1 2 3 4 5 6 \f\fa. Compute the value of the test statistic (to 2 decimals). Z=(phat-p)/sqrt(p*(1-p)/n) =(0.175-0.2)/sqrt(0.2*0.8/400) = -1.25 b. What is the p-value (to 4 decimals)? The p-value = 2*P(Z<-1.25) =0.2113 (from standard normal table) c. Using = .05, can you conclude that the population proportion is not equal to .20? Since the p-value is larger than 0.05, we do not reject Ho. So we cannot conclude that the population proportion is not equal to .20 Answer the next three questions using the critical value approach. d. Using = .05, what is the critical value for the test statistic? (+ or -) Given a=0.05, the critical values are Z(0.025)=-1.96 or 1.96 (from standard normal table) e. State the rejection rule: Reject H0 if z is less than the lower critical value or greater than the upper critical value. f. Using = .05, can you conclude that the population proportion is not equal to .20? No p0 = 0.125 a. H0: p = 0.125 H1: p > 0.125 b. x = 52 , n = 400 , p1=x/n = 0.13 standard error, SE = sqrt(p0(1-p0)/n) SE = sqrt(0.125(1-0.125) / 400) = 0.01654 . c test statistic, z = (p1-p0) / SE z = (0.13-0.125) / 0.01654) = 0.30 P-value = P(Z>0.3) = 1 - P(Z<0.3) = 1 - 0.6179 = 0.3821 Since P-value > 0.05, hence fail to reject null hypothesis. So, we conclude that union efforts to organize have increased union membership. \fPrice (cents per gallon) Price per Gallon (in dollars) 232.5 2.325 227.7 2.277 237.4 2.374 235.9 2.359 234.1 2.341 239.1 2.391 241.0 2.41 238.4 2.384 232.9 2.329 237.3 2.373 228.8 2.288 231.5 2.315 236.5 2.365 235.6 2.356 227.1 2.271 244.8 2.448 233.8 2.338 239.9 2.399 232.5 2.325 239.6 2.396 244.4 2.444 238.7 2.387 236.6 2.366 230.5 2.305 242.2 2.422 226.4 2.264 230.8 2.308 229.5 2.295 238.7 2.387 228.2 2.282 233.2 2.332 227.9 2.279 233.6 2.336 237.5 2.375 238.1 2.381 237.4 2.374 231.2 2.312 234.6 2.346 233.6 2.336 237.6 2.376 232.7 2.327 240.0 2.4 231.5 2.315 234.2 2.342 238.2 2.382 231.4 2.314 240.6 2.406 234.3 2.343 234.2 2.342 233.8 2.338 In this demo, we'll practice using Excel to compute p-values using the t distribution. The U.S. Energy Administration reported that the mean price for a gallon of regular gasoline in the United States was $2.36 (in 2006). We want to determine if gasoline prices in the Lower Atlantic states are lower than the national average. This file contains regular gasoline prices for a sample of 50 service stations from Lower Atlantic states. Use = 0.01 as your level of significance. (In this problem we do not know .) We'll be using the second column of numbers (the price in dollars). 1. Set up your hypotheses. H0: 2.36 Ha: < 2.36 This is a hypothesis test for an average, not a proportion. We also do not know , so we'll have to use the sample standard deviation s. 2. Use Excel to compute the sample mean (xbar). Use =AVERAGE(B2:B51) xbar = 2.3496 3. Use Excel to compute the standard deviation (s). Use =STDEV.S(B2:B51) s = 0.04437 4. Compute your test statistic using parts 2 and 3. = ( (/)=(2.34962.36)/(0.04437/50)=1.657 _0)/ 5. Compute the degrees of freedom (n - 1). n = 50, df = 50 - 1 = 49 6. To get the p-value for a lower tail test, type =1-T.DIST.RT( your test statistic, your degrees of freedom) in an empty cell. To get the p-value for an upper tail test, type =T.DIST.RT(your test statistic, your degrees of freedom) in an empty cell. To get the p-value for a two-tailed test: - If your test statistic is negative, type =2*(1-T.DIST.RT( your test statistic, degrees of freedom)) - If your test statistic is positive, type =2*T.DIST.RT( your test statistic, degrees of freedom) 7. What is the p-value for this test? We have a lower tail test, so in Excel we type: =1-T.DIST.RT( -1.657, 49) The p-value is 0.0520. 8. If your p-value is less than = 0.01 reject H0. If your p-value is greater than =0.01 do not reject H0. Our p-value is bigger than 0.01, so we do not reject H0. We do not have enough evidence to say that the average price of a gallon of gasoline is lower in the Lower Atlantic States. Sale Price 12400 10400 12100 10000 11000 8895 7675 9975 6350 10470 9895 11250 8795 12500 9340 10150 9200 9395 11000 10640 10000 7500 8000 10440 10200 10300 9740 9280 10930 8000 9000 7680 9400 10730 7350 12240 11970 8240 9910 10080 9440 8970 9500 10050 10130 11400 8500 7500 9090 10500 Excel Project 3 This Excel Project contains 7 questions. Work out the answers to the questions and then click on "Excel Project 3" to submit your answers. Directions for performing a hypothesis test for means in Excel are in the Excel Demo. **Give your answers to 2 decimal places!** According to the National Automobile Dealers Association, the mean price for used cars is $10,192. A manager of a Kansas City used car dealership claims that his prices are different than the national mean. To test his claim, he randomly selected 50 cars from his dealership. The sales prices for those 50 cars are in this data set. Use 0.05 as your level of significance. 1. What is the alternative hypothesis? 2. Compute the sample mean. 3. Compute the sample standard deviation. 4. Compute the test statistic. 5. What are the degrees of freedom? 6. Compute the p-value using Excel. 7. What is your decision? What does this decision mean in the context of this problem? Price (cents per gallon) Price per Gallon (in dollars) 232.5 2.325 227.7 2.277 237.4 2.374 235.9 2.359 234.1 2.341 239.1 2.391 241.0 2.41 238.4 2.384 232.9 2.329 237.3 2.373 228.8 2.288 231.5 2.315 236.5 2.365 235.6 2.356 227.1 2.271 244.8 2.448 233.8 2.338 239.9 2.399 232.5 2.325 239.6 2.396 244.4 2.444 238.7 2.387 236.6 2.366 230.5 2.305 242.2 2.422 226.4 2.264 230.8 2.308 229.5 2.295 238.7 2.387 228.2 2.282 233.2 2.332 227.9 2.279 233.6 2.336 237.5 2.375 238.1 2.381 237.4 2.374 231.2 2.312 234.6 2.346 233.6 2.336 237.6 2.376 232.7 2.327 240.0 2.4 231.5 2.315 234.2 2.342 238.2 2.382 231.4 2.314 240.6 2.406 234.3 2.343 234.2 2.342 233.8 2.338 In this demo, we'll practice using Excel to compute p-values using the t distribution. The U.S. Energy Administration reported that the mean price for a gallon of regular gasoline in the United States was $2.36 (in 2006). We want to determine if gasoline prices in the Lower Atlantic states are lower than the national average. This file contains regular gasoline prices for a sample of 50 service stations from Lower Atlantic states. Use = 0.01 as your level of significance. (In this problem we do not know .) We'll be using the second column of numbers (the price in dollars). 1. Set up your hypotheses. H0: 2.36 Ha: < 2.36 This is a hypothesis test for an average, not a proportion. We also do not know , so we'll have to use the sample standard deviation s. 2. Use Excel to compute the sample mean (xbar). Use =AVERAGE(B2:B51) xbar = 2.3496 3. Use Excel to compute the standard deviation (s). Use =STDEV.S(B2:B51) s = 0.04437 4. Compute your test statistic using parts 2 and 3. 2.34962.36 = / 0 = 0.04437/ 50 = 1.657 5. Compute the degrees of freedom (n - 1). n = 50, df = 50 - 1 = 49 6. To get the p-value for a lower tail test, type =1-T.DIST.RT( your test statistic, your degrees of freedom) in an empty cell. To get the p-value for an upper tail test, type =T.DIST.RT(your test statistic, your degrees of freedom) in an empty cell. To get the p-value for a two-tailed test: - If your test statistic is negative, type =2*(1-T.DIST.RT( your test statistic, degrees of freedom)) - If your test statistic is positive, type =2*T.DIST.RT( your test statistic, degrees of freedom) 7. What is the p-value for this test? We have a lower tail test, so in Excel we type: =1-T.DIST.RT( -1.657, 49) The p-value is 0.0520. 8. If your p-value is less than = 0.01 reject H0. If your p-value is greater than =0.01 do not reject H0. Our p-value is bigger than 0.01, so we do not reject H0. We do not have enough evidence to say that the average price of a gallon of gasoline is lower in the Lower Atlantic States. Sale Price 12400 10400 12100 10000 11000 8895 7675 9975 6350 10470 9895 11250 8795 12500 9340 10150 9200 9395 11000 10640 10000 7500 8000 10440 10200 10300 9740 9280 10930 8000 9000 7680 9400 10730 7350 12240 11970 8240 9910 10080 9440 8970 9500 10050 10130 11400 8500 7500 9090 10500 Excel Project 3 This Excel Project contains 7 questions. Work out the answers to the questions and then click on "Excel Project 3" to submit your answers. Directions for performing a hypothesis test for means in Excel are in the Excel Demo. **Give your answers to 2 decimal places!** According to the National Automobile Dealers Association, the mean price for used cars is $10,192. A manager of a Kansas City used car dealership claims that his prices are different than the national mean. To test his claim, he randomly selected 50 cars from his dealership. The sales prices for those 50 cars are in this data set. Use 0.05 as your level of significance. 1. What is the alternative hypothesis? 2. Compute the sample mean. 3. Compute the sample standard deviation. 4. Compute the test statistic. 5. What are the degrees of freedom? 6. Compute the p-value using Excel. 7. What is your decision? What does this decision mean in the context of this problem? \f\fa. Compute the value of the test statistic (to 2 decimals). Z=(phat-p)/sqrt(p*(1-p)/n) =(0.175-0.2)/sqrt(0.2*0.8/400) = -1.25 b. What is the p-value (to 4 decimals)? The p-value = 2*P(Z<-1.25) =0.2113 (from standard normal table) c. Using = .05, can you conclude that the population proportion is not equal to .20? Since the p-value is larger than 0.05, we do not reject Ho. So we cannot conclude that the population proportion is not equal to .20 Answer the next three questions using the critical value approach. d. Using = .05, what is the critical value for the test statistic? (+ or -) Given a=0.05, the critical values are Z(0.025)=-1.96 or 1.96 (from standard normal table) e. State the rejection rule: Reject H0 if z is less than the lower critical value or greater than the upper critical value. f. Using = .05, can you conclude that the population proportion is not equal to .20? No p0 = 0.125 a. H0: p = 0.125 H1: p > 0.125 b. x = 52 , n = 400 , p1=x/n = 0.13 standard error, SE = sqrt(p0(1-p0)/n) SE = sqrt(0.125(1-0.125) / 400) = 0.01654 . c test statistic, z = (p1-p0) / SE z = (0.13-0.125) / 0.01654) = 0.30 P-value = P(Z>0.3) = 1 - P(Z<0.3) = 1 - 0.6179 = 0.3821 Since P-value > 0.05, hence fail to reject null hypothesis. So, we conclude that union efforts to organize have increased union membership. \fPrice (cents per gallon) Price per Gallon (in dollars) 232.5 2.325 227.7 2.277 237.4 2.374 235.9 2.359 234.1 2.341 239.1 2.391 241.0 2.41 238.4 2.384 232.9 2.329 237.3 2.373 228.8 2.288 231.5 2.315 236.5 2.365 235.6 2.356 227.1 2.271 244.8 2.448 233.8 2.338 239.9 2.399 232.5 2.325 239.6 2.396 244.4 2.444 238.7 2.387 236.6 2.366 230.5 2.305 242.2 2.422 226.4 2.264 230.8 2.308 229.5 2.295 238.7 2.387 228.2 2.282 233.2 2.332 227.9 2.279 233.6 2.336 237.5 2.375 238.1 2.381 237.4 2.374 231.2 2.312 234.6 2.346 233.6 2.336 237.6 2.376 232.7 2.327 240.0 2.4 231.5 2.315 234.2 2.342 238.2 2.382 231.4 2.314 240.6 2.406 234.3 2.343 234.2 2.342 233.8 2.338 In this demo, we'll practice using Excel to compute p-values using the t distribution. The U.S. Energy Administration reported that the mean price for a gallon of regular gasoline in the United States was $2.36 (in 2006). We want to determine if gasoline prices in the Lower Atlantic states are lower than the national average. This file contains regular gasoline prices for a sample of 50 service stations from Lower Atlantic states. Use = 0.01 as your level of significance. (In this problem we do not know .) We'll be using the second column of numbers (the price in dollars). 1. Set up your hypotheses. H0: 2.36 Ha: < 2.36 This is a hypothesis test for an average, not a proportion. We also do not know , so we'll have to use the sample standard deviation s. 2. Use Excel to compute the sample mean (xbar). Use =AVERAGE(B2:B51) xbar = 2.3496 3. Use Excel to compute the standard deviation (s). Use =STDEV.S(B2:B51) s = 0.04437 4. Compute your test statistic using parts 2 and 3. = ( (/)=(2.34962.36)/(0.04437/50)=1.657 _0)/ 5. Compute the degrees of freedom (n - 1). n = 50, df = 50 - 1 = 49 6. To get the p-value for a lower tail test, type =1-T.DIST.RT( your test statistic, your degrees of freedom) in an empty cell. To get the p-value for an upper tail test, type =T.DIST.RT(your test statistic, your degrees of freedom) in an empty cell. To get the p-value for a two-tailed test: - If your test statistic is negative, type =2*(1-T.DIST.RT( your test statistic, degrees of freedom)) - If your test statistic is positive, type =2*T.DIST.RT( your test statistic, degrees of freedom) 7. What is the p-value for this test? We have a lower tail test, so in Excel we type: =1-T.DIST.RT( -1.657, 49) The p-value is 0.0520. 8. If your p-value is less than = 0.01 reject H0. If your p-value is greater than =0.01 do not reject H0. Our p-value is bigger than 0.01, so we do not reject H0. We do not have enough evidence to say that the average price of a gallon of gasoline is lower in the Lower Atlantic States. Sale Price 12400 10400 12100 10000 11000 8895 7675 9975 6350 10470 9895 11250 8795 12500 9340 10150 9200 9395 11000 10640 10000 7500 8000 10440 10200 10300 9740 9280 10930 8000 9000 7680 9400 10730 7350 12240 11970 8240 9910 10080 9440 8970 9500 10050 10130 11400 8500 7500 9090 10500 Excel Project 3 This Excel Project contains 7 questions. Work out the answers to the questions and then click on "Excel Project 3" to submit your answers. Directions for performing a hypothesis test for means in Excel are in the Excel Demo. **Give your answers to 2 decimal places!** According to the National Automobile Dealers Association, the mean price for used cars is $10,192. A manager of a Kansas City used car dealership claims that his prices are different than the national mean. To test his claim, he randomly selected 50 cars from his dealership. The sales prices for those 50 cars are in this data set. Use 0.05 as your level of significance. 1. What is the alternative hypothesis? 2. Compute the sample mean. 3. Compute the sample standard deviation. 4. Compute the test statistic. 5. What are the degrees of freedom? 6. Compute the p-value using Excel. 7. What is your decision? What does this decision mean in the context of this problem? Price (cents per gallon) Price per Gallon (in dollars) 232.5 2.325 227.7 2.277 237.4 2.374 235.9 2.359 234.1 2.341 239.1 2.391 241.0 2.41 238.4 2.384 232.9 2.329 237.3 2.373 228.8 2.288 231.5 2.315 236.5 2.365 235.6 2.356 227.1 2.271 244.8 2.448 233.8 2.338 239.9 2.399 232.5 2.325 239.6 2.396 244.4 2.444 238.7 2.387 236.6 2.366 230.5 2.305 242.2 2.422 226.4 2.264 230.8 2.308 229.5 2.295 238.7 2.387 228.2 2.282 233.2 2.332 227.9 2.279 233.6 2.336 237.5 2.375 238.1 2.381 237.4 2.374 231.2 2.312 234.6 2.346 233.6 2.336 237.6 2.376 232.7 2.327 240.0 2.4 231.5 2.315 234.2 2.342 238.2 2.382 231.4 2.314 240.6 2.406 234.3 2.343 234.2 2.342 233.8 2.338 In this demo, we'll practice using Excel to compute p-values using the t distribution. The U.S. Energy Administration reported that the mean price for a gallon of regular gasoline in the United States was $2.36 (in 2006). We want to determine if gasoline prices in the Lower Atlantic states are lower than the national average. This file contains regular gasoline prices for a sample of 50 service stations from Lower Atlantic states. Use = 0.01 as your level of significance. (In this problem we do not know .) We'll be using the second column of numbers (the price in dollars). 1. Set up your hypotheses. H0: 2.36 Ha: < 2.36 This is a hypothesis test for an average, not a proportion. We also do not know , so we'll have to use the sample standard deviation s. 2. Use Excel to compute the sample mean (xbar). Use =AVERAGE(B2:B51) xbar = 2.3496 3. Use Excel to compute the standard deviation (s). Use =STDEV.S(B2:B51) s = 0.04437 4. Compute your test statistic using parts 2 and 3. 2.34962.36 = / 0 = 0.04437/ 50 = 1.657 5. Compute the degrees of freedom (n - 1). n = 50, df = 50 - 1 = 49 6. To get the p-value for a lower tail test, type =1-T.DIST.RT( your test statistic, your degrees of freedom) in an empty cell. To get the p-value for an upper tail test, type =T.DIST.RT(your test statistic, your degrees of freedom) in an empty cell. To get the p-value for a two-tailed test: - If your test statistic is negative, type =2*(1-T.DIST.RT( your test statistic, degrees of freedom)) - If your test statistic is positive, type =2*T.DIST.RT( your test statistic, degrees of freedom) 7. What is the p-value for this test? We have a lower tail test, so in Excel we type: =1-T.DIST.RT( -1.657, 49) The p-value is 0.0520. 8. If your p-value is less than = 0.01 reject H0. If your p-value is greater than =0.01 do not reject H0. Our p-value is bigger than 0.01, so we do not reject H0. We do not have enough evidence to say that the average price of a gallon of gasoline is lower in the Lower Atlantic States. Sale Price 12400 10400 12100 10000 11000 8895 7675 9975 6350 10470 9895 11250 8795 12500 9340 10150 9200 9395 11000 10640 10000 7500 8000 10440 10200 10300 9740 9280 10930 8000 9000 7680 9400 10730 7350 12240 11970 8240 9910 10080 9440 8970 9500 10050 10130 11400 8500 7500 9090 10500 Excel Project 3 This Excel Project contains 7 questions. Work out the answers to the questions and then click on "Excel Project 3" to submit your answers. Directions for performing a hypothesis test for means in Excel are in the Excel Demo. **Give your answers to 2 decimal places!** According to the National Automobile Dealers Association, the mean price for used cars is $10,192. A manager of a Kansas City used car dealership claims that his prices are different than the national mean. To test his claim, he randomly selected 50 cars from his dealership. The sales prices for those 50 cars are in this data set. Use 0.05 as your level of significance. 1. What is the alternative hypothesis? 2. Compute the sample mean. 3. Compute the sample standard deviation. 4. Compute the test statistic. 5. What are the degrees of freedom? 6. Compute the p-value using Excel. 7. What is your decision? What does this decision mean in the context of this