The NWBC found that 16.5% of women-owned businesses did not pro- vide any employee ben- efits. What sample size could be 99% confident that the estimated (sam- ple) proportion is within 6 percentage points of the true population propor- tion? Question Suppose a market re- searcher wants to deter- mine the current percent- age of online shoppers who are males. How many online shoppers should the researcher survey in order to be 90% confident that the estimated (sam- ple) proportion is within 3 percentage points of the Correct an- swers:$254$254 EBP = 6%. ±=0.01T. he z-score is found 0.012=0.005=z±2=2.576 n=(z±2)2p(Ù2)(q2)EBP2 =(2.5762)(0.165)(0.835)0.06 n=253.96 Rounded up to 254 peo- ple. Correct an- swers:$752$752 Given the information in the question, EBP=0.03 since 3%=0.03 and z±2=- z0.05=1.645 because the confidence level is 90%. The values of p2 and q2 are unknown, but using a value of 0.5 for p2 will result in the largest pos- sible product of p2q2, and thus the largest possible n. Question A recent survey asked 1,379 top executives about business trends. The surveyed showed that 23% want to strengthen in- novation to capitalize on new opportunities. What is the value of q2 as a deci- mal? Round to the nearest hundredth. Answer Explanation Correct answers: 0.77 The value of q2 is: 1 p2=1 0.23=0.77 Correct an- swers:1$0.652$0.652 2$0. To form the sample pro- portion, take the number Question Alice wants to estimate the percentage of people who own a mountain bike. She surveys 230 individu- als and finds that 150 own a mountain bike. What are the sample proportions for of successes, and divide it by n, the number of tri- als. In this scenario, the number of successes is the number of people who own a mountain bike, 150. The total number of peo- ple surveyed was 230. p2=xn==0.652 successes, p2, and failures,So, p2=0.652 is the sample q2? Round your answers to three decimal places. proportion, which is the point estimate of the pop- ulation proportion. Since p2+q2=1, we can solve for q2.q2=1 p2=1 0.652=0.3 Correct Using p2=0.652, q2=0.348,an- and n=230, what is the 95% confidence interval for the proportion of the population who own a mountain bike? swers:1$0.590$0.590 2$0. We know n=230, p2=0.652, and q2=0.348. Since the confidence level is 95%, ±=1 0.95=0.0a5n,d ±2=0.025T. he z-value for Use the table of common z-scores above. Round z0.025 is 1.96. EBP=(z±2)(- p2q2n )=(1.96)((0 your answers to three dec- imal places. The Parameters for a Con- fidence Interval with a Pro- portion So we can write the confidence interval as (0.652 0.062,0.652+0.062) 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 look- ing for a range of numbers, written as: (point estimate - margin of error, point es- timate + margin of error), with (point estimate - mar- gin of error) as the lower value and (point estimate + margin of error) as the upper value of the confi- dence interval. Since we're looking for a point estimate for the true proportion and mar- gin of error for proportions here, instead of means, the point estimate will be the sample proportion, p2, Finding the Sample Pro- portion (p2) 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 under- lying distribution is a bino- mial distribution, and there is no mention of a mean or average. If X is the bino- mial random variable, then XB(n,p), where n is the number of trials and p is the probability of a suc- cess. 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 p2(read as "p prime"). p'=x/n p2 is the estimated propor- tion of successes, and will be used as the point esti- Question Suppose that a research firm wants to estimate the percent of families living in a city who have pets. Five hundred randomly se- lected families in this city are surveyed. Of the 500 families surveyed, 321 re- sponded "yes" - they own a pet. What is the value 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==0.642 of p2, the estimated propor-So, p2=0.642 is the sample tion of successes? proportion, which is the point estimate of the pop- ulation proportion. 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 Face- book. What is the value of Correct an- swers:$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 p2, the estimate proportionsurveyed was 1,597. of Facebook users in this research study? Answer in decimal form, rounding to the nearest hundredth. p'=xn==0.71 So, p2=0.71 is the sample proportion, which is the point estimate of the pop- ulation proportion This error bound formu- la is similar to the error bound formula for a mean, except that the "appropri- ate standard deviation" is different. For a mean, when the pop- ulation standard deviation is known, the appropriate standard deviation that we use is Ãn . For a proportion, however, the appropriate standard deviation is p2q2n . er, in the error bound for- mula, we use p2q2n p2 and q2 are sample pro- portion estimates of the unknown population pro- portions p and q. Find- ing the z-score to calcu- late the EBP will look the same as when we calcu- late the error bound for means (EBM). We need to find the value of z that puts an area equal to the confidence level (in deci- mal form) in the middle of the standard normal dis- tribution. The confidence level is CL=1-±,which is shown by the shaded area in the figure below. So, ± is the (non-shaded) area that is split equally be- tween the two tails. Since it is split equally, we can write each individual area (in the tails) as ±2S. o, in Find the Confidence In- terval Given a Population Proportion Finding the Error Bound for the Proportion (EBP) order to find the z-value for a certain confidence level, we will use the value for ±2. For example, Let's say we want to find the z-score that corresponds with a 95% confidence level. We can compute ±first. ±=1CL=1 0.95=0.05 This means that ±2=0.052=0.025W. e then use the Standard Normal Table to find the value that corresponds with z±2=- z0.025. We can see that z0.025=1.960. Once we determine the z-score, we Question Suppose that a research firm wants to estimate the 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 p2=0.642, and so we can find the value of q2, because q2=1 p2. So, q2=1 0.642=0.358. We are given the confi- dence level (CL), 98%, or 0.98. So, we can calculate alpha (±). ±=1CL =1 0.98 =0.02 Since ±=0.02w, e know that ±2=0.022=0.01S. o we will percent of families living in a city who have pets. Five hundred randomly se- lected families in this city are surveyed. Of the 500 families surveyed, 321 re- sponded "yes" - they own a pet. What is the er- ror bound for proportions 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=(z±2)p(2q2n (EBP) of a confidence in- =(2.326)((0.642)(0.358)(500) terval with a 98% confi- dence level? H(2.326)(0. H(2.326)(0.00046 H0.050 Question 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 p2=0.71, and so we can find the val- The Pew Social Media ue of q2, because q2=1 p2. Update 2014 surveyed 1,597 adult internet users on which social media websites they use. Of q2=1 0.71=0.29. We are given the confi- dence level (CL), 99%, or 0.99. So, we can calculate alpha (±). ±=1CL =1 0.99 =0.01 Since ±=0.01,we know that ±2=0.012=0.005S. o the users surveyed, 1,134 responded "yes" when asked if they use Face- book. From the ques- tion above, we know that p2=0.71. What is the er- ror bound for proportions (EBP) of a confidence in- terval with a 99% confi- dence level? 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=(z±2)p(2q2n =(2.576)((0.71)(0.29)(1597) H(2.576)(0. H(2.576)(0. H0.029 Find the Confidence In- terval Given a Population A confidence interval, cre- ated for an unknown pop- ulation 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 popu- lation parameter. So, when we are trying to estimate the population proportion, the point estimate is the sample proportion, p2. The margin of error is the error bound, found with the fol- lowing formula: Proportion Finding the Confidence In- terval With a Proportion IMPORTANT: When find- ing confidence inter- vals for proportions, they should only be used if the number of successes np2 and the number of failures EBP=(z±2)p(2q2n ...where z±2is the z-score that corresponds with the confidence level ±,p2 is the estimated proportion of successes, q2 is the es- timated proportion of fail- ures, and n is the sam- ple size. The z-scores for the five most common con- fidence intervals can be found in the portion of the Standard Normal Table be- low: So, using the sample pro- nq2 are both greater than on as the point esti- mate, and the EBP as the margin of error, the confi- dence interval at a confi- dence level, ±,can be writ- ten as: (p' EBP,p'+EBP) Solution Using z-scores: We can use the informa- tion in the question to de- fine some parameters: n=80 (sample size) p2 =6880=0.85 (sample proportion of successes) q2=1 p2=1 0.85=0.15 (sa ple proportion of failures) Next, we should check if np2 is greater than 5, so we know we should use this method of find- ing a confidence interval. np2=(80)(0.85)=68, so we can continue. To find the confidence in- terval, you need the sam- ple proportion, p2, and the EBP. We know p2=0.85. Question Professor Cope wants to gauge how many stu- dents passed his final exam. Students can either We can calculate the EBP with the formula: EBP=(z±2)p(qn ) We know the values for p2, q2, and n. We need to find the z-value before we can find EBP. Since the confidence lev- el is 95%, ±=1 0.95=0.05, and ±2=0.025W. e can use the table above to find the z-value for z0.025, which is 1.960. Now we can calculate the EBP. EBP=(z±2)p(2q2n =(1.960)((0.85)(0.15)80 H(1.960)(0.0016 score a "Pass" or "Fail" on their exam. He ran- domly selects 80 exams and finds that 68 students who passed. What is the 95% confidence interval for the population propor- tion of Professor Cope's students who passed their final exams. H0.0782 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) 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 EN- TER. Arrow to x, and enter 68. (number of successes) Arrow to n, and enter 80. Question The Pew Social Media Update 2014 surveyed Correct answer: (0.71 0.029,0.71+0.029)=( We can use the informa- tion in the question to de- fine some parameters: n=1597 (sample size) p2 ==0.71 (sam- ple proportion of success- es) 1,597 adult internet users q2=1 p2=1 0.71=0.29 (sa on which social media websites they use. Of the users surveyed, 1,134 ple proportion of failures) Since the confidence level is 99%, ±=1 0.99=0.0a1n,d responded "yes" when ±2=0.005T. he z-value for asked if they use Face- z0.005 is 2.576. Now we can calculate the EBP. EBP=(z±2)p(2q2n =(2.576)((0.71)(0.29)1597 H(2.576)(0. book. What is the confi- dence interval, at the 99% confidence level, for the H0.029 Confidence intervals are proportion of the popula- written as (p' EBP,p'+EBP). tion of households who So we can write this as: have more than one vehi- (0.71 0.029,0.71+0.029) or cle? (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. Correct an- Using p2=0.167, q2=0.833,swers:1$0.113$0.113 2$0. and n=180, what is the 95% confidence interval We know n=180, p2=0.167, and q2=0.833. Since the confidence level is 95%, ±=1 0.95=0.0a5n,d ±2=0.025T. he z-value for for the proportion of the population who prefer brand named items? Question Suppose a restaurant wants to determine the current percentage of cus- tomers who eat out more than twice a month. How many customers should the restaurant survey in or- der to be 98% confident that the estimated (sam- ple) proportion is within 4 percentage points of the true population proportion z0.025 is 1.96. EBP=(z±2)(- p2q2n )=(1.96)((0 So we can write the confidence interval as (0.167 0.054,0.167+0.054) Correct an- swers:$846$846 Given the information in the question, EBP=0.04 since 4%=0.04 and z±2=- z0.01=2.326 because the confidence level is 98%. The values of p2 and q2 are unknown, but using a value of 0.5 for p2 will result in the largest pos- sible product of p2q2, and thus the largest possible n. If p2=0.5, then q2=1 0.5=0 Therefore, of customers who eat out n=z2p2q2EBP2=2.3262(0.5) more than twice a month? Round the answer up to the next integer to be sure the sample size is Question Suppose a pharmaceuti- cal industry wants to de- termine the current per- centage of customers who are hospitals. How many customers should the in- dustry survey in order to be 92% confident that the estimated (sample) pro- portion is within 4 percent- Correct an- swers:$480$480 Given the information in the question, EBP=0.04 since 4%=0.04 and z±2=- z0.04=1.751 because the confidence level is 92%. The values of p2 and q2 are unknown, but using a value of 0.5 for p2 will result in the largest pos- sible product of p2q2, and thus the largest possible n. If p2=0.5, then q2=1 0.5=0 Therefore, age points of the true pop- n=z2p2q2EBP2=1.7512(0.5) ulation proportion of cus- tomers who are hospitals? Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 480 cus- tomers. The NWBC found that 42.1% of women-owned businesses provided re- tirement plans contribu- tions. What sample size could be 95% confident that the estimated (sam- ple) proportion is within 4 percentage points of the true population propor- tion? Correct an- swers:$586$586 EBP = 4%. ±=0.05T. he z-score is found 0.052=0.025=z±2=1.960 n=(z±2)2p(Ù2)(q2)EBP2 =(1.9602)(0.421)(0.579)0.04 n=585.27 Rounded up to 586 peo- ple. Solution In order to find the sample size, n, we need to plug values into the equation: n=(z±2)2p('q')(EBP2) From the problem, we know that EBP=0.03 (or 3%). We are also giv- en the confidence inter- val, 90%, so we can find the value for z±2from the table. ±=1CL=1 .90=0.10. So, ±2=0.102=0.05F. rom Question Suppose a mobile phone company wants to deter- mine the current percent- age of senior customers (those over the age of 50 years) who use text messaging. How many se- nior customers should the company survey in or- der to be 90% confident the table, z0.05=1.645. The only two parameters we are missing are p2 and q2 (the sample propor- tions for successes and failures). Remember that q'=1-p'. But, we do not know p2 yet. Since we mul- tiply p2 and q2together, we make them both equal to 0.5 because this will yield the largest possible prod- uct: p'q'=(0.5)(0.5)=0.25. (Try other prod- ucts: (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 pro- portion. To calculate the sample size, use the formula and make the substitutions. n=(z±2)2p('q')(EBP2) that the estimated (sam- ple) proportion is within 3 percentage points of the true population proportion of senior customers who actually use text messag- ing? =(1.645)2(0.5)(0.5)(0.032) H(2.706)(0.25)0.0009 H751.67 Because we're talking about surveying people, we need to round the an- swer to the next higher value. The sample size should be 752 senior cus- tomers. Answer Explanation Correct answer: 542 customers From the problem, we Question Suppose an internet mar- keting company wants to determine the current per- centage of customers who click on ads on their smart- phones. How many cus- tomers should the compa- ny survey in order to be 98% confident that the es- timated proportion is with- in 5 percentage points of the true population pro- portion of customers who click on ads on their smart- phones? know that EBP=0.05 (or 5%). We are also giv- en the confidence inter- val, 98%, so we can find the value for z±2from the table. ±=1CL=1 .98=0.02. So, ±2=0.022=0.01From the table, z0.01=2.326. We make p2 and q2 both equal to 0.5 because this will yield the largest possi- ble product. To calculate the sample size, use the formula and make the substitutions. n=(z±2)2p('q')(EBP2) =(2.326)2(0.5)(0.5)(0.052) H(5.410)(0.25)0.0025 H541.0276 Suppose a lawn and gar- den company wants to de- termine the current per- centage of customers who use fertilizer on their lawns. How many cus- tomers should the compa- ny survey in order to be 95% confident that the es- timated (sample) propor- tion is within 4 percent- age points of the true pop- ulation proportion of cus- tomers who use fertilizer on their lawns? Correct an- swers:$601$601 Given the information in the question, EBP=0.04 since 4%=0.04 and z±2=- z0.025=1.96 because the confidence level is 95%. The values of p2 and q2 are unknown, but using a value of 0.5 for p2 will result in the largest pos- sible product of p2q2, and thus the largest possible n. If p2=0.5, then q2=1 0.5=0 Therefore, n=z2p2q2EBP2=1.962(0.5)( Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 601 cus- tomers. Question Suppose a realtor wants to determine the current percentage of customers who have a family of five or more. How many cus- tomers should the realtor survey in order to be 98% confident that the estimat- ed (sample) proportion is within 2 percentage points of the true population pro- Correct an- swers:$3382$3382 Given the information in the question, EBP=0.02 since 2%=0.02 and z±2=- z0.01=2.326 because the confidence level is 98%. The values of p2 and q2 are unknown, but using a value of 0.5 for p2 will result in the largest pos- sible product of p2q2, and thus the largest possible n. If p2=0.5, then q2=1 0.5=0 Therefore, portion of customers who n=z2p2q2EBP2=2.3262(0.5) have a family of five or more? A company has conduct- ed a poll to know the awareness among con- Round the answer up to the next integer to be sure the sample size is large enough. The sample should include 3382 cus- tomers. sumers for its various product ranges. From a sample of 55 male respon- dents and 40 female re- spondents, the company found that the male re- spondents were aware of about 15 of its products and female respondents were aware of about 12 Correct an- of its products, on aver- swers:1$-1.29pm0.01$ 1.2 age. The population stan- dard deviation for male re- Consider the following val- spondents is 14 and for ues:x¯¯¯1=15x¯¯¯2=12zc=1 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 normal- ly distributed. Compute a 95% confidence interval for the difference in popu- Thus, the 95% confidence interval is ( 1.29,7.29). Question A college has found from past records that the wages among female stu- dents are $55 on aver- age and the wages among male students are $45 on average. The data was obtained from a sample Correct answers:$text{up- per bound=}15.79$upper bound=15.79 Consider the following val- ues: x¯¯¯1x¯¯¯2Ãz1cÃ2- of 105 female students n1n2=55=45=2.575=12=18= and 88 male students. The population standard deviations for female and male students' wages were $12 and $18, re- spectively. Compute 99% confidence intervals for the difference in popu- lation means. The confi- dence interval is approx- imately (4.2114, ). Assume that the distrib- utions of the wages of men and women are nor- The lower bound of the confidence interval is(x¯¯¯1 x¯¯¯Ã22)n11+zÃc 22- n2 =(55 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 news organization inter- ested in chronicling winter holiday travel trends con- ducted a survey. Of the Correct answer: ( 0.481, 0.144) In this scenario, the sam- ple proportion and sample size for the first sample are p^1=4296=0.4375 and n1=96, respectively. Then 96 people surveyed in the q^1=1 0.4375=0.5625. The eastern half of a country, 42 said they fly to visit fam- ily members for the winter holidays. Of the 108 peo- ple surveyed in the west- sample proportion and sample size for the second sample are p^2=81108=0.75 and n2=108, respectively. ern half of the country, 81 Then q^2=1 0.75=0.25. Ac- said they fly to visit fam- ily members for the win- ter holidays. Construct a 99% confidence interval for the difference in popu- cording to the table, for a 99% confidence inter- val zc=2.575. Substitute these values into the ex- pressions for the bounds of the confidence interval: (p^1 p^2) zcp^1q^1n1+p^ (p^1 p^2)+zcp^1q^1n1+p^2 lation proportions of peo- ple in the eastern half of a country who fly to vis- it family members for the winter holidays and peo- ple in the western half of a The 99% confidence in- terval is approximately ( 0.481, 0.144). This mea we are 99% confident that the true difference in pop- ulation proportions of peo- country who fly to visit fam- ple in the eastern half of ily members for the win- ter holidays. Assume that random samples are ob- tained and the samples are independent. (Round your answers to three dec- imal places.) a country who fly to vis- it family members for the winter holidays and peo- ple in the western half of a country who fly to vis- it family members for the winter holidays is between 48.1% and 14.4%. (Usin the TI-83 or TI-84 calcula- tors or Google Sheets pro- duces a similar result.) Correct answer: (0.029,0.051) In this scenario, the sam- ple proportion and sam- ple size for the first Question A recent survey on the usage of alcohol among youth provided the follow- ing data: sample are p^1=0.25 and n1=7000, respective- ly. Then q^1=1 0.25=0.75. The sample propor- tion and sample size for the second sam- ple are p^2=0.21 and n2=8500, respectively. Year: 2005; Sample size: Then q^2=1 0.21=0.79. Ac- 7000; Youth who are con- suming alcohol: 25%Year: 2015; Sample size: 8500; Youth who are consuming alcohol: 21% Construct a 90% confi- dence interval for the dif- cording to the table, for a 90% confidence inter- val zc=1.645. Substitute these values into the ex- pressions for the bounds of the confidence interval: ference in population pro- (p^1 p^2) zcp^1q^1n1+p^ portions of youth who were consuming alcohol in 2005 and youth who (p^1 p^2)+zcp^1q^1n1+p^2 were consuming alcohol in 2015. Assume that random samples are ob- The 90% confidence in- terval is approximately (0.029,0.051). This means we are 90% confident that the true difference tained and the samples are independent. (Round your answers to three dec- imal places.) A gubernatorial candidate wants to analyze how she is performing with differ- ent age groups. Her cam- paign staff surveyed 150 random constituents in the 18-24 age range and 175 constituents in the 65+ age range. Of those con- stituents, 67 of the people 18-24 years old and 90 of the people 65+ years old said they planned on vot- ing for the candidate for governor. Find the upper bound of the 99% confi- between population pro- portions 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.) Correct answers:$text{up- per bound=}0.075$upper bound=0.075 In this scenario, the sam- ple proportion and sam- ple size for the first sam- ple are p^1=67150H0.447 and n1=150, respective- ly. Then q^1H1 0.447=0.55 The sample propor- tion and sample size for the second sam- ple are p^2=90175H0.514 and n2=175, respectively. Then q^2=1 0.514=0.486. dence interval for the dif- ference in population pro- portions 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 According to the table, for a 99% confidence inter- val zc=2.576. Substitute these values into the ex- pression for the upper bound of the confidence interval: governor. Let the people (p^1 p^2)+zcp^1q^1n1+p^2 18-24 years old be sam- ple 1 and let the people 65+ years old be sample 2. Assume that the sam- ples are independent. En- ter your answer as a deci- mal rounded to three deci- mal places. A gubernatorial candidate wants to analyze how she is performing with differ- ent age groups. Her cam- paign staff surveyed 150 random constituents in the 18-24 age range and 175 constituents in the 65+ age range. Of those con- 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. Correct answer: We are 99% confident the difference in popula- tion proportions of people 18-24 years old who plan to vote for the candidate for governor and people 65+ years old who plan stituents, 67 of the people 18-24 years old and 90 of the people 65+ years old said they planned on vot- ing for the candidate for governor. Assuming that the samples are indepen- dent, the 99% confidence interval for the difference in population proportions of people 18-24 years old who plan to vote for the candidate for gover- nor and people 65+ years old who plan to vote for the candidate for governor 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 popula- tion parameter of interest. As such, we are 99% confi- dent the difference in pop- ulation proportions of peo- ple 18-24 years old who plan to vote for the candi- date for governor and peo- ple 65+ years old who plan to vote for the candidate is ( 0.210,0.075). Interpretfor governor is between this interval in context. 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 21.0% and 7.5%. 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 popula- tion that meets a certain criteria or test, represent- ed by p Question A factory QA supervisor examines two assembly percentage of a popula- tion that meets a certain criteria or test, represent- ed by p Correct answer: ( 0.028,0.053) In this scenario, the sam- ple proportion and sample size for the first sample are p^1=16256=0.0625 and n1=256, re- spectively. Then lines for defective prod- q^1=1 0.0625=0.9375. The ucts. A random sample of 256 products on one as- sembly line found that 16 were defective. A random sample of 240 products sample proportion and sample size for the second sample are p^2=12240=0.05 and n2=240, respectively. on another assembly line Then q^2=1 0.05=0.95. Ac- found that 12 were defec- tive. Construct a 95% con- cording to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expres- sions for the bounds of the confidence interval: (p^1 p^2) zcp^1q^1n1+p^ (p^1 p^2)+zcp^1q^1n1+p^2 fidence interval for the dif- ference in population pro- The 95% confidence in- terval is approximately portions of defective items ( 0.028,0.053). This means from the two assembly lines. Assume that the samples are independent. (Round your answers to three decimal places.) Question In the year 2004, a sur- vey was undertaken to we are 95% confident that the true difference in pop- ulation proportions of de- fective items from the two assembly lines is between 2.8% and 5.3%. (Using the TI-83 or TI-84 calcula- tors or Google Sheets pro- duces a similar result.) Correct answer: (0.001,0.079) In this scenario, the sam- ple proportion and sam- ple size for the first sample are p^1=0.95 and n1=300, respective- ly. Then q^1=1 0.95=0.05. The sample propor- find whether citizens of a certain political party sup- ported their party's can- tion and sample size for the second sam- ple are p^2=0.91 and n2=350, respectively. didate for governor. In a Then q^2=1 0.91=0.09. Ac- sample of 300 citizens of this party, 95% of them ex- pressed support for their party's candidate. A sim- ilar survey was conduct- ed four years later and cording to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expres- sions for the bounds of the confidence interval: showed that 91% of a (p^1 p^2) zcp^1q^1n1+p^ sample of 350 citizens of this party expressed sup- port for their party's can- (p^1 p^2)+zcp^1q^1n1+p^2 didate. Construct a 95% confidence interval for the difference in population proportions of citizens of this party who supported their gubernatorial candi- date in 2004 and citizens of this party who support- ed their gubernatorial can- The 95% confidence in- terval is approximately (0.001,0.079). This means we are 95% confident that the true difference in pop- ulation proportions of cit- izens of this party who supported their guberna- torial candidate in 2004 didate four years later. As- sume that random sam- ples are obtained and the samples are independent. (Round your answers to three decimal places.) Researchers wanted to compare whether a dri- ver's speed on the high- way 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 record- ed. 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 li- cense plates had an aver- and citizens of this par- ty who supported their gu- bernatorial candidate four years later is between 0.1% and 7.9%. (Using the TI-83 or TI-84 calcula- tors or Google Sheets pro- duces a similar result.) In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sample are x¯1=70.2,Ã1=3.4a, nd n1=72, respectively. The sample mean, population standard deviation, and sample size for the sec- ond sample are x¯2=68.6, Ã2=4.2a, nd n2=53, respec- tively. According to tech- nology or a z-table, for a 95% confidence interval, age speed of 68.6 miles per hour. Assume the population standard devi- ations of the speeds of cars with Connecticut li- cense plates and the cars zc=1.96. Substitute these values into the expression for the lower bound of the confidence interval: with out-of-state license (x¯1 x¯2Ã) 2n1z1c+Ã22- plates are 3.4 and 4.2 miles per hour, respective- ly. Find the 95% confi- dence interval for the dif- ference in mean speed be- tween the cars with Con- necticut license plates and the cars with out-of-state license plates. Let the cars with Connecticut license plates be the first sam- ple and let the cars with out-of-state license plates be the second sample. As- sume that both the popu- lation distributions are nor- mally distributed. Round your answer to two deci- n2 =(70. Substitute these values into the expression for the upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(70. Using TI-83 or TI-84 cal- culators or Google Sheets to find the exact inter- val with no intermediate rounding also produces a confidence interval of (0.22,2.98). Correct an- Brent, a pet supply store swers:1$-5.92pm0.01$ 5.9 owner, is comparing the weights of two different breeds of dogs. A sam- ple of 41 adult male Pem- broke 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. As- sume that the popula- tion standard deviations for the weights of Pem- broke Welsh corgis and border collies are 1.35 kilograms and 2.12 kilo- grams, respectively. Find the 90% confidence in- terval for the difference In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sam- ple are x¯1=11.85,Ã1=1.35, and n1=41, respectively. The sample mean, pop- ulation standard devia- tion, and sample size for the second sample are x¯2=17.11,Ã2=2.12a, nd n2=38, respectively. Ac- cording to technology or a z-table, for a 90% con- fidence interval, zc=1.645. Substitute these values into the expression for the lower bound of the confi- dence interval: (x¯1 x¯2Ã) 2n1z1c+Ã22- in mean weights between Pembroke Welsh corgis and border collies. Let the Pembroke Welsh cor- gis be the first sample, and let the border collies be the second sample. As- sume the samples are ran- dom and independent. As- sume that both the popu- lation distributions are nor- mally distributed. Round your answer to two deci- mal places. n2 =(11. Substitute these values into the expression for the upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(11. Using TI-83 or TI-84 cal- culators or Google Sheets to find the exact inter- val with no intermediate rounding also produces a confidence interval of ( 5.92, 4.60). A confidence interval for the difference between two population means, ¼1 and ¼2is, a confidence in- terval for the difference ¼1 The confidence interval for the difference in means makes use of a normal distribution and critical val- ues zc to construct the confidence interval. (This assumes that the popu- lation standard deviations are known for each group of interest.) In order to proceed with the confidence interval cal- culation, the following con- ditions must be met: The samples must be ran- domly selected. The samples must be in- dependent. The population standard deviations, Ã1and Ã2m, be known. ust Either both distributions must be known to be nor- mally distributed or both n1e30 and n2e30. Once these conditions are verified, then the sam- pling distribution for x¯1 x¯ (the difference between the sample means) will fol- low a normal distribution with mean =¼1an¼d 2stan- dard error =Ã2n11+Ã2n22 To form the confidence in- terval for the difference in means, we will also need to determine the critical value zc, which is identi- cal to the critical values discussed in previous con- tent related to confidence intervals for one mean. The value of zc is the z-score that cuts off an area of ±2in each tail of a normal distribution for a (1±)Å100c%onfidence in- terval. For example, when constructing a 95% confi- dence interval, the corre- sponding 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-val- ues for typical confidence levels of 90% confidence, 95% confidence, and 99% confidence are provided in the table below. Once the requirements cit- ed above are met, then the confidence interval for the difference between two population means ¼1is ¼2 given as follows: (x¯1 x¯2Ã) 2n1z1c+Ã2n22 ¼1 x¼¯12(x¯2)+Ãz2cn11+Ã22- n2 The TI-83 and TI-84 graph- ing calculators can also be used to assist with the cal- culations for a confidence interval for the difference between two population proportions. Select STAT, then TESTS, then 2-Sam- pZInt. Choose the Stats in- put (instead of the Data input). Enter values for Ã1, Compute Confidence In- tervals for the Difference in Population Means Confidence Intervals for the Difference Between Population Means Ã2x, ¯1, n1, x¯2, and n2, and then select the Con- fidence Level. Then select Calculate. Google Sheets can also be used to assist with the calculations for a confi- dence interval for the dif- ference between two pop- ulation proportions. Enter the sample mean, population standard devia- tion, and sample size for the first sample in cells A1, A2, and A3, respectively. Enter the sample mean, population standard devia- tion, and sample size for the second sample in cells B1, B2, and B3, respec- tively. Enter the decimal form of the confidence lev- el into cell C1. Enter =1 (1 C1)/2 into cell C2. E ter =NORMSINV(C2) into Key Terms Confidence level: proba- bility that the population mean or proportion falls within the confidence in- terval 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: proba- bility that the population mean or proportion falls within the confidence in- terval Confidence interval: the region formed between a point estimate minus the margin or error and the point estimate plus the margin of error Let the male nurse salaries be the first sam- ple and let the female nurse salaries be the During a recent survey, the average salary for male nurses was deter- mined to be $59,500, based on a sample of 45 male nurses. A sur- vey 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. As- sume the population stan- dard deviation for male nurse salaries is $7000, and assume the popula- tion standard deviation for female nurse salaries is $6500. Construct a 95% second sample. Then, in this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sample are x¯1=59,500,Ã1=7000, and n1=45, respectively. The sample mean, pop- ulation standard devia- tion, and sample size for the second sample are x¯2=57,200,Ã2=6500a, nd n2=42, respectively. Ac- cording to the table, for a 95% confidence interval zc=1.96. Substitute these values into the expres- sions for the bounds of the confidence interval: (x¯1 x¯2Ã) 2n1z1c+Ã22- n2 =(59, (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(59, confidence interval for the difference between male nurse salaries and female nurse salaries. (Round your answers to the near- est whole number.) The 95% confidence in- terval is approximately ( 537,5137). This means we are 95% confident that the true difference be- tween population means of male and female nurse salaries is between $537 and $5137. (Using the TI-83 or TI-84 calculators or Google Sheets pro- duces a similar result.) Correct an- swers:1$-5.01pm0.01$ 5.0 From a sample of 225 peo- ple in Pennsylvania, the mean age is 41 years. From a sample of 300 peo- ple in New Jersey, the mean age is 45 years. As- sume the population stan- In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sam- ple are x¯1=41,Ã1=6a, nd n1=225, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=45,Ã2=8, dard deviations for the ages of people from Penn- sylvania and New Jersey are 6 years and 8 years, respectively. Construct a 90% confidence interval for the difference in aver- age age between Pennsyl- vania residents and New Jersey residents. Enter your answer in the for- mat (lower , upper) where "lower" is the lower con- fidence 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 sam- ple. Round your answers to two decimal places. Do not include units in your answer. and n2=300, respectively. According to the table, for a 90% confidence inter- val, zc=1.645. Substitute these values into the ex- pressions for the bounds of the confidence interval: (x¯1 x¯2Ã) 2n1z1c+Ã22- n2 =(41 (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(41 The 90% confidence in- terval is approximately ( 5.01, 2.99). This means we are 90% confident that the true difference be- tween the mean Pennsyl- vania age and the mean New Jersey age is be- tween 5.01 years and 2. years. (Using the TI-83 or TI-84 calculators or A school has found from past records that the av- erage 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 devi- Correct answers:$text{up- per bound=}2.49$upper bound=2.49 Consider the following val- ations for girls' and boys' ues:x¯¯¯1=75x¯¯¯2=85zc=2 scores were 25 and 17, respectively. The 99% con- fidence interval for the difference of the popu- lation means (girls mi- nus boys)is approximate- Note that using a TI-83/84 calculator or Google Sheets to find the up- per bound directly results in a value of approxi- mately 2.494. Rounding to ly ( 22.49, ). Assumetwo decimal places yields that the distributions of the scores of boys and girls are normally distributed. Round your answers to two decimal places. 2.49. Correct an- An upscale restaurant has establishments in Boston and Chicago. The manag- er of the Boston restau- rant samples 55 random checks and finds the av- erage tip percentage, as a decimal, is 0.25. The manager of the Chica- go restaurant samples 45 random checks and finds the average tip percent- age, as a decimal, is 0.23. Assume the popu- lation standard deviations for the tip percentages in the Boston restaurant and the Chicago restau- rant are 0.028 and 0.024, respectively. Find the 90% confidence interval for the difference in tip percent- ages at the two restau- swers:1$0.01pm0.01$0.01± In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sam- ple are x¯1=0.25,Ã1=0.028, and n1=55, respectively. The sample mean, pop- ulation standard devia- tion, and sample size for the second sample are x¯2=0.23,Ã2=0.024a, nd n2=45, respectively. Ac- cording to technology or a z-table, for a 90% con- fidence interval, zc=1.645. Substitute these values into the expression for the lower bound of the confi- dence interval: (x¯1 x¯2Ã) 2n1z1c+Ã22- n2 =(0.2 Substitute these values into the expression for the rants. Let the Boston tips be the first sample, and let the Chicago tips be the second sample. As- sume the samples are ran- dom and independent. As- sume that both the popu- lation distributions are nor- mally distributed. Round your answer to two deci- mal places. An upscale restaurant has establishments in Boston and Chicago. The manag- er of the Boston restau- rant samples 55 random checks and finds the av- erage tip percentage is 0.25. The manager of the Chicago restaurant sam- ples 45 random checks and finds the average tip percentage is 0.23. As- sume the population stan- upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(0.2 Using TI-83 or TI-84 cal- culators or Google Sheets to find the exact inter- val with no intermediate rounding also produces a confidence interval of (0.01,0.03). Correct answers:$text{up- per bound=}0.03%$up- per bound=0.03% In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sam- ple are x¯1=0.25,Ã1=0.028, and n1=55, respectively. The sample mean, pop- ulation standard devia- tion, and sample size for dard deviations for the tip percentages in the Boston restaurant and the Chica- go restaurant are 0.028 and 0.024, respectively. Find the upper bound of the 90% confidence in- terval for the difference in tip percentages at the two restaurants. Let the Boston tips be the first sample, and let the Chica- go tips be the second sample. Assume the sam- ples are random and in- dependent. Assume that both the population distrib- utions are normally distrib- uted. Round your answer to two decimal places. A college has found from past records that the aver- age wage among female graduates is $55,000 per year and the average wage among male gradu- the second sample are x¯2=0.23,Ã2=0.024a, nd n2=45, respectively. Ac- cording to technology or a z-table, for a 90% con- fidence interval, zc=1.645. Substitute these values into the expression for the upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(0.2 Using TI-83 or TI-84 calcu- lators or Google Sheets to find the exact value with no intermediate rounding also produces a value of 0.03. ates is $57,000 per year. The data were obtained from a sample of 105 female students and 88 male students. The pop- ulation standard devia- tions for female and male students' wages were $12,000 and $18,000, re- spectively. Assume that the distributions of the Correct an- swers:1$-7790pm2$ 7790 Consider the following val- wages of men and women ues:x¯¯¯1=55,000x¯¯¯2=57 are normally distributed. Let the wages of the fe- male students be the first sample, and let the wages of the male students be the second sample. Com- pute a 99% confidence interval for the difference in population mean wages per year, rounding to the nearest dollar. Do not in- clude the dollar sign in the answer Note that when using a TI-83, TI-83 Plus, or TI-84 calculator or Google Sheets to find the confi- dence interval, the lower bound is $7,790 and the upper bound is $3,790. Researchers asked two groups of people to solve a crossword puzzle. One group of 36 people, who were provided a dictionary and thesaurus while work- ing, had a mean time of 57.2 minutes to solve the puzzle. A second group of 39 people, working with- out aid, had a mean time of 75.1 minutes to solve the puzzle. Assume the population standard devia- tions for the solution times for the people working with a dictionary and the- saurus and those given no aid were 8.1 minutes and 13.5 minutes, respectively. Find the 95% confidence interval for the population mean difference in solu- tion times between people working with a dictionary Correct an- swers:1$-22.90pm0.01$ 2 In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sample are x¯1=57.2,Ã1=8.1a, nd n1=36, respectively. The sample mean, population standard deviation, and sample size for the sec- ond sample are x¯2=75.1, Ã2=13.5a, nd n2=39, re- spectively. 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¯2Ã) 2n1z1c+Ã22- n2 =(57. Substitute these values and thesaurus and peo- ple working without aid. Let the people working with a dictionary and the- saurus be the first sample, and let the people work- ing without aid be the sec- ond sample. Assume the samples are random and independent. Assume that both the population distrib- utions are normally distrib- uted. Round your answer to two decimal places. A school has found from past records that the aver- age 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 stan- into the expression for the upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(57. Using TI-83 or TI-84 cal- culators or Google Sheets to find the exact val- ue with no intermediate rounding also produces a confidence interval of ( 22.90, 12.90). Correct an- swers:1$-17.98pm0.01$ 1 Consider the following val- dard deviations for boys' ues:x¯¯¯1=75x¯¯¯2=85zc=1 and girls' scores were 25 and 17, respectively. As- sume that the distributions of the scores of boys and Note that using a TI-83/84 girls are normally distrib- uted, and let the first sam- ple be the boys' scores and let the second sample be the girls' scores. Find the 90% confidence inter- val for the difference of the population means, round- ing to two decimal places. Researchers asked two groups of people to con- struct a jigsaw puzzle. One group of 55 people, who listened to soft clas- sical music while working, had a mean construction time of 45 minutes. Anoth- er group of 45 people, who worked in silence, had a mean construction time of 55 minutes. Assume the population standard devi- ations for the construc- tion times for people lis- tening to classical music and people working in si- calculator or Google Sheets to find the confi- dence interval directly re- sults in the approximate in- terval ( 17.98, 2.02). Correct an- swers:1$-15.36pm0.01$ 1 In this scenario, the sam- ple mean, population stan- dard deviation, and sam- ple size for the first sam- ple are x¯1=45,Ã1=9.5a, nd n1=55, respectively. The sample mean, population standard deviation, and sample size for the second sample are x¯2=55,Ã2=11, and n2=45, respectively. According to technology or a z-table, for a 99% con- fidence interval, zc=2.576. lence are 9.5 minutes and 11 minutes, respectively. Find the 99% confidence interval for the difference in population mean con- Substitute these values into the expression for the lower bound of the confi- dence interval: struction times between (x¯1 x¯2Ã) 2n1z1c+Ã22- people listening to classi- cal music as they work and people working in si- lence. Let the people who lis- tened to classical music be the first sample and let the people working in silence be the second sample. Assume the sam- ples are random and in- dependent. Assume that both the population distrib- utions are normally distrib- uted. Round your answer to two decimal places. n2 =(45 Substitute these values into the expression for the upper bound of the confi- dence interval: (x¯1 x¯2)+Ãz2cn11+Ã22- n2 =(45 Using TI-83 or TI-84 cal- culators or Google Sheets to find the exact val- ue with no intermediate rounding also produces a confidence interval of ( 15.36, 4.64). Correct answers:$left(1071.77, 1477.33right)$(1071.77, 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 ¼. Round your answers to two decimal places and use increasing order. 1477.33) Use Excel to calculate the 99% confidence interval, where ±=0.01and 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, us- ing the STDEV.S function. Thus, the sample mean, rounded to two decimal places, is 1274.55 and the sample standard devi- ation, rounded to two dec- imal places, is 335.93. 2. Click on any empty cell, enter =CONFI- DENCE.T(0.01,335.93,22), a 3. The margin of error, rounded to two decimal places, is 202.78. The confidence interval for the population mean has a A 98% confidence inter- val for ¼is (x¯ ±/z2nà ±/,2xï-+z Assume the distribution of commute times to a ma- jor city follows the nor- mal probability distribution and the standard devia- tion is 4.5 minutes. A ran- dom sample of 38 com- mute times is given below in minutes. Use Excel to find the 98% confidence interval for the mean trav- el time in minutes. Round your answers to one deci- n ).±H=e0r.e0,2, Ã=4.a5n, d n=38. Use Excel to calcu- late 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 dec- imal place, is x¯=27.6. 2. Click on any empty cell, enter =CONFI- DENCE.NORM(0.02,4.5,38) 3. The margin of error, rounded to one decimal In order to determine the Correct answers:$left(15.36, 19.28right)$(15.36, 19.28) A 95% confidence inter- average weight of car- val for ¼is (x¯ ±/z2nà ±/,2xï-+z ry-on luggage by passen- gers in airplanes, a sam- ple of 25 pieces of car- ry-on luggage was col- lected and weighed in pounds. Assume that the population is normally dis- tributed with a standard n ).±H=e0r.e0,5, Ã=a5n, d n=25. Use Excel to calcu- late 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 =CONFI- DENCE.NORM(0.05,5,25), a 3. The margin of er- deviation of 5 pounds. Find the 95% confidence interval of the mean weight in pounds. Round your answers to two dec- imal places and use as- cending order. ror, rounded to two decimal places, is z±/2Ã- n H1.96. The confiden terval for the popula- tion 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 ¼is (15.36, 19.28). Correct an- swers:1$0.3409$0.3409 2$ The confidence interval for the unknown population proportion p is (p^ zÆp^(1 p^)n 1. Identify ±C. lick on cell A1 and enter =1 0.95 and press ENTER. 2. Thus, ±=0.05E. nter the number of successes, A large company is con- cerned about the com- mute times of its employ- ees. Of the 333 employ- ees who were surveyed, 131 said that they had a x=131, and sample size, n=333, in the Excel sheet in cells A2 and A3. To find the proportion of success- es, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^H0.3934. Use the NORM.S.INV func- tion in Excel to find zÆ. Click on cell A5, en- ter =NORM.S.INV(1 A1/2), and press ENTER. daily commute longer than 4. The answer for zÆ, round- 30 minutes. Create a 95% ed to two decimal places, confidence interval for the is zÆH1.96. To calculate the proportion of employees who have a daily commute standard error, p^(1 p^)n click on cell A6 and enter =SQRT(A4 (1 A4)/A3) an press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1 p^)n late the margin of error, zÆp^(1 p^)n longer than 30 minutes. Use Excel to create the confidence interval, round- ing to four decimal places. A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is zÆp^(1 p^)n confidence interval for the population proportion has a lower limit of A4 A7=0.3409 and an up- per 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). Correct answers:$left(91.31, 95.17right)$(91.31, 95.17) Use Excel to calculate the 95% confidence interval, where ±=0.05and n=21. The heart rates for a group of 21 students taking a fi- nal 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. 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, us- ing the STDEV.S function. Thus, the sample mean is 93.24 and the sample standard deviation, round- ed to two decimal places, is 4.24. 2. Click on any empty cell, enter =CONFI- DENCE.T(0.05,4.24,21), and press ENTER. 3. The margin of error, rounded to two decimal places, is 1.93. The confi- dence interval for the pop- ulation 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 Answer Explanation (1$$, 2$$) Correct an- swers:1$0.6942$0.6942 2$ The confidence interval for the unknown population proportion p is (p^ zÆp^(1 p^)n 1. Identify ±C. lick on cell A1 and enter =1 0.95 and press ENTER. 2. Thus, ±=0.05E. nter 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 success- es, p^, click on cell A4 and enter =722/1000 and press ENTER. 3. Thus, p^=0.722. Use the NORM.S.INV func- tion in Excel to find zÆ. Click on cell A5, en- ter =NORM.S.INV(1 A1/2), and press ENTER. 4. The answer for zÆ, round- ed to two decimal places, is zÆH1.96. To calculate the In a survey of 1,000 adults in a country, 722 said standard error, p^(1 p^)n click on cell A6 and enter that they had eaten fast =SQRT(A4 (1 A4)/A3) an 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 cre- ate the confidence inter- val, rounding to four deci- mal places. press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1 p^)n late the margin of error, zÆp^(1 p^)n A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is zÆp^(1 p^)n confidence interval for the population proportion has a lower limit of A4 A7H0.6942 and an up- The yearly incomes, in thousands, for 24 random married couples living in a city are given below. As- sume the yearly incomes are approximately normal- Correct answers:$left(58.984, 59.026right)$(58.984, 59.026) Use Excel to calculate the 95% confidence interval, where ±=0.05and n=24. 1. Open Excel and en- ter the given data in col- umn A. Find the sample mean, x¯, using the AV- ERAGE function and the sample standard devia- tion, s, using the STDEV.S function. Thus, the sam- ple mean is 59.005 and the sample standard de- viation, rounded to three decimal places, is 0.049. 2. Click on any empty cell, enter =CONFI- DENCE.T(0.05,0.049,24), an ly 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. 3. The margin of error, rounded to three dec- imal places, is 0.021. The confidence inter- val for the population mean has a lower lim- it 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). Correct answers:$left(153.21, 161.43right)$(153.21, 161.43) Use Excel to calculate the 90% confidence interval, where ±=0.1and n=22. 1. Open Excel and enter In a city, 22 coffee shops are randomly selected, and the temperature of the coffee sold at each shop is noted. Use Ex- cel to find the 90% confi- dence interval for the pop- ulation mean temperature. Assume the temperatures are approximately normal- ly distributed. Round your answers to two decimal places and use increasing order. the given data in column A. Find the sample mean, x¯, using the AVERAGE function and the sample standard deviation, s, us- ing the STDEV.S function. Thus, the sample mean is 157.32, and the sample standard deviation, round- ed to two decimal places, is 11.21. 2. Click on any empty cell, enter =CONFI- DENCE.T(0.1,11.21,22), and press ENTER. 3. The margin of er- ror, rounded to two dec- imal places, is 4.11. The confidence inter- val for the population mean has a lower lim- it of 157.32 4.11=153.21 and an upper limit of 157.32+4.11=161.43. Thus, the 90% confidence The amounts of time that Correct answers:$left(44.89, 57.27right)$(44.89, 57.27) A 99% confidence inter- customers stay in a cer- val for ¼is (x¯ ±/z2nà ±/,2xï-+z tain restaurant for lunch is normally distributed with a standard deviation of 17 minutes. A random sam- ple of 50 lunch customers was taken at this restau- rant. Construct a 99% con- fidence interval for the true average amount of time customers spend in the restaurant for lunch. Round your answers to n ).±H=e0r.e0,1, Ã=1a7n, d n=50. Use Excel to calcu- late 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 =CONFI- DENCE.NORM(0.01,17,50), 3. The margin of er- ror, rounded to two decimal places, is z±/2Ã- two decimal places and use ascending order. The number of hours n H6.19. The confiden terval for the popula- tion 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 ¼is (44.89, 57.27). Correct answers:$left(2,090.03, 2,106.09right)$(2,090.03, 2, A 98% confidence inter- worked per year per adult val for ¼is (x¯ ±/z2nà ±/,2xï-+z in a state is normally dis- tributed with a standard deviation of 37. A sample of 115 adults is selected at random, and the num- ber of hours worked per year per adult is given be- low. Use Excel to calculate n ).±H=e0r.e0,2, Ã=3a7n, d n=115. Use Excel to cal- culate 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 the 98% confidence inter- val for the mean hours worked per year for adults in this state. Round your answers to two decimal places and use ascending order. =CONFI- DENCE.NORM(0.02,37,115 3. The margin of er- ror, rounded to two decimal places, is z±/2Ã- n H8.03. The confiden terval 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 ¼is (2,090.03, 2,106.09). Correct an- swers:1$0.5091$0.5091 2$ The confidence interval for the unknown population proportion p is (p^ zÆp^(1 p^)n 1. Identify ±C. lick on cell A1 and enter =1 0.95 and press ENTER. 2. Thus, ±=0.05E. nter the number of successes, A city is holding a referen- dum on increasing proper- ty taxes to pay for a new high school. In a survey of 458 likely voters, 254 said that they would vote "yes" x=254, and sample size, n=458, in the Excel sheet in cells A2 and A3. To find the proportion of success- es, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^H0.5546. Use the NORM.S.INV func- tion in Excel to find zÆ. Click on cell A5, en- ter =NORM.S.INV(1 A1/2), and press ENTER. on the referendum. Create 4. The answer for zÆ, round- a 95% confidence interval ed to two decimal places, for the proportion of likely is zÆH1.96. To calculate the voters who will vote "yes" on the referendum. Use standard error, p^(1 p^)n click on cell A6 and enter =SQRT(A4 (1 A4)/A3) an press ENTER. 5. The answer for the standard error, rounded to four decimal places, is p^(1 p^)n late the margin of error, zÆp^(1 p^)n Excel to create the con- fidence interval, rounding to four decimal places. A7 and enter =A5*A6 and press ENTER. 6. The answer for the margin of error, rounded to four decimal places, is zÆp^(1 p^)n confidence interval for the population proportion has a lower limit of A4 A7=0.5091 and an up- per 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). Correct an- swers:1$0.4734$0.4734 2$ The confidence interval for the unknown population proportion p is (p^ zÆp^(1 p^)n 1. Identify ±C. lick on cell A1 and enter =1 0.95 and press ENTER. 2. Thus, ±=0.05E. nter 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% confi- dence interval for the pro- portion of fans who bought 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 success- es, p^, click on cell A4 and enter =A2/A3 and press ENTER. 3. Thus, p^H0.5257. Use the NORM.S.INV func- tion in Excel to find zÆ. Click on cell A5, en- ter =NORM.S.INV(1 A1/2), and press ENTER. 4.
