Alta - Chapter 8 - Confidence Intervals - Part 2

Suppose a repairman wants to determine the current percentage of customers who keep up with regular house maintenance. How many customers should the repairman survey in order to be 98% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who keep up with regular house maintenance? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table of values above. Round only at the final step.

$542$542​ Given the information in the question, EBP=0.05 since 5%=0.05 and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=2.3262(0.5)(0.5)0.052=541.03 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 542 customers.

The NWBC found that 67.6% of women-owned businesses provided employees health insurance. What sample size could be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion?

$659$659​ EBP = 3%. α=0.10. The z-score is found 0.102=0.05=zα2=1.645 n=(zα2)2(˙p′)(q′)EBP2 =(1.6452)(0.676)(0.324)0.032 n=658.54 Rounded up to 659 people.

A tax assessor wants to assess the mean property tax bill for all homeowners in a certain state. From a survey ten years ago, a sample of 28 property tax bills is given below. Assume the property tax bills are approximately normally distributed. Use Excel to construct a 95% confidence interval for the population mean property tax bill. Round your answers to two decimal places and use increasing order. Property Tax Bill 1715.35, 1157.82, 651.10, 1093.36, 1609.22, 1391.47, 1398.89, 402.99, 1028.76, 1571.23, 529.42, 1579.48, 699.56, 2222.16, 838.60, 866.41, 1066.60, 1498.64, 1784.28, 2016.99, 2465.65, 1098.81, 1346.65, 1574.00, 1447.78, 2018.33, 2156.75, 1710.81

$\left(1185.91,\ 1595.59\right)$(1185.91, 1595.59)​ Use Excel to calculate the 95% confidence interval, where α=0.05 and n=28. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 1390.75 and the sample standard deviation, rounded to two decimal places, is 528.27. 2. Click on any empty cell, enter =CONFIDENCE.T(0.05,528.27,28), and press ENTER. 3. The margin of error, rounded to two decimal places, is 204.84. The confidence interval for the population mean has a lower limit of 1390.75−204.84=1185.91 and an upper limit of 1390.75+204.84=1595.59. Thus, the 95% confidence interval for the population mean property tax bill is (1185.91, 1595.59).

A company wants to determine a confidence interval for the average CPU time of its teleprocessing transactions. A sample of 70 random transactions in milliseconds is given below. Assume that the transaction times follow a normal distribution with a standard deviation of 600 milliseconds. Use Excel to determine a 98% confidence interval for the average CPU time in milliseconds. Round your answers to the nearest integer and use ascending order.

$\left(5,906,\ 6,240\right)$(5,906, 6,240)​ A 98% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.02, σ=600, and n=70. Use Excel to calculate the 98% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to the nearest integer, is x¯=6,073. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,600,70), and press ENTER. 3. The margin of error, rounded to the nearest integer, is zα/2σn−−√≈167. The confidence interval for the population mean has a lower limit of 6,073−167=5,906 and an upper limit of 6,073+167=6,240. Thus, the 98% confidence interval for μ is (5,906, 6,240).

A recent survey asked 1,379 top executives about business trends. The surveyed showed that 23% want to strengthen innovation to capitlize on new opportunities. What is the confidence interval at the 99% level?

1$0.201$0.201​ 2$0.259$0.259​ First find the z-score. For the 99% confidence level, it is z0.012=z0.005=2.576. Then calculate the Error Bound for Proportion. EBP = z-score times the square root of (p′⋅q′) divided by the sample size n. EBP=2.576⋅(0.23)(0.77)1379−−−−−−−−−−√=0.029 The confidence interval:0.23±0.029=(0.201,0.259)

A recent survey had 38% of the 1,379 surveyed CEOs that were very confident about their company's prospects for revenue growth over the next 12 months. What is the confidence interval at the 90% level? z0.10z0.05z0.025z0.01z0.0051.2821.6451.9602.3262.576 Use the table of common z-scores above. Round your answers to three decimal places.

1$0.358$0.358​ 2$0.402$0.402​ First find the z-score. For the 90% confidence level, it is z0.102=z0.05=1.645. Then calculate the Error Bound for Proportion. EBP = z score times the square root of (p′⋅q′) divided by the sample size n. EBP=1.645⋅(0.38)(0.62)1379−−−−−−−−−−√=0.022 The confidence level:0.38±0.022=(0.358,0.402)

Using p′=0.684, q′=0.316, and n=380, what is the 90% confidence interval for the proportion of the population who pay for Company ABC's healthcare?

1$0.645$0.645​ 2$0.723$0.723​ We know n=380, p′=0.684, and q′=0.316. Since the confidence level is 90%, α=1−0.9=0.1, and α2=0.05. The z-value for z0.05 is 1.645. EBP=(zα2)(p′q′n−−−−√)=(1.645)((0.684)(0.316)380−−−−−−−−−−−−√)≈(1.645)(0.000569−−−−−−−√)≈0.039 So we can write the confidence interval as (0.684−0.039,0.684+0.039)=(0.645,0.723). We estimate with 90% confidence that the true population proportion of people who pay for Company ABC's healthcare is between 0.645 and 0.723.

Researchers wanted to compare the time taken to complete a task using the non-dominant hand for right-handed people and left-handed people. One group of 72 right-handed people took an average of 25 seconds to write a provided short sentence using their left hand. Another group of 62 left-handed people took an average of 19 seconds to write the same sentence using their right hand. Assume that the population standard deviations of times taken to write the sentence are 1.8 seconds for the righties and 1.5 seconds for the lefties. Find the 95% confidence interval for the difference in population mean time to write the sentence between righties and lefties. Let the right-handed people be the first sample, and let the left-handed people be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

1$5.44\pm0.01$5.44±0.01​ 2$6.56\pm0.01$6.56±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=25, σ1=1.8, and n1=72, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=19, σ2=1.5, and n2=62, respectively. According to technology or a z-table, for a 95% confidence interval, zc=1.96. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(25−19)−1.961.8272+1.5262−−−−−−−−−−√≈5.44 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(25−19)+1.961.8272+1.5262−−−−−−−−−−√≈6.56 Using TI-83 or TI-84 calculators or Google Sheets to find the exact interval with no intermediate rounding also produces a confidence interval of (5.44,6.56).

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 time in minutes. Round your answers to one decimal place and use ascending order.

A 98% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.02, σ=4.5, and n=38. Use Excel to calculate the 98% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to one decimal place, is x¯=27.6. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,4.5,38), and press ENTER. 3. The margin of error, rounded to one decimal place, is zα/2σn−−√≈1.7. The confidence interval for the population mean has a lower limit of 27.6−1.7=25.9 and an upper limit of 27.6+1.7=29.3. Thus, the 98% confidence interval for μ is (25.9, 29.3).

Question Suppose an internet marketing company wants to determine the current percentage of customers who click on ads on their smartphones. How many customers should the company survey in order to be 98% confident that the estimated proportion is within 5 percentage points of the true population proportion of customers who click on ads on their smartphones?

Answer Explanation Correct answer: 542 customers From the problem, we know that EBP=0.05 (or 5%). We are also given the confidence interval, 98%, so we can find the value for zα2 from the table. α=1−CL=1−.98=0.02. So, α2=0.022=0.01 From the table, z0.01=2.326. We make p′ and q′ both equal to 0.5 because this will yield the largest possible product. To calculate the sample size, use the formula and make the substitutions. n=(zα2)2(p'q')(EBP2) =(2.326)2(0.5)(0.5)(0.052) ≈(5.410)(0.25)0.0025 ≈541.0276 We will round up here, so the sample size should be 542 customers.

A news organization interested in chronicling winter holiday travel trends conducted a survey. Of the 96 people surveyed in the eastern half of a country, 42 said they fly to visit family members for the winter holidays. Of the 108 people surveyed in the western half of the country, 81 said they fly to visit family members for the winter holidays. Construct a 99% confidence interval for the difference in population proportions of people in the eastern half of a country who fly to visit family members for the winter holidays and people in the western half of a country who fly to visit family members for the winter holidays. Assume that random samples are obtained and the samples are independent. (Round your answers to three decimal places.)

Correct answer: (−0.481,−0.144) In this scenario, the sample proportion and sample size for the first sample are p^1=4296=0.4375 and n1=96, respectively. Then q^1=1−0.4375=0.5625. The sample proportion and sample size for the second sample are p^2=81108=0.75 and n2=108, respectively. Then q^2=1−0.75=0.25. According to the table, for a 99% confidence interval zc=2.575. Substitute these values into the expressions for the bounds of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.4375−0.75)−2.5750.4375(0.5625)96+0.75(0.25)108−−−−−−−−−−−−−−−−−−−−−−−√≈−0.481 (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.4375−0.75)+2.5750.4375(0.5625)96+0.75(0.25)108−−−−−−−−−−−−−−−−−−−−−−−√≈−0.144 The 99% confidence interval is approximately (−0.481,−0.144). This means we are 99% confident that the true difference in population proportions of people in the eastern half of a country who fly to visit family members for the winter holidays and people in the western half of a country who fly to visit family members for the winter holidays is between −48.1% and −14.4%. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

Question The Pew Social Media Update 2014 surveyed 1,597 adult internet users on which social media websites they use. Of the users surveyed, 1,134 responded "yes" when asked if they use Facebook. From the question above, we know that p′=0.71. What is the error bound for proportions (EBP) of a confidence interval with a 99% confidence level?

Correct answer: 0.029 We know the sample size, n, is the entire sample of users surveyed. So, n=1597. We also know (from the first example) that p′=0.71, and so we can find the value of q′, because q′=1−p′. So, q′=1−0.71=0.29. We are given the confidence level (CL), 99%, or 0.99. So, we can calculate alpha (α). α=1−CL =1−0.99 =0.01 Since α=0.01, we know that α2=0.012=0.005. So we will look for the value of z0.005. From the table above, z0.005=2.576. Now we can substitute the values into the formula to find the error bound. EBP=(zα2)(p′q′n−−−−√) =(2.576)((0.71)(0.29)(1597)−−−−−−−−−−√) ≈(2.576)(0.20591597−−−−−−√) ≈(2.576)(0.000129−−−−−−−√) ≈0.029 So, the error bound (EBP) is 0.029.

Question Suppose a pharmaceutical industry wants to determine the current percentage of customers who are hospitals. How many customers should the industry survey in order to be 92% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who are hospitals?

Correct answers:$480$480​ Given the information in the question, EBP=0.04 since 4%=0.04 and zα2=z0.04=1.751 because the confidence level is 92%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=1.7512(0.5)(0.5)0.042=479.1 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 480 customers.

The NWBC found that 42.1% of women-owned businesses provided retirement plans contributions. What sample size could be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion?

Correct answers:$586$586​ EBP = 4%. α=0.05. The z-score is found 0.052=0.025=zα2=1.960 n=(zα2)2(˙p′)(q′)EBP2 =(1.9602)(0.421)(0.579)0.042 n=585.27 Rounded up to 586 people.

In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Excel to find the 90% confidence interval for the population mean temperature. Assume the temperatures are approximately normally distributed. Round your answers to two decimal places and use increasing order.

Correct answers:$\left(153.21,\ 161.43\right)$(153.21, 161.43)​ Use Excel to calculate the 90% confidence interval, where α=0.1 and n=22. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean is 157.32, and the sample standard deviation, rounded to two decimal places, is 11.21. 2. Click on any empty cell, enter =CONFIDENCE.T(0.1,11.21,22), and press ENTER. 3. The margin of error, rounded to two decimal places, is 4.11. The confidence interval for the population mean has a lower limit of 157.32−4.11=153.21 and an upper limit of 157.32+4.11=161.43. Thus, the 90% confidence interval for the population mean temperature is (153.21, 161.43).

The number of hours worked per year per adult in a state is normally distributed with a standard deviation of 37. A sample of 115 adults is selected at random, and the number of hours worked per year per adult is given below. Use Excel to calculate the 98% confidence interval for the mean hours worked per year for adults in this state. Round your answers to two decimal places and use ascending order.

Correct answers:$\left(2,090.03,\ 2,106.09\right)$(2,090.03, 2,106.09)​ A 98% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.02, σ=37, and n=115. Use Excel to calculate the 98% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=2,098.06. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,37,115), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn−−√≈8.03. The confidence interval for the population mean has a lower limit of 2,098.06−8.03=2,090.03 and an upper limit of 2,098.06+8.03=2,106.09. Thus, the 98% confidence interval for μ is (2,090.03, 2,106.09).

The amounts of time that customers stay in a certain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sample of 50 lunch customers was taken at this restaurant. Construct a 99% confidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to two decimal places and use ascending order.

Correct answers:$\left(44.89,\ 57.27\right)$(44.89, 57.27)​ A 99% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.01, σ=17, and n=50. Use Excel to calculate the 99% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=51.08. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.01,17,50), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn−−√≈6.19. The confidence interval for the population mean has a lower limit of 51.08−6.19=44.89 and an upper limit of 51.08+6.19=57.27. Thus, the 99% confidence interval for μ is (44.89, 57.27).

Using p′=0.167, q′=0.833, and n=180, what is the 95% confidence interval for the proportion of the population who prefer brand named items?

Correct answers:1$0.113$0.113​2$0.221$0.221​ We know n=180, p′=0.167, and q′=0.833. Since the confidence level is 95%, α=1−0.95=0.05, and α2=0.025. The z-value for z0.025 is 1.96. EBP=(zα2)(p′q′n−−−−√)=(1.96)((0.167)(0.833)180−−−−−−−−−−−−√)≈(1.960)(0.000772−−−−−−−√)≈0.054 So we can write the confidence interval as (0.167−0.054,0.167+0.054)=(0.113,0.221). We estimate with 95% confidence that the true population proportion of people who prefer brand named items is between 0.113 and 0.221.

Question Suppose that a research firm wants to estimate the percent of families living in a city who have pets. Five hundred randomly selected families in this city are surveyed. Of the 500 families surveyed, 321 responded "yes" - they own a pet. What is the value of p′, the estimated proportion of successes?

Solution To form a proportion, take the number of successes, and divide it by n, the number of trials. In this scenario, the number of successes is the number of families who have pets - 321. The total number of families surveyed was 500. p'=xn=321500=0.642 So, p′=0.642 is the sample proportion, which is the point estimate of the population proportion.

Finding the Sample Proportion (p′)

To form a proportion, take X, the random variable for the number of successes, and divide it by n, the number of trials. We write this "new" random variable as p′(read as "p prime"). p'=x/n p′ is the estimated proportion of successes, and will be used as the point estimate when we write confidence intervals. In the next section, we will use q′, which is the estimated proportion of failures, where q′=1−p′.

Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who are over the age of forty?

$423$423​ Given the information in the question, EBP=0.04 since 4%=0.04 and zα2=z0.05=1.645 because the confidence level is 90%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=1.6452(0.5)(0.5)0.042=422.8 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 423 customers.

Confidence interval: the region formed between a point estimate minus the margin or error and the point estimate plus the margin of error Population proportion: the percentage of a population that meets a certain criteria or test, represented by p

Confidence interval: the region formed between a point estimate minus the margin or error and the point estimate plus the margin of error Population proportion: the percentage of a population that meets a certain criteria or test, represented by p

Question The Pew Social Media Update 2014 surveyed 1,597 adult internet users on which social media websites they use. Of the users surveyed, 1,134 responded "yes" when asked if they use Facebook. What is the confidence interval, at the 99% confidence level, for the proportion of the population of households who have more than one vehicle?

Correct answer: (0.71−0.029,0.71+0.029)=(0.681,0.739) We can use the information in the question to define some parameters: n=1597 (sample size) p′ =11341597=0.71 (sample proportion of successes) q′=1−p′=1−0.71=0.29 (sample proportion of failures) Since the confidence level is 99%, α=1−0.99=0.01, and α2=0.005. The z-value for z0.005 is 2.576. Now we can calculate the EBP. EBP=(zα2)(p′q′n−−−−√) =(2.576)((0.71)(0.29)1597−−−−−−−−−−√) ≈(2.576)(0.000129−−−−−−−√) ≈0.029 Confidence intervals are written as (p'−EBP,p'+EBP). So we can write this as: (0.71−0.029,0.71+0.029) or (0.681,0.739). So, we estimate with 99% confidence that the true population proportion of households who have more than one vehicle is between 0.681 and 0.739.

The NWBC found that 16.5% of women-owned businesses did not provide any employee benefits. What sample size could be 99% confident that the estimated (sample) proportion is within 6 percentage points of the true population proportion?

Correct answers:$254$254​ EBP = 6%. α=0.01. The z-score is found 0.012=0.005=zα2=2.576 n=(zα2)2(˙p′)(q′)EBP2 =(2.5762)(0.165)(0.835)0.062 n=253.96 Rounded up to 254 people.

Suppose a lawn and garden company wants to determine the current percentage of customers who use fertilizer on their lawns. How many customers should the company survey in order to be 95% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who use fertilizer on their lawns?

Correct answers:$601$601​ Given the information in the question, EBP=0.04 since 4%=0.04 and zα2=z0.025=1.96 because the confidence level is 95%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=1.962(0.5)(0.5)0.042=600.2 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 601 customers.

Question Suppose a market researcher wants to determine the current percentage of online shoppers who are males. How many online shoppers should the researcher survey in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of online shoppers who are males? z0.101.282z0.051.645z0.0251.960z0.012.326z0.0052.576 Use the table of values above. Round only at the final step.

Correct answers:$752$752​ Given the information in the question, EBP=0.03 since 3%=0.03 and zα2=z0.05=1.645 because the confidence level is 90%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=1.6452(0.5)(0.5)0.032=751.7 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 752 online shoppers.

The National Women's Business Council (NWBC) in January 2012 published a report on women-owned businesses in the US. Given that they found that 74.3% provided employees paid holidays, vacation, and/or sick leave, what sample size could be 80% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion?

$126$126​ EBP = 5%. α=0.20. The z-score is found 0.202=0.10=zα2=1.282 n=(zα2)2(˙p′)(q′)EBP2 =(1.2822)(0.743)(0.257)0.052 n=125.53 Rounded up to 126 people.

A college admissions director wishes to estimate the mean age of all students currently enrolled. The age of a random sample of 23 students is given below. Assume the ages are approximately normally distributed. Use Excel to construct a 90% confidence interval for the population mean age. Round your answers to two decimal places and use increasing order. 25.8, 22.2, 22.5, 22.8, 24.6, 24, 22.6, 23.6, 22.8, 23.1, 21.5, 21.4, 22.5, 24.5, 21.5, 22.5, 20.5, 23, 25.1, 25.2, 23.8, 21.8, 24.1

$\left(22.61,\ 23.59\right)$(22.61, 23.59)​ Use Excel to calculate the 90% confidence interval, where α=0.1 and n=23. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 23.10 and the sample standard deviation, rounded to two decimal places, is 1.37. 2. Click on any empty cell, enter =CONFIDENCE.T(0.1,1.37,23), and press ENTER. 3. The margin of error, rounded to two decimal places, is 0.49. The confidence interval for the population mean has a lower limit of 23.10−0.49=22.61 and an upper limit of 23.10+0.49=23.59. Thus, the 90% confidence interval for the mean is (22.61, 23.59).

Assume that farm sizes in a particular region are normally distributed with a population standard deviation of 150 acres. A random sample of 50 farm sizes in this region is given below in acres. Estimate the mean farm size for this region with 90% confidence. Round your answers to two decimal places and use ascending order.

$\left(474.87,\ 544.65\right)$(474.87, 544.65)​ A 90% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.1, σ=150, and n=50. Use Excel to calculate the 90% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=509.76. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.1,150,50), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn−−√≈34.89. The confidence interval for the population mean has a lower limit of 509.76−34.89=474.87 and an upper limit of 509.76+34.89=544.65. Thus, the 90% confidence interval for μ is (474.87, 544.65).

Alison, the owner of a regional chain of pizza stores, is trying to decide whether to add calzones to the menu. She conducts a survey of 700 people in the region and asks whether they would order calzones if they were on the menu. Of the 700 responses, 46 people responded "Yes." Create a 90% confidence interval for the proportion of people in the region who would order calzones if they were on the menu. Round your answer to four decimal places.

1$0.0503$0.0503​ 2$0.0811$0.0811​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.90 and press ENTER. 2. Thus, α=0.1. Enter the number of successes, x=46, and sample size, n=700, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.0657. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.64. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0094. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0154. The confidence interval for the population proportion has a lower limit of A4−A7=0.0503 and an upper limit of A4+A7=0.0811. Thus, the 90% confidence interval for the population proportion of people in the region who would order calzones if they were on the menu, based on this sample, is approximately (0.0503, 0.0811).

Question A recent survey asked 1,379 top executives about business trends. The surveyed showed that 23% want to strengthen innovation to capitalize on new opportunities. What is the value of q′ as a decimal? Round to the nearest hundredth.

Answer Explanation Correct answers: 0.77 The value of q′ is: 1−p′=1−0.23=0.77

Question A factory QA supervisor examines two assembly lines for defective products. A random sample of 256 products on one assembly line found that 16 were defective. A random sample of 240 products on another assembly line found that 12 were defective. Construct a 95% confidence interval for the difference in population proportions of defective items from the two assembly lines. Assume that the samples are independent. (Round your answers to three decimal places.)

Correct answer: (−0.028,0.053) In this scenario, the sample proportion and sample size for the first sample are p^1=16256=0.0625 and n1=256, respectively. Then q^1=1−0.0625=0.9375. The sample proportion and sample size for the second sample are p^2=12240=0.05 and n2=240, respectively. Then q^2=1−0.05=0.95. According to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expressions for the bounds of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.0625−0.05)−1.960.0625(0.9375)256+0.05(0.95)240−−−−−−−−−−−−−−−−−−−−−−−√≈−0.028 (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.0625−0.05)+1.960.0625(0.9375)256+0.05(0.95)240−−−−−−−−−−−−−−−−−−−−−−−√≈0.053 The 95% confidence interval is approximately (−0.028,0.053). This means we are 95% confident that the true difference in population proportions of defective items from the two assembly lines is between −2.8% and 5.3%. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

A gubernatorial candidate wants to analyze how she is performing with different age groups. Her campaign staff surveyed 150 random constituents in the 18-24 age range and 175 constituents in the 65+ age range. Of those constituents, 67 of the people 18-24 years old and 90 of the people 65+ years old said they planned on voting for the candidate for governor. Assuming that the samples are independent, the 99% confidence interval for the difference in population proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor is (−0.210,0.075). Interpret this interval in context.

Correct answer: We are 99% confident the difference in population proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor is between −21.0% and 7.5%. A confidence interval is an interval of values that we are confident (but not sure) contain the population parameter of interest. As such, we are 99% confident the difference in population proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor is between −21.0% and 7.5%.

The Pew Social Media Update 2014 surveyed 1,597 adult internet users on which social media websites they use. Of the users surveyed, 1,134 responded "yes" when asked if they use Facebook. What is the value of p′, the estimate proportion of Facebook users in this research study? Answer in decimal form, rounding to the nearest hundredth.

Correct answers:$0.71$0.71​ To form a proportion, take the number of successes, and divide it by n, the number of trials. In this scenario, the number of successes is the number of internet users who use Facebook, 1,134. The total number of internet users surveyed was 1,597. p'=xn=11341597=0.71 So, p′=0.71 is the sample proportion, which is the point estimate of the population proportion

Question Suppose a realtor wants to determine the current percentage of customers who have a family of five or more. How many customers should the realtor survey in order to be 98% confident that the estimated (sample) proportion is within 2 percentage points of the true population proportion of customers who have a family of five or more?

Correct answers:$3382$3382​ Given the information in the question, EBP=0.02 since 2%=0.02 and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=2.3262(0.5)(0.5)0.022=3381.4 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 3382 customers.

A sample of 22 test-tubes tested for number of times they can be heated on a Bunsen burner before they crack is given below. Assume the counts are normally distributed. Use Excel to construct a 99% confidence interval for μ. Round your answers to two decimal places and use increasing order.

Correct answers:$\left(1071.77,\ 1477.33\right)$(1071.77, 1477.33)​ Use Excel to calculate the 99% confidence interval, where α=0.01 and n=22. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 1274.55 and the sample standard deviation, rounded to two decimal places, is 335.93. 2. Click on any empty cell, enter =CONFIDENCE.T(0.01,335.93,22), and press ENTER. 3. The margin of error, rounded to two decimal places, is 202.78. The confidence interval for the population mean has a lower limit of 1274.55−202.78=1071.77 and an upper limit of 1274.55+202.78=1477.33. Thus, the 99% confidence interval for the mean is (1071.77, 1477.33).

A school has found from past records that the average score for boys is 75 in a sample of 45 boys and the average score for girls is 85 in a sample of 30 girls. The population standard deviations for boys' and girls' scores were 25 and 17, respectively. Assume that the distributions of the scores of boys and girls are normally distributed, and let the first sample be the boys' scores and let the second sample be the girls' scores. Find the 90% confidence interval for the difference of the population means, rounding to two decimal places.

Correct answers:1$-17.98\pm0.01$−17.98±0.01​2$-2.02\pm0.01$−2.02±0.01​ Consider the following values:x¯¯¯1=75x¯¯¯2=85zc=1.645σ1=25σ2=17n1=45n2=30Calculate the lower bound of the confidence interval as follows:(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(75−85)−1.645(25)245+(17)230−−−−−−−−−√≈−10−1.645(4.850)≈−10−7.978≈−17.978Calculate the upper bound of the confidence interval as follows:(x¯¯¯1−x¯¯¯2)+zcσ21n1+σ22n2−−−−−−−√=(75−85)+1.645(25)245+(17)230−−−−−−−−−√≈−10+1.645(4.850)≈−10+7.978≈−2.022 Note that using a TI-83/84 calculator or Google Sheets to find the confidence interval directly results in the approximate interval (−17.98,−2.02).

A large company is concerned about the commute times of its employees. Of the 333 employees who were surveyed, 131 said that they had a daily commute longer than 30 minutes. Create a 95% confidence interval for the proportion of employees who have a daily commute longer than 30 minutes. Use Excel to create the confidence interval, rounding to four decimal places.

Correct answers:1$0.3409$0.3409​2$0.4459$0.4459​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=131, and sample size, n=333, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.3934. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0268. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0525. The confidence interval for the population proportion has a lower limit of A4−A7=0.3409 and an upper limit of A4+A7=0.4459. Thus, the 95% confidence interval for the population proportion of employees who have a daily commute longer than 30 minutes , based on this sample, is approximately (0.3409, 0.4459).

Question Professor Cope wants to gauge how many students passed his final exam. Students can either score a "Pass" or "Fail" on their exam. He randomly selects 80 exams and finds that 68 students who passed. What is the 95% confidence interval for the population proportion of Professor Cope's students who passed their final exams.

Solution Using z-scores: We can use the information in the question to define some parameters: n=80 (sample size) p′ =6880=0.85 (sample proportion of successes) q′=1−p′=1−0.85=0.15 (sample proportion of failures) Next, we should check if np′ is greater than 5, so we know we should use this method of finding a confidence interval. np′=(80)(0.85)=68, so we can continue. To find the confidence interval, you need the sample proportion, p′, and the EBP. We know p′=0.85. We can calculate the EBP with the formula: EBP=(zα2)(pqn−−−√) We know the values for p′, q′, and n. We need to find the z-value before we can find EBP. Since the confidence level is 95%, α=1−0.95=0.05, and α2=0.025. We can use the table above to find the z-value for z0.025, which is 1.960. Now we can calculate the EBP. EBP=(zα2)(p′q′n−−−−√) =(1.960)((0.85)(0.15)80−−−−−−−−−−√) ≈(1.960)(0.0016−−−−−√) ≈0.0782 Confidence intervals are written as (p'−EBP,p'+EBP). So we can write this confidence interval as: (0.85−0.0782,0.85+0.0782) or (0.772,0.928). So, we estimate with 95% confidence that the true population proportion of students who passed their final exams is between 0.77 and 0.93, or between approximately 77% and 93% of students passed. Using a calculator: Press STAT and arrow over to TESTS. Arrow down to A:1 - PropZint, and press ENTER. Arrow to x, and enter 68. (number of successes) Arrow to n, and enter 80. (sample size) Arrow to C-Level and enter 0.95. (confidence level) Arrow down to Calculate, and press ENTER.

A recent survey had 31% of the 1,379 surveyed CEOs that were extremely concerned about the availability of key skills. What is the Error Bound for Proportions (EBP) at the 95% level?

$\text{EBP=}0.024$EBP=0.024​ First find the z-score. For the 95% confidence level, it is z0.052=z0.025=1.960. Then calculate the Error Bound for Proportion. EBP = z score time the square root of (p′⋅q′) divided by the sample size n. EBP=1.960⋅(0.31)(0.69)1379−−−−−−−−−−√=0.024

A business magazine conducted a survey of 751 employees who had been at their current employer for 5 or more years. Of these employees, 295 responded that they were bored in their current position. Create a 99% confidence interval for the proportion of employees who have been with their current employer for 5 or more years and reported being bored in their current position. Use Excel to create the confidence interval, rounding to four decimal places.

1$0.3469$0.3469​ 2$0.4387$0.4387​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.99 and press ENTER. 2. Thus, α=0.01. Enter the number of successes, x=295, and sample size, n=751, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.3928. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈2.58. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0178. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0459. The confidence interval for the population proportion has a lower limit of A4−A7=0.3469 and an upper limit of A4+A7=0.4387. Thus, the 99% confidence interval for the population proportion of employees who have been with their current employer for 5 or more years and reported being bored in their current position, based on this sample, is approximately (0.3469, 0.4387).

A school has found from past records that the average score for 8th graders on a standardized math exam is 87 in a sample of 34 8th graders and the average score for 7th graders is 72 in a sample of 49 7th graders. The population standard deviations for 8th graders' scores and 7th graders' scores were 25 and 17, respectively. Assume that the distributions of the scores of 8th graders and 7th graders are normally distributed. Let the first sample be the 8th graders' scores, and let the second sample be the 7th graders' scores. Find the 99% confidence interval for the difference of the population means, rounding to two decimal places.

1$2.31\pm0.01$2.31±0.01​ 2$27.69\pm0.01$27.69±0.01​ Consider the following values:x¯¯¯1=87x¯¯¯2=72zc=2.576σ1=25σ2=17n1=34n2=49The lower bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(87−72)−2.576(25)234+(17)249−−−−−−−−−√≈15−2.576(4.928)≈15−12.695≈2.31The upper bound of the confidence interval is(x¯¯¯1−x¯¯¯2)+zcσ21n1+σ22n2−−−−−−−√=(87−72)+2.576(25)234+(17)249−−−−−−−−−√≈15+2.576(4.928)≈15+12.695≈27.70 Thus, the 99% confidence interval is (2.31,27.70).

Find the Confidence Interval Given a Population Proportion Finding the Confidence Interval With a Proportion IMPORTANT: When finding confidence intervals for proportions, they should only be used if the number of successes np′ and the number of failures nq′ are both greater than 5.

A confidence interval, created for an unknown population proportion, 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 proportion, the point estimate is the sample proportion, p′. The margin of error is the error bound, found with the following formula: EBP=(zα2)(p′q′n−−−−√) ...where zα2 is the z-score that corresponds with the confidence level α, p′ is the estimated proportion of successes, q′ is the estimated proportion of failures, and n is the sample size. The z-scores for the five most common confidence intervals can be found in the portion of the Standard Normal Table below: So, using the sample proportion as the point estimate, and the EBP as the margin of error, the confidence interval at a confidence level, α, can be written as: (p'−EBP,p'+EBP) *We can also use a calculator to find confidence intervals. The steps for using a calculator are below in the example.

Question A recent survey on the usage of alcohol among youth provided the following data: Year: 2005; Sample size: 7000; Youth who are consuming alcohol: 25%Year: 2015; Sample size: 8500; Youth who are consuming alcohol: 21% Construct a 90% confidence interval for the difference in population proportions of youth who were consuming alcohol in 2005 and youth who were consuming alcohol in 2015. Assume that random samples are obtained and the samples are independent. (Round your answers to three decimal places.)

Correct answer: (0.029,0.051) In this scenario, the sample proportion and sample size for the first sample are p^1=0.25 and n1=7000, respectively. Then q^1=1−0.25=0.75. The sample proportion and sample size for the second sample are p^2=0.21 and n2=8500, respectively. Then q^2=1−0.21=0.79. According to the table, for a 90% confidence interval zc=1.645. Substitute these values into the expressions for the bounds of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.25−0.21)−1.6450.25(0.75)7000+0.21(0.79)8500−−−−−−−−−−−−−−−−−−−√≈0.029 (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.25−0.21)+1.6450.25(0.75)7000+0.21(0.79)8500−−−−−−−−−−−−−−−−−−−√≈0.051 The 90% confidence interval is approximately (0.029,0.051). This means we are 90% confident that the true difference between population proportions of youth who were consuming alcohol in 2005 and youth who were consuming alcohol in 2015 is between 2.9% and 5.1%. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

Question Suppose a restaurant wants to determine the current percentage of customers who eat out more than twice a month. How many customers should the restaurant survey in order to be 98% confident that the estimated (sample) proportion is within 4 percentage points of the true population proportion of customers who eat out more than twice a month?

Correct answers:$846$846​ Given the information in the question, EBP=0.04 since 4%=0.04 and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=2.3262(0.5)(0.5)0.042=845.4 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 846 customers.

In order to determine the average weight of carry-on luggage by passengers in airplanes, a sample of 25 pieces of carry-on luggage was collected and weighed in pounds. Assume that the population is normally distributed with a standard deviation of 5 pounds. Find the 95% confidence interval of the mean weight in pounds. Round your answers to two decimal places and use ascending order. Weights 17, 25, 15, 11, 12, 17, 8, 30, 14, 18, 17, 15, 26, 18, 18, 17, 11, 27, 12, 39, 14, 8, 4, 20, 20

Correct answers:$\left(15.36,\ 19.28\right)$(15.36, 19.28)​ A 95% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.05, σ=5, and n=25. Use Excel to calculate the 95% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=17.32. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.05,5,25), and press ENTER. 3. The margin of error, rounded to two decimal places, is zα/2σn−−√≈1.96. The confidence interval for the population mean has a lower limit of 17.32−1.96=15.36 and an upper limit of 17.32+1.96=19.28. Thus, the 95% confidence interval for μ is (15.36, 19.28).

The yearly incomes, in thousands, for 24 random married couples living in a city are given below. Assume the yearly incomes are approximately normally distributed. Use Excel to find the 95% confidence interval for the true mean, in thousands. Round your answers to three decimal places and use increasing order.

Correct answers:$\left(58.984,\ 59.026\right)$(58.984, 59.026)​ Use Excel to calculate the 95% confidence interval, where α=0.05 and n=24. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean is 59.005 and the sample standard deviation, rounded to three decimal places, is 0.049. 2. Click on any empty cell, enter =CONFIDENCE.T(0.05,0.049,24), and press ENTER. 3. The margin of error, rounded to three decimal places, is 0.021. The confidence interval for the population mean has a lower limit of 59.005−0.021=58.984 and an upper limit of 59.005+0.021=59.026. Thus, the 95% confidence interval for the true mean, in thousands, is (58.984, 59.026).

Brent, a pet supply store owner, is comparing the weights of two different breeds of dogs. A sample of 41 adult male Pembroke Welsh corgis had an average weight of 11.85 kilograms. A sample of 38 adult male border collies had an average weight of 17.11 kilograms. Assume that the population standard deviations for the weights of Pembroke Welsh corgis and border collies are 1.35 kilograms and 2.12 kilograms, respectively. Find the 90% confidence interval for the difference in mean weights between Pembroke Welsh corgis and border collies. Let the Pembroke Welsh corgis be the first sample, and let the border collies be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

Correct answers:1$-5.92\pm0.01$−5.92±0.01​2$-4.60\pm0.01$−4.60±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=11.85, σ1=1.35, and n1=41, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=17.11, σ2=2.12, and n2=38, respectively. According to technology or a z-table, for a 90% confidence interval, zc=1.645. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(11.85−17.11)−1.6451.35241+2.12238−−−−−−−−−−−−√≈−5.92 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(11.85−17.11)+1.6451.35241+2.12238−−−−−−−−−−−−√≈−4.60 Using TI-83 or TI-84 calculators or Google Sheets to find the exact interval with no intermediate rounding also produces a confidence interval of (−5.92,−4.60).

Using p′=0.652, q′=0.348, and n=230, what is the 95% confidence interval for the proportion of the population who own a mountain bike? Use the table of common z-scores above. Round your answers to three decimal places.

Correct answers:1$0.590$0.590​2$0.714$0.714​ We know n=230, p′=0.652, and q′=0.348. Since the confidence level is 95%, α=1−0.95=0.05, and α2=0.025. The z-value for z0.025 is 1.96. EBP=(zα2)(p′q′n−−−−√)=(1.96)((0.652)(0.348)230−−−−−−−−−−−−√)≈(1.960)(0.000986−−−−−−−√)≈0.062 So we can write the confidence interval as (0.652−0.062,0.652+0.062)=(0.590,0.714). We estimate with 95% confidence that the true population proportion of people who own a mountain bike is between 0.590 and 0.714.

Question Suppose a mobile phone company wants to determine the current percentage of senior customers (those over the age of 50 years) who use text messaging. How many senior customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of senior customers who actually use text messaging?

Solution In order to find the sample size, n, we need to plug values into the equation: n=(zα2)2(p'q')(EBP2) From the problem, we know that EBP=0.03 (or 3%). We are also given the confidence interval, 90%, so we can find the value for zα2 from the table. α=1−CL=1−.90=0.10. So, α2=0.102=0.05. From the table, z0.05=1.645. The only two parameters we are missing are p′ and q′ (the sample proportions for successes and failures). Remember that q'=1-p'. But, we do not know p′ yet. Since we multiply p′ and q′together, we make them both equal to 0.5 because this will yield the largest possible product: p'q'=(0.5)(0.5)=0.25. (Try other products: (0.6)(0.4)=0.24 or (0.3)(0.7)=0.21, and so on). The largest possible product will give us the largest possible n, which will give us a large enough sample so that we can be 90% confident that we are within 3 percentage points of the true population proportion. To calculate the sample size, use the formula and make the substitutions. n=(zα2)2(p'q')(EBP2) =(1.645)2(0.5)(0.5)(0.032) ≈(2.706)(0.25)0.0009 ≈751.67 Because we're talking about surveying people, we need to round the answer to the next higher value. The sample size should be 752 senior customers.

Question Suppose that a research firm wants to estimate the percent of families living in a city who have pets. Five hundred randomly selected families in this city are surveyed. Of the 500 families surveyed, 321 responded "yes" - they own a pet. What is the error bound for proportions (EBP) of a confidence interval with a 98% confidence level?

Solution We know the sample size, n, is the entire sample of families surveyed. So, n=500. We also know (from the first example) that p′=0.642, and so we can find the value of q′, because q′=1−p′. So, q′=1−0.642=0.358. We are given the confidence level (CL), 98%, or 0.98. So, we can calculate alpha (α). α=1−CL =1−0.98 =0.02 Since α=0.02, we know that α2=0.022=0.01. So we will look for the value of z0.01. From the table above, z0.01=2.326. Now we can substitute the values into the formula to find the error bound. EBP=(zα2)(p′q′n−−−−√) =(2.326)((0.642)(0.358)(500)−−−−−−−−−−−−√) ≈(2.326)(0.230500−−−−−√) ≈(2.326)(0.00046−−−−−−√) ≈0.050 So, the error bound (EBP) is 0.050.

Suppose a hotel wants to determine the current percentage of customers who are over the age of twenty-seven. How many customers should the hotel survey in order to be 98% confident that the estimated (sample) proportion is within 3 percentage points of the true population proportion of customers who are over the age of twenty-seven?

$1503$1503​ Given the information in the question, EBP=0.03 since 3%=0.03 and zα2=z0.01=2.326 because the confidence level is 98%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=2.3262(0.5)(0.5)0.032=1502.9 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 1503 customers.

The NWBC found that 13% of women-owned businesses provided profit-sharing and/or stock options. What sample size could be 98% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion?

$245$245​ EBP = 5%. α=0.02. The z-score is found 0.022=0.01=zα2=2.326 n=(zα2)2(p′)(q′)EBP2 =(2.3262)(0.13)(0.87)0.052 n=244.76 Rounded up to 245 people.

The NWBC found that 16.5% of women-owned businesses did not provide any employee benefits. What sample size could be 99% confident that the estimated (sample) proportion is within 6 percentage points of the true population proportion?

$254$254​ EBP = 6%. α=0.01. The z-score is found 0.012=0.005=zα2=2.576 n=(zα2)2(˙p′)(q′)EBP2 =(2.5762)(0.165)(0.835)0.062 n=253.96 Rounded up to 254 people.

Suppose a clothing store wants to determine the current percentage of customers who are over the age of forty. How many customers should the company survey in order to be 92% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who are over the age of forty?

$307$307​ Given the information in the question, EBP=0.05 since 5%=0.05 and zα2=z0.04=1.751 because the confidence level is 92%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p′q′, and thus the largest possible n. If p′=0.5, then q′=1−0.5=0.5. Therefore, n=z2p′q′EBP2=1.7512(0.5)(0.5)0.052=306.6 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 307 customers.

A study was conducted to estimate the mean age when people buy their first new car. The ages of purchase for 22 randomly selected people are given below. Assume the ages are approximately normally distributed. Use Excel to determine the 99% confidence interval for the mean. Round your answers to two decimal places and use increasing order. Age 19, 25, 27, 23, 15, 21, 22, 15, 18, 26, 18, 22, 28, 25, 23, 19, 20, 24, 20, 20, 22, 20

$\left(19.33,\ 23.57\right)$(19.33, 23.57)​ Use Excel to calculate the 99% confidence interval, where α=0.01 and n=22. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 21.45 and the sample standard deviation, rounded to two decimal places, is 3.51. 2. Click on any empty cell, enter =CONFIDENCE.T(0.01,3.51,22), and press ENTER. 3. The margin of error, rounded to two decimal places, is 2.12. The confidence interval for the population mean has a lower limit of 21.45−2.12=19.33 and an upper limit of 21.45+2.12=23.57. Thus, the 99% confidence interval for the mean is (19.33, 23.57).

Weights, in pounds, of ten-year-old girls are collected from a neighborhood. A sample of 26 is given below. Assuming normality, use Excel to find the 98% confidence interval for the population mean weight μ. Round your answers to three decimal places and use increasing order. 66.4 86.3 71.3 52.8 68 85 66.2 79.2 93.5 84.5 71.1 74.5 65 58.5 59.8 80.2 69.2 92.9 78.9 59.4 63.6 66.5 60.7 80.1 60.4 74.5

$\left(66.496,\ 77.234\right)$(66.496, 77.234)​ Use Excel to calculate the 98% confidence interval, where α=0.02 and n=26. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean, rounded to three decimal places, is 71.865 and the sample standard deviation, rounded to three decimal places, is 11.016. 2. Click on any empty cell, enter =CONFIDENCE.T(0.02,11.016,26), and press ENTER. 3. The margin of error, rounded to three decimal places, is 5.369. The confidence interval for the population mean has a lower limit of 71.865−5.369=66.496 and an upper limit of 71.865+5.369=77.234. Thus, the 98% confidence interval for the mean weight is (66.496, 77.234).

The following data represent a sample of the assets (in millions of dollars) of 28 credit unions in a state. Assume that the population in this state is normally distributed with σ=3.5 million dollars. Use Excel to find the 99% confidence interval of the mean assets in millions of dollars. Round your answers to three decimal places and use ascending order. Assets 12.23, 2.89, 13.19, 73.25, 11.59, 8.74, 7.92, 40.22, 5.01, 2.27, 16.56, 1.24, 9.16, 1.91, 6.69, 3.17, 4.78, 2.42, 1.47, 12.77, 4.39, 2.17, 1.42, 14.64, 1.06, 18.13, 16.85, 21.58

$\left(9.643,\ 13.051\right)$(9.643, 13.051)​ A 99% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.01, σ=3.5, and n=28. Use Excel to calculate the 99% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean, rounded to three decimal places, is x¯=11.347. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.01,3.5,28), and press ENTER. 3. The margin of error, rounded to three decimal places, is zα/2σn−−√≈1.704. The confidence interval for the population mean has a lower limit of 11.347−1.704=9.643 and an upper limit of 11.347+1.704=13.051. Thus, the 99% confidence interval for μ is (9.643, 13.051).

The monthly incomes from a random sample of 20 workers in a factory is given below in dollars. Assume the population has a normal distribution and has standard deviation $518. Compute a 98% confidence interval for the mean of the population. Round your answers to the nearest dollar and use ascending order. 11621 12367 12030 12135 12561 11453 12313 11561 12424 12346 11591 12032 11562 12224 12850 11802 12360 12519 11960 12337

$\left(\$11,833,\ \$12,372\right)$($11,833, $12,372)​ A 98% confidence interval for μ is (x¯−zα/2σn−−√,x¯+zα/2σn−−√). Here, α=0.02, σ=518, and n=20. Use Excel to calculate the 98% confidence interval. 1. Open Excel, enter the given data in column A, and find the sample mean, x¯, using the AVERAGE function. Thus, the sample mean is x¯=12,102.4. 2. Click on any empty cell, enter =CONFIDENCE.NORM(0.02,518,20), and press ENTER. 3. The margin of error, rounded to the nearest tenth, is zα/2σn−−√≈269.5. The confidence interval for the population mean has a lower limit of 12,102.4−269.5=11,832.9 and an upper limit of 12,102.4+269.5=12,371.9. Thus, the 98% confidence interval for μ, rounded to the nearest integer, is (11,833, 12,372).

A gubernatorial candidate wants to analyze how she is performing with different age groups. Her campaign staff surveyed 150 random constituents in the 18-24 age range and 175 constituents in the 65+ age range. Of those constituents, 67 of the people 18-24 years old and 90 of the people 65+ years old said they planned on voting for the candidate for governor. Find the lower bound of the 99% confidence interval for the difference in population proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor. Let the people 18-24 years old be sample 1 and let the people 65+ years old be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

$\text{lower bound=}-0.210$lower bound=−0.210​ In this scenario, the sample proportion and sample size for the first sample are p^1=67150≈0.447 and n1=150, respectively. Then q^1≈1−0.447=0.553. The sample proportion and sample size for the second sample are p^2=90175≈0.514 and n2=175, respectively. Then q^2≈1−0.514=0.486. According to the table, for a 99% confidence interval zc=2.576. Substitute these values into the expression for the lower bound of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.447−0.514)−2.5760.447(0.553)150+0.514(0.486)175−−−−−−−−−−−−−−−−−−−−−−−√≈−0.210 Using the TI-83 or TI-84 calculators, Excel, or Google Sheets to find the exact value with no intermediate rounding also produces a value of −0.210.

Researchers asked two groups of people to construct a jigsaw puzzle. One group of 55 people, who listened to soft classical music while working, had a mean construction time of 45 minutes. Another group of 45 people, who worked in silence, had a mean construction time of 55 minutes. Assume the population standard deviations for the construction times for people listening to classical music and people working in silence are 9.5 minutes and 11 minutes, respectively. Find the lower bound of the 99% confidence interval for the difference in population construction times between people listening to classical music as they work and people working in silence. Let the people who listened to classical music be the first sample and let the people working in silence be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

$\text{lower bound=}-15.36$lower bound=−15.36​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=45, σ1=9.5, and n1=55, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=55, σ2=11, and n2=45, respectively. According to technology or a z-table, for a 99% confidence interval, zc=2.575. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(45−55)−2.5759.5255+11245−−−−−−−−−√≈−15.36 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a value of −15.36.

A random sample of 48 professional baseball players yielded a mean weight of 195 pounds. A random sample of 54 professional hockey players yielded a mean weight of 210 pounds. Assume the population standard deviations for the baseball players and the hockey players are 32 pounds and 28.5 pounds, respectively. Find the lower bound of the 99% confidence interval for the difference in population mean weights between baseball players and hockey players. Let the baseball players be the first sample and let the hockey players be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

$\text{lower bound=}-30.54$lower bound=−30.54​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=195, σ1=32, and n1=48, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=210, σ2=28.5, and n2=54, respectively. According to technology or a z-table, for a 99% confidence interval, zc=2.575. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(195−210)−2.57632248+28.5254−−−−−−−−−−√≈−30.54 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding produces a value of −30.54.

A survey asked 286 married men and 307 single men if they regularly watch do-it-yourself house design TV shows. Of those men surveyed, 157 married men and 102 single men responded positively. Find the lower bound of the 90% confidence interval for the difference in population proportions of married men who regularly watch do-it-yourself house design TV shows and single men who regularly watch do-it-yourself house design TV shows. Let the married men be sample 1 and let the single men be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

$\text{lower bound=}0.151$lower bound=0.151​ In this scenario, the sample proportion and sample size for the first sample are p^1=157286≈0.549 and n1=286, respectively. Then q^1≈1−0.549=0.451. The sample proportion and sample size for the second sample are p^2=102307≈0.332 and n2=307, respectively. Then q^2≈1−0.332=0.668. According to the table, for a 90% confidence interval zc=1.645. Substitute these values into the expression for the lower bound of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.549−0.332)−1.6450.549(0.451)286+0.332(0.668)307−−−−−−−−−−−−−−−−−−−−−−−√≈0.151 Using the TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a value of 0.151.

An aquarium manager wants to study gift shop browsing. She randomly observes 120 couples that visit the aquarium with children and finds that 107 enter the gift shop at the end of their visit. She randomly observes 76 couples that visit the aquarium with no children and finds that 59 enter the gift shop at the end of their visit. Find the upper bound of the 95% confidence interval for the difference in population proportions of couples with children that enter the gift shop and couples without children that enter the gift shop. Let the couples with children be sample 1 and let the couples without children be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

$\text{upper bound=}0.224$upper bound=0.224​ In this scenario, the sample proportion and sample size for the first sample are p^1=107120≈0.892 and n1=120, respectively. Then q^1≈1−0.892=0.108. The sample proportion and sample size for the second sample are p^2=5976≈0.776 and n2=76, respectively. Then q^2≈1−0.776=0.224. According to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expressions for the upper bound of the confidence interval: (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.892−0.776)+1.960.892(0.108)120+0.776(0.224)76−−−−−−−−−−−−−−−−−−−−−−−√≈0.225 Using the TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding produces a value of 0.224.

A survey asked 286 married men and 307 single men if they regularly watch do-it-yourself house design TV shows. Of those men surveyed, 157 married men and 102 single men responded positively. Find the upper bound of the 90% confidence interval for the difference in population proportions of married men who regularly watch do-it-yourself house design TV shows and single men who regularly watch do-it-yourself house design TV shows. Let the married men be sample 1 and let the single men be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

$\text{upper bound=}0.282$upper bound=0.282​ In this scenario, the sample proportion and sample size for the first sample are p^1=157286≈0.549 and n1=286, respectively. Then q^1≈1−0.549=0.451. The sample proportion and sample size for the second sample are p^2=102307≈0.332 and n2=307, respectively. Then q^2≈1−0.332=0.668. According to the table, for a 90% confidence interval zc=1.645. Substitute these values into the expression for the upper bound of the confidence interval: (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.549−0.332)+1.6450.549(0.451)286+0.332(0.668)307−−−−−−−−−−−−−−−−−−−−−−−√≈0.283 Using the TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding produces a value of 0.282.

A random sample of 200 sales at a video game store in the northern end of a large city found that the average sale amount was $64.52. A random sample of 250 sales at a video game store in the southern end of the same city found that the average sale amount was $76.96. Assume the population standard deviations for sale totals in the northern store and the southern store are $12.88 and $20.07, respectively. Find the 95% confidence interval for the difference in population mean sale totals at the two stores. Let the northern store sales be the first sample and let the southern store sales be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

1$-15.50\pm0.01$−15.50±0.01​ 2$-9.38\pm0.01$−9.38±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=64.52, σ1=12.88, and n1=200, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=76.96, σ2=20.07, and n2=250, respectively. According to technology or a z-table, for a 95% confidence interval, zc=1.95. Substitute these values into the expressions for the bounds of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(64.52−76.96)−1.9612.882200+20.072250−−−−−−−−−−−−−√≈−15.50 (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(64.52−76.96)+1.9612.882200+20.072250−−−−−−−−−−−−−√≈−9.38 Using TI-83 or TI-84 calculators or Google Sheets to find the exact interval with no intermediate rounding also produces values of −15.50 and −9.38. So, the 95% confidence interval for the difference in population mean sale totals at the two stores is (−$15.50,−$9.38).

In a random sample of 2,282 college students, 356 reported getting 8 or more hours of sleep per night. Create a 95% confidence interval for the proportion of college students who get 8 or more hours of sleep per night. Use Excel to create the confidence interval, rounding to four decimal places.

1$0.1411$0.1411​ 2$0.1709$0.1709​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=356, and sample size, n=2282, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.156. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0076. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0149. The confidence interval for the population proportion has a lower limit of A4−A7=0.1411 and an upper limit of A4+A7=0.1709. Thus, the 95% confidence interval for the population proportion of college students who get 8 or more hours of sleep per night, based on this sample, is approximately (0.1411, 0.1709).

Gary is studying the prevalence of regional dialects. He surveyed 351 people in a region and asked what words they used for certain things. Gary found that 292 of the respondents referred to soft drinks as "pop," as opposed to "soda" or "cola." Create a 90% confidence interval for the proportion of people in this region who refer to soft drinks as "pop." Use Excel to create the confidence interval, rounding to four decimal places.

1$0.7991$0.7991​ 2$0.8647$0.8647​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.90 and press ENTER. 2. Thus, α=0.1. Enter the number of successes, x=292, and sample size, n=351, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.8319. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.64. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0200. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0328. The confidence interval for the population proportion has a lower limit of A4−A7=0.7991 and an upper limit of A4+A7=0.8647. Thus, the 90% confidence interval for the population proportion of people in this region who refer to soft drinks as "pop," based on this sample, is approximately (0.7991, 0.8647).

The mean bowling average achieved by 35 children who trained with a professional for a year was 130. The study also indicated that the mean bowling average achieved by 40 children who practiced on their own over the same year was 95. The population standard deviations for the averages for children who train with the pro and children who practice on their own are 24 and 30, respectively. Find the 90% confidence interval for the difference in population mean bowling averages between children who train with the pro and children who practice on their own. Let the pro-trained children be the first sample, and let the solo practice children be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

1$24.73\pm0.01$24.73±0.01​ 2$45.27\pm0.01$45.27±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=130, σ1=24, and n1=35, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=95, σ2=30, and n2=40, respectively. According to technology or a z-table, for a 90% confidence interval, zc=1.645. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(130−95)−1.64524235+30240−−−−−−−−−√≈24.73 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(130−95)+1.64524235+30240−−−−−−−−−√≈45.27 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a confidence interval of (24.73,45.27).

Suppose an automotive repair company wants to determine the current percentage of customers who keep up with regular vehicle maintenance. How many customers should the company survey in order to be 90% confident that the estimated (sample) proportion is within 5 percentage points of the true population proportion of customers who keep up with regular vehicle maintenance?

271 Given the information in the question, EBP=0.05 since 5%=0.05 and α2=z0.05=1.645 because the confidence level is 90%. The values of p′ and q′ are unknown, but using a value of 0.5 for p′ will result in the largest possible product of p'q', and thus the largest possible n. If p'=0.5, then q'=1−0.5=0.5. Therefore, n=(zα2)2(p'q')(EBP2) =(1.645)2(0.5)(0.5)(0.052) =270.6 Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 271 customers.

Compute Confidence Intervals for the Difference in Population Means Confidence Intervals for the Difference Between Population Means

A confidence interval for the difference between two population means, μ1 and μ2, is a confidence interval for the difference μ1−μ2. The confidence interval for the difference in means makes use of a normal distribution and critical values zc to construct the confidence interval. (This assumes that the population standard deviations are known for each group of interest.) In order to proceed with the confidence interval calculation, the following conditions must be met: The samples must be randomly selected. The samples must be independent. The population standard deviations, σ1 and σ2, must be known. Either both distributions must be known to be normally distributed or both n1≥30 and n2≥30. Once these conditions are verified, then the sampling distribution for x¯1−x¯2 (the difference between the sample means) will follow a normal distribution with mean =μ1−μ2 and standard error =σ21n1+σ22n2−−−−−−−√. To form the confidence interval for the difference in means, we will also need to determine the critical value zc, which is identical to the critical values discussed in previous content related to confidence intervals for one mean. The value of zc is the z-score that cuts off an area of α2 in each tail of a normal distribution for a (1−α)⋅100% confidence interval. For example, when constructing a 95% confidence interval, the corresponding zc-value would be the z-score that cuts off an area of 2.5% in each tail of the normal curve. A summary table of zc-values for typical confidence levels of 90% confidence, 95% confidence, and 99% confidence are provided in the table below. Once the requirements cited above are met, then the confidence interval for the difference between two population means μ1−μ2 is given as follows: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√<μ1−μ2<(x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√ The TI-83 and TI-84 graphing calculators can also be used to assist with the calculations for a confidence interval for the difference between two population proportions. Select STAT, then TESTS, then 2-SampZInt. Choose the Stats input (instead of the Data input). Enter values for σ1, σ2, x¯1, n1, x¯2, and n2, and then select the Confidence Level. Then select Calculate. Google Sheets can also be used to assist with the calculations for a confidence interval for the difference between two population proportions. Enter the sample mean, population standard deviation, and sample size for the first sample in cells A1, A2, and A3, respectively. Enter the sample mean, population standard deviation, and sample size for the second sample in cells B1, B2, and B3, respectively. Enter the decimal form of the confidence level into cell C1. Enter =1−(1−C1)/2 into cell C2. Enter =NORMSINV(C2) into cell C3. This value is the value of zc. Note that this value will be much more precise than the values from the table above. In two blank cells, enter =(A1−B1)−C3*SQRT(((A2^2)/A3)+((B2^2)/B3)) and =(A1−B1)+C3*SQRT(((A2^2)/A3)+((B2^2)/B3)). These are the lower and upper bounds of the confidence interval, respectively.

In a survey of 1,000 adults in a country, 722 said that they had eaten fast food at least once in the past month. Create a 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month. Use Excel to create the confidence interval, rounding to four decimal places.

Answer Explanation (1$$, 2$$) Correct answers:1$0.6942$0.6942​2$0.7498$0.7498​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=722, and sample size, n=1000, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =722/1000 and press ENTER. 3. Thus, p^=0.722. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0142. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0278. The confidence interval for the population proportion has a lower limit of A4−A7≈0.6942 and an upper limit of A4+A7≈0.7498. Thus, the 95% confidence interval for the population proportion of adults who ate fast food at least once in the past month, based on this sample, is approximately (0.6942, 0.7498).

Question In the year 2004, a survey was undertaken to find whether citizens of a certain political party supported their party's candidate for governor. In a sample of 300 citizens of this party, 95% of them expressed support for their party's candidate. A similar survey was conducted four years later and showed that 91% of a sample of 350 citizens of this party expressed support for their party's candidate. Construct a 95% confidence interval for the difference in population proportions of citizens of this party who supported their gubernatorial candidate in 2004 and citizens of this party who supported their gubernatorial candidate four years later. Assume that random samples are obtained and the samples are independent. (Round your answers to three decimal places.)

Correct answer: (0.001,0.079) In this scenario, the sample proportion and sample size for the first sample are p^1=0.95 and n1=300, respectively. Then q^1=1−0.95=0.05. The sample proportion and sample size for the second sample are p^2=0.91 and n2=350, respectively. Then q^2=1−0.91=0.09. According to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expressions for the bounds of the confidence interval: (p^1−p^2)−zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.95−0.91)−1.960.95(0.05)300+0.91(0.09)350−−−−−−−−−−−−−−−−−−−√≈0.001 (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.95−0.91)+1.960.95(0.05)300+0.91(0.09)350−−−−−−−−−−−−−−−−−−−√≈0.078 The 95% confidence interval is approximately (0.001,0.079). This means we are 95% confident that the true difference in population proportions of citizens of this party who supported their gubernatorial candidate in 2004 and citizens of this party who supported their gubernatorial candidate four years later is between 0.1% and 7.9%. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

The heart rates for a group of 21 students taking a final exam are given below. Assume the heart rates are normally distributed. Use Excel to find the 95% confidence interval for the true mean. Round your answers to two decimal places and use increasing order.

Correct answers:$\left(91.31,\ 95.17\right)$(91.31, 95.17)​ Use Excel to calculate the 95% confidence interval, where α=0.05 and n=21. 1. Open Excel and enter the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, using the STDEV.S function. Thus, the sample mean is 93.24 and the sample standard deviation, rounded to two decimal places, is 4.24. 2. Click on any empty cell, enter =CONFIDENCE.T(0.05,4.24,21), and press ENTER. 3. The margin of error, rounded to two decimal places, is 1.93. The confidence interval for the population mean has a lower limit of 93.24−1.93=91.31 and an upper limit of 93.24+1.93=95.17. Thus, the 95% confidence interval for the true mean is (91.31, 95.17).

An upscale restaurant has establishments in Boston and Chicago. The manager of the Boston restaurant samples 55 random checks and finds the average tip percentage is 0.25. The manager of the Chicago restaurant samples 45 random checks and finds the average tip percentage is 0.23. Assume the population standard deviations for the tip percentages in the Boston restaurant and the Chicago restaurant are 0.028 and 0.024, respectively. Find the upper bound of the 90% confidence interval for the difference in tip percentages at the two restaurants. Let the Boston tips be the first sample, and let the Chicago tips be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

Correct answers:$\text{upper bound=}0.03\%$upper bound=0.03%​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=0.25, σ1=0.028, and n1=55, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=0.23, σ2=0.024, and n2=45, respectively. According to technology or a z-table, for a 90% confidence interval, zc=1.645. Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(0.25−0.23)+1.6450.028255+0.024245−−−−−−−−−−−−−√≈0.02856 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a value of 0.03.

A gubernatorial candidate wants to analyze how she is performing with different age groups. Her campaign staff surveyed 150 random constituents in the 18-24 age range and 175 constituents in the 65+ age range. Of those constituents, 67 of the people 18-24 years old and 90 of the people 65+ years old said they planned on voting for the candidate for governor. Find the upper bound of the 99% confidence interval for the difference in population proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan to vote for the candidate for governor. Let the people 18-24 years old be sample 1 and let the people 65+ years old be sample 2. Assume that the samples are independent. Enter your answer as a decimal rounded to three decimal places.

Correct answers:$\text{upper bound=}0.075$upper bound=0.075​ In this scenario, the sample proportion and sample size for the first sample are p^1=67150≈0.447 and n1=150, respectively. Then q^1≈1−0.447=0.553. The sample proportion and sample size for the second sample are p^2=90175≈0.514 and n2=175, respectively. Then q^2=1−0.514=0.486. According to the table, for a 99% confidence interval zc=2.576. Substitute these values into the expression for the upper bound of the confidence interval: (p^1−p^2)+zcp^1q^1n1+p^2q^2n2−−−−−−−−−−−√=(0.447−0.514)+2.5760.447(0.553)150+0.514(0.486)175−−−−−−−−−−−−−−−−−−−−−−−√≈0.076 Using the TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding produces a value of 0.075.

Question A college has found from past records that the wages among female students are $55 on average and the wages among male students are $45 on average. The data was obtained from a sample of 105 female students and 88 male students. The population standard deviations for female and male students' wages were $12 and $18, respectively. Compute 99% confidence intervals for the difference in population means. The confidence interval is approximately (4.2114, _____). Assume that the distributions of the wages of men and women are normally distributed. Round your answer to two decimal places.

Correct answers:$\text{upper bound=}15.79$upper bound=15.79​ Consider the following values: x¯¯¯1x¯¯¯2zcσ1σ2n1n2=55=45=2.575=12=18=105=88 The lower bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−−√=(55−45)−2.575(12)2105+(18)288−−−−−−−−−−−−√=10−2.575(2.248)=10−5.7886=4.2114The upper bound of the confidence interval isl(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−−√=(55−45)+2.575(12)2105+(18)288−−−−−−−−−−−−√=10+2.575(2.248)=10+5.7886=15.7886 Note that using a TI-83/84 calculator or Google Sheets to find the upper bound directly results in a value of approximately 15.79.

A school has found from past records that the average scores for boys is 85 in a sample of 30 boys and the average scores for girls is 75 in a sample of 45 girls. The population standard deviations for girls' and boys' scores were 25 and 17, respectively. The 99% confidence interval for the difference of the population means (girls minus boys)is approximately (−22.49, _____). Assume that the distributions of the scores of boys and girls are normally distributed. Round your answers to two decimal places.

Correct answers:$\text{upper bound=}2.49$upper bound=2.49​ Consider the following values:x¯¯¯1=75x¯¯¯2=85zc=2.576σ1=25σ2=17n1=45n2=30The lower bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(75−85)−2.576(25)245+(17)230−−−−−−−−−√=−10−2.576(4.850)=−10−12.494=−22.494The upper bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(75−85)+2.576(25)245+(17)230−−−−−−−−−√=−10+2.576(4.850)=−10+12.494=2.494 Note that using a TI-83/84 calculator or Google Sheets to find the upper bound directly results in a value of approximately 2.494. Rounding to two decimal places yields 2.49.

A company has conducted a poll to know the awareness among consumers for its various product ranges. From a sample of 55 male respondents and 40 female respondents, the company found that the male respondents were aware of about 15 of its products and female respondents were aware of about 12 of its products, on average. The population standard deviation for male respondents is 14 and for female respondents is 7. Let the male respondents be the first sample and let the female respondents be the second sample. Assume the distributions of the male and female respondents are normally distributed. Compute a 95% confidence interval for the difference in population means, rounding to two decimal places.

Correct answers:1$-1.29\pm0.01$−1.29±0.01​2$7.29\pm0.01$7.29±0.01​ Consider the following values:x¯¯¯1=15x¯¯¯2=12zc=1.96σ1=14σ2=7n1=55n2=40The lower bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(15−12)−1.96(14)255+(7)240−−−−−−−−−√≈3−1.96(2.1883)≈3−4.289≈−1.289The upper bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(15−12)+1.96(14)255+(7)240−−−−−−−−−√≈3+1.96(2.1883)≈3+4.289≈7.289 Thus, the 95% confidence interval is (−1.29,7.29).

Researchers asked two groups of people to construct a jigsaw puzzle. One group of 55 people, who listened to soft classical music while working, had a mean construction time of 45 minutes. Another group of 45 people, who worked in silence, had a mean construction time of 55 minutes. Assume the population standard deviations for the construction times for people listening to classical music and people working in silence are 9.5 minutes and 11 minutes, respectively. Find the 99% confidence interval for the difference in population mean construction times between people listening to classical music as they work and people working in silence. Let the people who listened to classical music be the first sample and let the people working in silence be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

Correct answers:1$-15.36\pm0.01$−15.36±0.01​2$-4.64\pm0.01$−4.64±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=45, σ1=9.5, and n1=55, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=55, σ2=11, and n2=45, respectively. According to technology or a z-table, for a 99% confidence interval, zc=2.576. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(45−55)−2.5769.5255+11245−−−−−−−−−√≈−15.36 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(45−55)+2.5769.5255+11245−−−−−−−−−√≈−4.64 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a confidence interval of (−15.36,−4.64).

Researchers asked two groups of people to solve a crossword puzzle. One group of 36 people, who were provided a dictionary and thesaurus while working, had a mean time of 57.2 minutes to solve the puzzle. A second group of 39 people, working without aid, had a mean time of 75.1 minutes to solve the puzzle. Assume the population standard deviations for the solution times for the people working with a dictionary and thesaurus and those given no aid were 8.1 minutes and 13.5 minutes, respectively. Find the 95% confidence interval for the population mean difference in solution times between people working with a dictionary and thesaurus and people working without aid. Let the people working with a dictionary and thesaurus be the first sample, and let the people working without aid be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

Correct answers:1$-22.90\pm0.01$−22.90±0.01​2$-12.90\pm0.01$−12.90±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=57.2, σ1=8.1, and n1=36, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=75.1, σ2=13.5, and n2=39, respectively. According to technology or a z-table, for a 95% confidence interval, zc=1.96. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(57.2−75.1)−1.968.1236+13.5239−−−−−−−−−−−√≈−22.90 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(57.2−75.1)+1.968.1236+13.5239−−−−−−−−−−−√≈−12.90 Using TI-83 or TI-84 calculators or Google Sheets to find the exact value with no intermediate rounding also produces a confidence interval of (−22.90,−12.90).

From a sample of 225 people in Pennsylvania, the mean age is 41 years. From a sample of 300 people in New Jersey, the mean age is 45 years. Assume the population standard deviations for the ages of people from Pennsylvania and New Jersey are 6 years and 8 years, respectively. Construct a 90% confidence interval for the difference in average age between Pennsylvania residents and New Jersey residents. Enter your answer in the format (lower , upper) where "lower" is the lower confidence limit and "upper" is the upper confidence limit. Let the Pennsylvania ages be the first sample and let the New Jersey ages be the second sample. Round your answers to two decimal places. Do not include units in your answer.

Correct answers:1$-5.01\pm0.01$−5.01±0.01​2$-2.99\pm0.01$−2.99±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=41, σ1=6, and n1=225, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=45, σ2=8, and n2=300, respectively. According to the table, for a 90% confidence interval, zc=1.645. Substitute these values into the expressions for the bounds of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(41−45)−1.64562225+82300−−−−−−−−−√≈−5.01 (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(41−45)+1.64562225+82300−−−−−−−−−√≈−2.99 The 90% confidence interval is approximately (−5.01,−2.99). This means we are 90% confident that the true difference between the mean Pennsylvania age and the mean New Jersey age is between −5.01 years and −2.99 years. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

A college has found from past records that the average wage among female graduates is $55,000 per year and the average wage among male graduates is $57,000 per year. The data were obtained from a sample of 105 female students and 88 male students. The population standard deviations for female and male students' wages were $12,000 and $18,000, respectively. Assume that the distributions of the wages of men and women are normally distributed. Let the wages of the female students be the first sample, and let the wages of the male students be the second sample. Compute a 99% confidence interval for the difference in population mean wages per year, rounding to the nearest dollar. Do not include the dollar sign in the answer

Correct answers:1$-7790\pm2$−7790±2​2$3790\pm2$3790±2​ Consider the following values:x¯¯¯1=55,000x¯¯¯2=57,000zc=2.576σ1=12,000σ2=18,000n1=105n2=88The lower bound of the confidence interval is(x¯¯¯1−x¯¯¯2)−zcσ21n1+σ22n2−−−−−−−√=(55,000−57,000)−2.576(12,000)2105+(18,000)288−−−−−−−−−−−−−−√≈−2000−2.576(2247.943)≈−2000−5791≈−7791The upper bound of the confidence interval is(x¯¯¯1−x¯¯¯2)+zcσ21n1+σ22n2−−−−−−−√=(55,000−57,000)+2.576(12,000)2105+(18,000)288−−−−−−−−−−−−−−√≈−2000+2.576(2247.943)≈−2000+5791≈3791 Note that when using a TI-83, TI-83 Plus, or TI-84 calculator or Google Sheets to find the confidence interval, the lower bound is −$7,790 and the upper bound is $3,790.

An upscale restaurant has establishments in Boston and Chicago. The manager of the Boston restaurant samples 55 random checks and finds the average tip percentage, as a decimal, is 0.25. The manager of the Chicago restaurant samples 45 random checks and finds the average tip percentage, as a decimal, is 0.23. Assume the population standard deviations for the tip percentages in the Boston restaurant and the Chicago restaurant are 0.028 and 0.024, respectively. Find the 90% confidence interval for the difference in tip percentages at the two restaurants. Let the Boston tips be the first sample, and let the Chicago tips be the second sample. Assume the samples are random and independent. Assume that both the population distributions are normally distributed. Round your answer to two decimal places.

Correct answers:1$0.01\pm0.01$0.01±0.01​2$0.03\pm0.01$0.03±0.01​ In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=0.25, σ1=0.028, and n1=55, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=0.23, σ2=0.024, and n2=45, respectively. According to technology or a z-table, for a 90% confidence interval, zc=1.645. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(0.25−0.23)−1.6450.028255+0.024245−−−−−−−−−−−−−√≈0.0114 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(0.25−0.23)+1.6450.028255+0.024245−−−−−−−−−−−−−√≈0.0286 Using TI-83 or TI-84 calculators or Google Sheets to find the exact interval with no intermediate rounding also produces a confidence interval of (0.01,0.03).

The owner of a restaurant is reviewing customer complaints. In a random sample of 227 complaints, 57 complaints were about the slow speed of the service. Create a 95% confidence interval for the proportion of complaints that were about the slow speed of the service. Use Excel to create the confidence interval, rounding to four decimal places.

Correct answers:1$0.1947$0.1947​2$0.3075$0.3075​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=57, and sample size, n=227, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.2511. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0288. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0564. The confidence interval for the population proportion has a lower limit of A4−A7=0.1947 and an upper limit of A4+A7=0.3075. Thus, the 95% confidence interval for the population proportion of complaints that were about the slow speed of the service, based on this sample, is approximately (0.1947, 0.3075).

In a random sample of 350 attendees of a minor league baseball game, 184 said that they bought food from the concession stand. Create a 95% confidence interval for the proportion of fans who bought food from the concession stand. Use Excel to create the confidence interval, rounding to four decimal places.

Correct answers:1$0.4734$0.4734​2$0.5780$0.5780​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=184, and sample size, n=350, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.5257. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0267. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0523. The confidence interval for the population proportion has a lower limit of A4−A7=0.4734 and an upper limit of A4+A7=0.5780. Thus, the 95% confidence interval for the population proportion of fans who bought food from the concession stand, based on this sample, is approximately (0.4734, 0.5780).

A city is holding a referendum on increasing property taxes to pay for a new high school. In a survey of 458 likely voters, 254 said that they would vote "yes" on the referendum. Create a 95% confidence interval for the proportion of likely voters who will vote "yes" on the referendum. Use Excel to create the confidence interval, rounding to four decimal places.

Correct answers:1$0.5091$0.5091​2$0.6001$0.6001​ The confidence interval for the unknown population proportion p is (p^−z⋆p^(1−p^)n−−−−−−−−√,p^−z⋆p^(1−p^)n−−−−−−−−√). The confidence interval can be calculated using Excel. 1. Identify α. Click on cell A1 and enter =1−0.95 and press ENTER. 2. Thus, α=0.05. Enter the number of successes, x=254, and sample size, n=458, in the Excel sheet in cells A2 and A3. To find the proportion of successes, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^≈0.5546. Use the NORM.S.INV function in Excel to find z⋆. Click on cell A5, enter =NORM.S.INV(1−A1/2), and press ENTER. 4. The answer for z⋆, rounded to two decimal places, is z⋆≈1.96. To calculate the standard error, p^(1−p^)n−−−−−−−−√, click on cell A6 and enter =SQRT(A4∗(1−A4)/A3) and press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1−p^)n−−−−−−−−√≈0.0232. To calculate the margin of error, z⋆p^(1−p^)n−−−−−−−−√, click on cell A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is z⋆p^(1−p^)n−−−−−−−−√≈0.0455. The confidence interval for the population proportion has a lower limit of A4−A7=0.5091 and an upper limit of A4+A7=0.6001. Thus, the 95% confidence interval for the population proportion of likely voters who will vote "yes" on the referendum, based on this sample, is approximately (0.5091, 0.6001).

Question Alice wants to estimate the percentage of people who own a mountain bike. She surveys 230 individuals and finds that 150 own a mountain bike. What are the sample proportions for successes, p′, and failures, q′? Round your answers to three decimal places.

Correct answers:1$0.652$0.652​2$0.348$0.348​ To form the sample proportion, take the number of successes, and divide it by n, the number of trials. In this scenario, the number of successes is the number of people who own a mountain bike, 150. The total number of people surveyed was 230. p′=xn=150230=0.652 So, p′=0.652 is the sample proportion, which is the point estimate of the population proportion. Since p′+q′=1, we can solve for q′.q′=1−p′=1−0.652=0.348

Researchers wanted to compare whether a driver's speed on the highway was affected by whether the driver was a resident of the state. The speeds of cars on a stretch of Interstate 95 with a speed limit of 65 miles per hour in central Connecticut were recorded. A sample of 72 cars with Connecticut license plates had an average speed of 70.2 miles per hour. A sample of 53 cars with out-of-state license plates had an average speed of 68.6 miles per hour. Assume the population standard deviations of the speeds of cars with Connecticut license plates and the cars with out-of-state license plates are 3.4 and 4.2 miles per hour, respectively. Find the 95% confidence interval for the difference in mean speed between the cars with Connecticut license plates and the cars with out-of-state license plates. Let the cars with Connecticut license plates be the first sample and let the cars with out-of-state license plates be the second sample. Assume that both the population distributions are normally distributed. Round your answer to two decimal places. Do not include units in your answer.

In this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=70.2, σ1=3.4, and n1=72, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=68.6, σ2=4.2, and n2=53, respectively. According to technology or a z-table, for a 95% confidence interval, zc=1.96. Substitute these values into the expression for the lower bound of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(70.2−68.6)−1.963.4272+4.2253−−−−−−−−−−√≈0.22 Substitute these values into the expression for the upper bound of the confidence interval: (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(70.2−68.6)+1.963.4272+4.2253−−−−−−−−−−√≈2.98 Using TI-83 or TI-84 calculators or Google Sheets to find the exact interval with no intermediate rounding also produces a confidence interval of (0.22,2.98).

Key Terms Confidence level: probability that the population mean or proportion falls within the confidence interval Confidence interval: the region formed between a point estimate minus the margin or error and the point estimate plus the margin of error

Key Terms Confidence level: probability that the population mean or proportion falls within the confidence interval Confidence interval: the region formed between a point estimate minus the margin or error and the point estimate plus the margin of error

During a recent survey, the average salary for male nurses was determined to be $59,500, based on a sample of 45 male nurses. A survey was also conducted for the average salary of female nurses, which was determined to be $57,200, based on a sample of 42 female nurses. Assume the population standard deviation for male nurse salaries is $7000, and assume the population standard deviation for female nurse salaries is $6500. Construct a 95% confidence interval for the difference between male nurse salaries and female nurse salaries. (Round your answers to the nearest whole number.)

Let the male nurse salaries be the first sample and let the female nurse salaries be the second sample. Then, in this scenario, the sample mean, population standard deviation, and sample size for the first sample are x¯1=59,500, σ1=7000, and n1=45, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=57,200, σ2=6500, and n2=42, respectively. According to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expressions for the bounds of the confidence interval: (x¯1−x¯2)−zcσ21n1+σ22n2−−−−−−−−√=(59,500−57,200)−1.967000245+6500242−−−−−−−−−−−−√≈−537 (x¯1−x¯2)+zcσ21n1+σ22n2−−−−−−−−√=(59,500−57,200)+1.967000245+6500242−−−−−−−−−−−−√≈5137 The 95% confidence interval is approximately (−537,5137). This means we are 95% confident that the true difference between population means of male and female nurse salaries is between −$537 and $5137. (Using the TI-83 or TI-84 calculators or Google Sheets produces a similar result.)

As the error bound of the proportion (EBP) increases, what is the effect on the sample size?

Sample size decreases. As the EBP increases, the sample size decreases.

The Parameters for a Confidence Interval with a Proportion

The procedure to find the confidence interval for a proportion is similar to that for the population mean, but the formulas are a bit different. We are still looking for a range of numbers, written as: (point estimate - margin of error, point estimate + margin of error), with (point estimate - margin of error) as the lower value and (point estimate + margin of error) as the upper value of the confidence interval. Since we're looking for a point estimate for the true proportion and margin of error for proportions here, instead of means, the point estimate will be the sample proportion, p′, and the margin of error will be the error bound for the proportion (EBP). How do you know you are dealing with a proportion problem? First, the underlying distribution is a binomial distribution, and there is no mention of a mean or average. If X is the binomial random variable, then X∼B(n,p), where n is the number of trials and p is the probability of a success.

Find the Confidence Interval Given a Population Proportion Finding the Error Bound for the Proportion (EBP)

This error bound formula is similar to the error bound formula for a mean, except that the "appropriate standard deviation" is different. For a mean, when the population standard deviation is known, the appropriate standard deviation that we use is σn√. For a proportion, however, the appropriate standard deviation is p′q′n−−−√. However, in the error bound formula, we use p′q′n−−−√, where p′ and q′ are sample proportion estimates of the unknown population proportions p and q. Finding the z-score to calculate the EBP will look the same as when we calculate the error bound for means (EBM). We need to find the value of z that puts an area equal to the confidence level (in decimal form) in the middle of the standard normal distribution. The confidence level is CL=1-α, which is shown by the shaded area in the figure below. So, α is the (non-shaded) area that is split equally between the two tails. Since it is split equally, we can write each individual area (in the tails) as α2. So, in order to find the z-value for a certain confidence level, we will use the value for α2. For example, Let's say we want to find the z-score that corresponds with a 95% confidence level. We can compute α first. α=1−CL=1−0.95=0.05 This means that α2=0.052=0.025. We then use the Standard Normal Table to find the value that corresponds with zα2=z0.025. We can see that z0.025=1.960. Once we determine the z-score, we can substitute the values - p′, q′, n, and zα2 - into the formula to calculate the EBP. We will examine this calculation in detail in the examples below.