=(tα2)(sn−−√) The question tells us s=6 and n=27. We need to find the t-value before we can find EBM. The degrees of freedom aredf=n−1=27−1=26The confidence level is 98%, so α=1−0.98=0.02, and α2=0.01. Now, we can use the table above to find the t-value for t0.01 with 26 degrees of freedom, which is 2.479. Now we can calculate the EBM.EBM=(tα2)(sn−−√)=(2.479)(627−−√)≈(2.479)(1.155)≈2.86

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant is looking to expand her business. She looks at the data set about cheese establishments in six categories, in which the sample mean is 3,324.3 and the sample standard deviation is 2,463.8. Find the lowest level of the 68% confidence interval estimate. Round your answer to ONE decimal place.

Correct answers:$\text{lowest level of the 68% confidence interval estimate=}860.5$lowest level of the 68% confidence interval estimate=860.5​ The lower bound for 68% is one standard deviation away from the mean. EBM=σ x¯−EBM=3324.3−2463.8=860.5

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant examines the data set about the product sales of cheese as a % of total sales, in which the sample mean is 15.8 and the sample standard deviation is 8.9. What is the rounded upper level of the 95% confidence interval estimate.

Correct answers:$\text{rounded upper level of the 95% confidence interval estimate=}33.6$rounded upper level of the 95% confidence interval estimate=33.6​ The upper bound for 95% is two standard deviations away from the mean. EBM=2⋅σ EBM=2⋅8.9=17.8 x¯+EBM=15.8+17.8=33.6

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant looks again at the data set about total cheese sales, in which the sample mean is 164.5 and the sample standard deviation is 103. Find the rounded upper level of the 68% confidence interval estimate. Round your answer to ONE decimal place.

Correct answers:$\text{the rounded upper level of the 68% confidence interval estimate=}267.5$the rounded upper level of the 68% confidence interval estimate=267.5​ The upper bound for 68% is one standard deviation away from the mean. EBM=σ x¯+EBM=164.5+103=267.5

The following data set provides information on wholesale sales by establishments and by total sales. A cheese merchant is looking to expand her business. She looks at the data set about cheese establishments in six categories, in which the sample mean is 3,324.3 and the sample standard deviation is 2,463.8. Find the rounded upper level of the 68% confidence interval estimate. Round your answer to ONE decimal place.

Correct answers:$\text{the rounded upper level of the 68% confidence interval estimate=}5788.1$the rounded upper level of the 68% confidence interval estimate=5788.1​ The upper bound for 68% is one standard deviation away from the mean. EBM=σ x¯+EBM=3324.3+2463.8=5788.1

The population mean of a set of data is unknown. The sample mean is 34, and the error bound for the mean is 4, at a 95% confidence level. (So, x¯=34 and EBM = 4.) Find and interpret the confidence interval estimate.

Correct answers:$\text{we can estimate with 95% confidence that the true value of the population mean is between }\ 30\ \text{ and }38$we can estimate with 95% confidence that the true value of the population mean is between 30 and 38​ We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. The EBM is dependent on the confidence level, which is 95%. This corresponds to the range of data that is two standard deviations away from the mean. Here σ=2 and μ=34. So the confidence interval estimate, at the 95% confidence level, is (34−4,34+4), and calculating these values gives the interval (30,38).

Suppose net gain, in dollars, of the departments for an industry per day are normally distributed and have a known population standard deviation of 325 dollars and an unknown population mean. A random sample of 20 departments is taken and gives a sample mean of 1640 dollars. Find the confidence interval for the population mean with a 98% confidence level. Round your answer to TWO decimal places.

Correct answers:1$1470.96$1470.96​2$1809.04$1809.04​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=1640. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(2.326)(32520−−√)≈(2.326)(72.672)≈169.04 So, the error bound (EBM) is 169.04. So we can write this confidence interval as: (1640−169.04,1640+169.04) or (1470.96,1809.04).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 325 for σ (standard deviation), 1640 for x¯ (sample mean), 20 for n (sample size), and 0.98 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 98% confidence that the true population mean is between 1470.96 and 1809.04 dollars.

Question The overtime, in hours, per month of managers is normally distributed with an unknown population mean and standard deviation. A random sample of 27 managers is taken and results in a sample mean of 20 hours per month and sample standard deviation of 6 hours per month. (a) The EBM, margin of error, for a 98% confidence interval estimate for the population mean using the Student's t-distribution is 2.86. (b) Find a 98% confidence interval estimate for the population mean using the Student's t-distribution. Round the final answers to two decimal places.

Correct answers:1$17.14$17.14​2$22.86$22.86​ As calculated above, the EBM is EBM=(tα2)(sn−−√)=(2.479)(627−−√)≈(2.479)(1.155)≈2.86 So we can write this confidence interval as: (20−2.86,20+2.86) or (17.14,22.86).Using a TI-83, 83+, or 84+ calculator, press STAT and arrow over to TESTS. Arrow down to 8:TInterval. Press ENTER. Arrow to Stats and press ENTER. Arrow down and enter 6 for Sx, 20 for xb, 27 for n, and 0.98 for C-level. Arrow down to Calculate and press ENTER. So, we estimate with 98% confidence that the true population mean is between 17.14 and 22.86 pounds.

Question Suppose heights, in inches, of orangutans are normally distributed and have a known population standard deviation of 4 inches and an unknown population mean. A random sample of 16 orangutans is taken and gives a sample mean of 56 inches. Find the confidence interval for the population mean with a 95% confidence level. Round your answer to TWO decimals.

Correct answers:1$54.04$54.04​2$57.96$57.96​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=56. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(1.96)(416−−√)≈(1.96)(1.000)≈1.96 So, the error bound (EBM) is 1.960. So we can write this confidence interval as: (56−1.96,56+1.96) or (54.04,57.96).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 56 for x¯ (sample mean), 16 for n (sample size), and 0.95 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 95% confidence that the true population mean is between 54.04 and 57.96 inches.

Suppose overtime of employees at Company XYZ are normally distributed and have a known population standard deviation of 4 hours per month and an unknown population mean. A random sample of 21 employees is taken and gives a sample mean of 61 hours per month. Find the confidence interval for the population mean with a 98% confidence level. Round your answer to TWO decimal places.

Correct answers:1$58.97$58.97​2$63.03$63.03​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=61. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(2.326)(421−−√)≈(2.326)(0.873)≈2.03 So, the error bound (EBM) is 2.03. So we can write this confidence interval as: (61−2.03,61+2.03) or (58.97,63.03).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 61 for x¯ (sample mean), 21 for n (sample size), and 0.98 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 98% confidence that the true population mean is between 58.97 and 63.03 hours per month.

The following data set provides information on energy usage in the City of Chicago during 2010. Your start-up business is looking into the energy usage in the community of South Lawndale. Your sample size is 12 months. Use the sample mean and standard deviation in killowatts (KWH) for the community South Lawndale. What is the confidence interval at the 95% level? Use the portion of the table below. Round the final answers to two decimal places.

Correct answers:1$5900.84$5900.84​2$8254.16$8254.16​ df=12−1=11 EBM=(tα2)⋅sn−−√ To find the t-value, we need to use:tα2=0.052=t0.025=2.201EBM=2.201⋅1851.9112−−√≈2.201⋅534.60≈1176.66Next, we add and subtract the EBM from the mean to find the 95% confidence interval:7077.5±1176.66=(5900.84,8254.16)

Suppose scores of a standardized test are normally distributed and have a known population standard deviation of 6 points and an unknown population mean. A random sample of 22 scores is taken and gives a sample mean of 92 points. Find the confidence interval for the population mean with a 98% confidence level. Round your answer to TWO decimal places.

Correct answers:1$89.02$89.02​2$94.98$94.98​ Confidence intervals are written as (x¯−EBM,x¯+EBM), so we need the sample mean, x¯, and the EBM. We know x¯=92. We can use the formula to find the error bound: EBM=(zα2)(σn−−√)=(2.326)(622−−√)≈(2.326)(1.279)≈2.98 So, the error bound (EBM) is 2.975. So we can write this confidence interval as: (92−2.98,92+2.98) or (89.02,94.98).Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 7: ZInterval, and press ENTER. Arrow to Stats, and press ENTER. Arrow down and enter 6 for σ (standard deviation), 92 for x¯ (sample mean), 22 for n (sample size), and 0.98 for the C-Level (confidence level). Arrow down to Calculate, and press ENTER. So we estimate with 98% confidence that the true population mean is between 89.02 and 94.98 points.

Which of the following terms is not a synonym for the other three terms?

Critical value times standard error of the mean Error bound of the population mean Margin of error Correct answer: Error bound of the sample mean Choices A, C, and D are synonyms. An error bound of the mean (EBM) deals with the population mean, not the sample mean.

Find and Interpret Confidence Interval Estimates in Business Examples Using the Empirical Rule Understanding Confidence Intervals Confidence Intervals In business, the population size could be unknown, affecting the accuracy of data interpretation. The use of confidence intervals provides an estimate with a set of values for a better estimation instead of a single point. The applications in business related fields are substantial from market research to financial investors to risk management. Market researchers can estimate a set of values a product could sell for with a predetermined percentage of confidence.

Introduction and Point Estimates In inferential statistics, we use sample data to make generalizations about an unknown population. The sample data helps us to make an estimate of a population parameter. Suppose you were trying to determine the mean rent of a two-bedroom apartment in your city. You might look in the newspaper or on the internet, write down several rent prices, and then find their average. This is called a point estimate of the true mean. You can do the same with a proportion as well. If you are trying to determine the percentage of times you make a shot when shooting a basketball, you could count the number of shots you make and divide that by the total number of shots you took. This is called a point estimate of the true proportion. The sample mean, x¯, is the point estimate for the population mean, μ. The sample standard deviation, s, is the point estimate for the population standard deviation, σ. Each of the point estimates are statistics, because they are data about the sample. We realize that the point estimate is usually not the exact value of the population parameter, but it is very close to it, and we can use it to make decisions about the population. After calculating point estimates, we can construct interval estimates, called confidence intervals. A confidence interval is another type of estimate but, instead of being just one number, it is an interval of numbers. The interval of numbers is a range of values calculated from a given set of sample data. It is important to keep in mind that the confidence interval is a random variable, and it's the population parameter that is fixed. Margins of Error A confidence interval is created for an unknown population parameter, so confidence intervals for some parameters have the form: (point estimate - margin of error, point estimate + margin of error) ...with the value of (point estimate - margin of error) as the lower value, and the (point estimate + margin of error) as the upper value. The margin of error here, depends on the confidence level (the percentage of confidence). The empirical rule, which applies to normal distributions, says that an approximate percentage of the samples will lie within a certain amount of standard deviations from the population mean, μ. This determines the confidence level. For example, if the standard deviation of a normal distribution is 3, then we can estimate that 95% of the sample data lie within 6 units of the mean. This means that the margin of error would be 6 at a 95% confidence level. The margin of error is called the error bound for the population mean, abbreviated as EBM. Writing a Confidence Interval Suppose a marketing department of an entertainment company is interested in the mean number of songs a customer downloads per month from iTunes. They want to conduct a survey and calculate the sample mean number of songs, x¯, in order to make an estimate for the population mean, μ, which is unknown. The sample mean, x¯, is the point estimate for the population mean, μ. As stated above, a confidence interval is written as: (point estimate - margin of error, point estimate + margin of error), or in symbols... (x¯−EBM,x¯+EBM) So, we need a point estimate for μ (because it is unknown), and the margin of error, which is dependent on the confidence level, in order to write the confidence interval. Let's suppose we know the sample mean, x¯=2, the sample size, n=100, and the population standard deviation, σ=1. We can use the central limit theorem, to estimate the standard deviation for the sample mean. The CLT for Means says that σ=σXn√=1100√=0.1. So one standard deviation is equal to 0.1. Now that we know the standard deviation, we need to define the confidence level, and use the Empirical Rule, to determine the confidence interval. Let's assume that the marketing company wants a 95% confidence level. How To The Empirical Rule states that approximately: 68% of the data lies within one standard deviation of the mean. 95% of the data is within two standard deviations of the mean. 99.7% of the data is within three standard deviations of the mean. So, in this example, two standard deviations is 2σ=(2)(0.1)=0.2. So, the error bound for the population mean (EBM) is 0.2. This means that the sample mean, x¯ is likely to be within 0.2 units of μ. So, now that we have a point estimate for the mean, x¯=2, and the EBM=0.2 , we can write a confidence interval. The population mean, μ, is contained in an interval whose lower value is x¯−EBM, and the upper value is x¯+EBM. (x¯−EBM,x¯+EBM) (2−0.2,2+0.2) (1.8,2.2) So, we can say that the 95% confidence interval is (1.8,2.2). We are 95% confident that the unknown population mean number of songs downloaded from iTunes per month (μ) is between 1.8 and 2.2.

Question The population standard deviation for the age of Foothill College students is 15 years. If we want to be 90% confident that the sample mean age is within 2 years of the true population mean age of Foothill College students, how many randomly selected Foothill College students must be surveyed?

Solution From the problem, we know that σ=15 and EBM=2. Because we are looking at a 90% confidence level, we can compute α and find the appropriate z-value. α=1−0.90=0.10, which means α2=0.05. Referencing the table above, we can see that z0.05=1.645. In order to find the number of students who must be surveyed (sample size), we need to use the formula and substitute values into the equation to solve for n. n=(z2α2σ2)(EBM2) =(1.6452)(152)(22) ≈(2.706)(225)4 ≈608.854 ≈152.21 So, using the sample size equation, n=152.21. But since we are talking about numbers of students, we cannot use a decimal. So, we will use n=153: Always round the answer UP to the next higher integer to ensure that the sample size is large enough. Therefore, 153 Foothill College students should be surveyed in order to be 90% confident that we are within two years of the true population mean age of Foothill College students.

Question Dr. Lasso performs a study to determine how effective acupuncture is in relieving pain. He measures sensory rates for a random sample of 22 patients. Suppose the sensory rates are normally distributed, with a sample mean of 8 and sample standard deviation of 1.7. Find a 99% confidence interval estimate for the population mean using the Student's t-distribution.

Solution Using t-scores: To find the confidence interval, you need the sample mean, x¯, and the EBM. The problem gives us the sample mean: x¯=8. We can calculate the EBM with the formula: EBM=(tα2)(sn−−√) We know the values for s and n. The question tells us s=1.7 and n=22. We need to find the t-value before we can find EBM. The t-value is dependent on the degrees of freedom and the confidence level. The degrees of freedom are 1 less than the sample size, which is 22: df=n−1=22−1=21. The confidence level is 99%, α=1−0.99=0.01, and α2=0.005. So, we can use the table above to find the t-value for t0.005 with 21 degrees of freedom, which is 2.831. Now we can calculate the EBM. EBM=(tα2)(sn−−√) =(2.831)(1.722−−√) ≈(2.831)(0.362) ≈1.026 Confidence intervals are written as (x¯−EBM,x¯+EBM). So we can write this confidence interval as: (8−1.026,8+1.026) or (6.974,9.026). So, we estimate with 99% confidence that the true population mean sensory rate for all patients is between 6.974 and 9.026. Using a calculator: Press STAT and arrow over to TESTS. Arrow down to 8:TInterval and press ENTER. Arrow to Data and press ENTER. Arrow down to List and enter the list name where you put the data. (There should be a 1 after Freq.) Arrow down to C-level and enter 0.99, or the desired confidence level. Arrow down to Calculate and press ENTER. The confidence interval to three decimal places is (6.974,9.026). This means that we estimate with 99% confidence that the true population mean sensory rate for all patients is between 6.974 and 9.026.

Question The population mean of a set of data is unknown. The sample mean is 7, and the error bound for the mean is 2.5, at a 95% confidence level. (So, x¯=7 and EBM=2.5.) Find and interpret the confidence interval estimate.

Solution We can find the range for the confidence interval with the point estimate for the mean, and the error bound for the mean. The confidence interval estimate ranges from the value of the point estimate - EBM as the lower value, to the point estimate + EBM as the upper value. So the confidence interval estimate is (7-2.5,7+2.5), and calculating the values gives (4.5,9.5). Because the confidence level is 95%, this means that we can estimate, with 95% confidence that the true value of the population mean is between 4.5 and 9.5.

The following data sets provide borough populations in New York City for 2000 and 2010 and information of the 20 largest cities by population to compare 2000 to 2010. Why would the minimum sample size from the New York City data set sample calculation in 2010 be so much bigger than the 20 largest cities sample size calculation in 2000?

The first set of data has a larger population standard deviation. The first set of data has a larger population standard deviation, due to a larger total population. The first set had a larger, not smaller, EBM. The confidence level was set lower in the first set, but that did not have an effect of increasing the sample size. It kept the minimum sample size smaller than it could have been.

The following data set provides information on energy usage in the City of Chicago during 2010. Your start-up business is looking into the energy usage in the community of South Lawndale. Your sample size is 12 months. Use the sample mean and standard deviation in killowatts (KWH) for the community South Lawndale. What is the confidence interval at the 80% level? Use the portion of the table below. Round the final answers to two decimal places.

df=12−1=11 EBM=(tα2)⋅sn−−√ To find the t-value, we need to use: tα2=0.202=t0.10=1.363 EBM=1.363⋅1851.9112−−√ ≈1.363⋅534.60 ≈728.66 Next, we add and subtract the EBM from the mean to find the80%confidence interval: 7077.5±728.66=(63,48.84 , 78,06.16)