FIND BOTH T STATISTIC AND P VALUE Calculate both R and R^2 1. One manufacturer has developed a quantitative index of the "sweetness" of orange juice. (The higher the index, the sweeter the juice). Is there a relationship between the sweetness index and a chemical measure such as the amount of watersoluble pectin (parts per million) in the orange juice? Data collected on these two variables for 24 production runs at a juice manufacturing plant are shown in the accompanying table. Suppose a manufacturer wants to use simple linear regression to predict the sweetness (y) from the amount of pectin (x). a. Find the least squares line for the data. Y= + ()x (Round to four decimal places as needed.) NOTE: The Image below is the same question (just a screenshot if that helps you better to understand the question). 2. Use the following pairs of observations to construct an 80% and a 98% confidence interval for beta 11. 3. If you pay more in tuition to go to a top business school, will it necessarily result in a higher probability of a job offer at graduation? Let yequals=percentage of graduates with job offers and xequals=tuition cost; then fit the simple linear model, Upper E left parenthesis y right parenthesis equals beta 0 plus beta 1 xE(y)= 0+1x, to the data below. Is there sufficient evidence (at alphaequals=0.050.05) of a positive linear relationship between y and x? A. Find the test statistic and pvalue of the test, part A. B. Make the appropriate conclusion at alphaequals=0.100.10. If the pvalue is less than alpha there is sufficient evidence to reject Upper H 0H0 and conclude that there exists a positive linear relationship between y and x. Otherwise, there is insufficient evidence to reject Upper H 0H 0. 2 4. Construct a scattergram for each data set. Then calculate r and r squaredr for each data set. Interpret their values. Complete parts a A. Calculate r B. Calculate r squaredr2. 5. Each week coaches in a certain football league face a decision during the game. On fourthdown, should the team punt the ball or go for a firstdown? To aid in the decisionmaking process, statisticians at a particular university developed a regression model for predicting the number of points scored (y) by a team that has a firstdown with a given number of yards (x) from the opposing goal line. One of the models fit to data collected on five league teams from a recent season was the simple linear regression model, Upper E left parenthesis y right parenthesis equals beta 0 plus beta 1xE(y)=0+1x. The regression yielded the following results: Modifying Above y with caret equals 4.27 minus 0.45 2 x<="">y=4.270.45x, r squared equals 0.19r =0.19. Complete parts a and b below. 2 B. Compute the value of the coefficient of correlation, r, from the value of r squaredr . Is the value of r positive or negative? Why? Select the correct choice below and fill in the answer box within your choice.(Round to three decimal places as needed.) 6. To study how social media may influence the products consumers buy, researchers collected the opening weekend box office revenue (in millions of dollars) for 23 recent movies and the social media message rate (average number of messages referring to the movie per hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue (y) and message rate (x). 1 One manufacturer has developed a quantitative index of the "sweetness" of orange juice. (The higher the index, the sweeter the juice). Is there a relationship between the sweetness index and a chemical measure such as the amount of water-soluble pectin (parts per million) in the orange juice? Data collected on these two variables for 24 production runs at a juice manufacturing plant are shown in the accompanying table. Suppose a manufacturer wants to use simple linear regression to predict the sweetness (y) from the amount of pectin (x). There are 3 more parts to this question after I answer question a. Data regarding orange juice sweetness Run Sweetness Index Pectin(ppm) 1 5.2 219 2 5.4 225 3 6.1 261 4 5.8 211 5 5.8 224 6 5.9 214 7 5.8 231 8 5.7 270 9 5.6 239 10 5.8 210 11 5.4 411 12 5.5 255 13 5.8 308 14 5.6 262 15 5.4 281 16 5.3 379 17 5.7 273 18 5.5 261 19 5.8 226 20 5.2 262 21 5.9 233 22 5.8 218 23 5.8 245 24 5.9 241 a. Find the least squares line for the data. Y^= 6.1491 + ( - 0.0019 )x (round 4 decimal places) 2 Use the following pairs of observations to construct an 80% and a 98% confidence interval for b1. X 0 5 3 1 4 2 6 Y 2 6 2 2 3 3 9 A 100(1-)% confidence interval for b1 is: b^ 1t ( 2 , n2) b^ 1=1.035714286 n=7 n2=72 n2=5 ^ 2= SS E S yyb^ 1 S xy = n2 n2 S yy =42.857142857 S xy =29 ^ 2 ^ 2 b 1 b^ 1 +t S xx ( 2 ,n2) S xx ^ 2= S yy b^ 1 S xy 42.85714286( 1.035714286 ) ( 29 ) 42.8571428630.03571428 = = n2 5 5 12.82142857 ^ 2 2 ^ = =2.564285715 5 S xx =28 a. The 80% confidence interval is left parenthesis is: 100 (1 ) =80 100 ( 1 )=80 1= 80 1 =0.8 10.8= 100 =0.2 0.2 = =0.1 2 2 2 t t t =t ( 0.1,5) tdistributionTable : t ( 0.1,5 )=1.476 t ( 2 ,n2) ) ( 2 ,n2) =1.476 ( 2 ,n2) ^ 2 2.564285715 =1.476 =1.476 0.091581633=1.476 ( 0.302624574 ) S xx 28 ( 2 ,n2 ^ 2 =0.446673871 S xx b^ 1t ( 2 , n2) ^ 2 ^ 2 ^ b b +t S xx 1 1 ( 2 ,n2) S xx 1.0357142860.446673871 b1 1.035714286+0.446673871 0.589040415 b 1 1.482388157 0.59 b1 1.48 (_0.59_, _1.48_) round 2 decimal places There is 1 more part to this question after I answer question a. 3 If you pay more in tuition to go to a top business school, will it necessarily result in a higher probability of a job offer at graduation? Let y = percentage of graduates with job offers and x = tuition cost; then fit the simple linear model, E(y) = bo + b1x, to the data below. Is there sufficient evidence (at a = 0.05 ) of a positive linear relationship between y and x? School Annual Tuition ($) % with job offer 1 39,793 99 2 39,391 88 3 39,239 98 4 38,495 86 5 38,064 87 6 37,899 83 7 37,838 87 8 37,486 95 9 37,258 95 10 37,138 89 Give the null and alternative hypotheses for testing whether there exists a positive linear relationship between y and x? There are 3 more parts after I complete this part. A. B. C. D. E. F. H0:b1=0 Ho: b1=0 Ho:bo=0 Ho:b0=0 Ho:b1=0 H0:bo=0 ha:b1 0 ha:b1<0 ha:b0>0 ha:b0<0 ha:b1>0 ha:bo 0 Answer: Option E. H0: b1=0 Ha: b1>0 5 Complete parts a and b below. 1 part after I complete part a. a. Give a practical interpretation of the coefficient of determination, r2. Choose the correct answer below. A. There is a positive linear relationship between numbers of yards to the opposing goal line and numbers of points scored because 0.29 is positive. B. There is little or no relationship between numbers of yards to the opposing goal line and numbers of points scored because 0.29 is near to zero. C. Sample variations in the numbers of yards to the opposing goal line explain 29% of the sample variation in the numbers of points scored using the least squares line. D. Sample variations in the numbers of yards to the opposing goal line explain 71% of the sample variation in the numbers of points scored using the least squares line. Answer: Option C. Sample variations in the numbers of yards to the opposing goal line explain 29% of the sample variation in the numbers of points scored using the least squares line. 6 To study how social media may influence the products consumers buy, researchers collected the opening weekend box office revenue (in millions of dollars) for 23 recent movies and the social media message rate (average number of messages referring to the movie per hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue (y) and message rate (x). The least squares regression equation is y^= _3.994_ + (_0.083 ) x. Round three decimal places Revenue and Message Rate for recent movies Message rate Revenue($millions) 1367.9 149 1215.6 80 678.8 67 581.4 33 459.8 31 410.4 29 310.8 24 292.5 23 254.8 23 160.1 22 150.7 22 146.2 21 144.8 21 125.9 21 119.4 20 108.6 20 104.5 19 98.3 17 85.6 14 73.8 7 55.7 5 38.1 2 2.8 1 I have a few parts to this question. 7 Run Pectin (ppm) Sweetness Index xi yi xi2 yi2 i 1 219 5.2 47961 27.04 2 225 5.4 50625 29.16 3 261 6.1 68121 37.21 4 211 5.8 44521 33.64 5 224 5.8 50176 33.64 6 214 5.9 45796 34.81 7 231 5.8 53361 33.64 8 270 5.7 72900 32.49 9 239 5.6 57121 31.36 10 210 5.8 44100 33.64 11 411 5.4 168921 29.16 12 255 5.5 65025 30.25 13 308 5.8 94864 33.64 14 262 5.6 68644 31.36 15 281 5.4 78961 29.16 16 379 5.3 143641 28.09 17 273 5.7 74529 32.49 18 261 5.5 68121 30.25 19 226 5.8 51076 33.64 20 262 5.2 68644 27.04 21 233 5.9 54289 34.81 22 218 5.8 47524 33.64 23 245 5.8 60025 33.64 24 241 5.9 58081 34.81 _^2 = _ = _ = _^2 = n=24 6159 135.7 1637027 768.61 256.62500 5.6541667 37,933,281 1,580,553.375 x iyi 1,138.8 1,215.0 1,592.1 1,223.8 1,299.2 1,262.6 1,339.8 1,539.0 1,338.4 1,218.0 2,219.4 1,402.5 1,786.4 1,467.2 1,517.4 2,008.7 1,556.1 1,435.5 1,310.8 1,362.4 1,374.7 1,264.4 1,421.0 1,421.9 _ _ = 34,715.1 56,473.625 =(_ )/=6,159/24 =256.625 =(_ )/=135,7/24 =5.654166667 _=_^2 (_ ) ^2/=1,637,027(6,159)^2/24=1,637,02737,933,281/24=1,637,0271,580,553.375 _=5 _= _ _ (_ )(_ )/=34,715,1(6,159) (135.7)/24=34,715,1835,776.3/24=34,715.134,824.0125_=108.9125 _1=_/_ =(108.9125)/56,473.625_1=0.001928555_1= 0.0019 _1=_/_ =(108.9125)/56,473.625_1=0.001928555_1= 0.0019 _0= _1 =5.654166667(0.001928555)(256.625)=5.654166667+0.494915464 _0=6.14908 = _0+_1 =6.14910.0019 _ _ = (_ )^2 1,416 1,000 19 2,082 1,064 1,817 657 179 311 2,174 23,832 3 2,639 29 594 14,976 268 19 938 29 558 1,492 135 244 _=(_56,474 )^2 = 835,776.3 ,637,0271,580,553.375 _=56,473.625 _=108.9125 (_ )^2 0.2062673611 0.0646006944 0.1987673611 0.0212673611 0.0212673611 0.0604340278 0.0212673611 0.0021006944 0.0029340278 0.0212673611 0.0646006944 0.0237673611 0.0212673611 0.0029340278 0.0646006944 0.1254340278 0.0021006944 0.0237673611 0.0212673611 0.2062673611 0.0604340278 0.0212673611 0.0212673611 0.0604340278 _=(_1.3395833333 )^2 = 34,824.0125 (_ )^2 (_ -195.6500 -170.7750 26.6875 -264.6250 -189.2250 -251.4875 -148.6250 76.2375 -98.7000 -270.4250 833.6250 -8.9375 297.9750 30.1000 131.6250 648.5875 93.3375 24.0625 -177.6250 27.9500 -139.3875 -224.0250 -67.4250 -92.1875 _= (_ ) _= -108.9125 -0.3959770596 =_/(_ _ )= -108.9125 6667+0.494915464 _0=6.149082131_06.1491 (_ )^2 (_ )_ _ ) _= /(_ _ )= -0.001928555 -0.494915464 6.149082131 xi i 1 2 3 4 5 6 7 yi 0 5 3 1 4 2 6 n=7 _ = 21 xi2 2 6 2 2 3 3 9 _^2 _ = 27 3.00000 3.8571429 =(_ )/=21/7 =3 xiyi yi2 0 25 9 1 16 4 36 = _^2 91 441 (_ )^2 4 0 9 36 30 4 4 6 0 4 2 4 9 12 1 9 6 1 81 54 9 = _ _ = _=(_ )^2 = _=(_ )^ 147 110 28 63.000 28.000 567.0 =(_ )/=27/7 =3.857142857 _=_^2 (_ ) ^2/=91(21)^2/3=91441/3=9163_=28 _= _ _ (_ )(_ )/=110(21) (27)/7=110567/7=11081_=29 _1=_/_ =29/28_1=1.035714286 _0= _1 =3.857142857(1.035714286)(3)=3.8571433.107143_0=0.75 = _0+_1 =0.751.0357 (_ )^2 (_ )^2 (_ )_ 3.4489795918 -6 4.5918367347 12 3.4489795918 0 3.4489795918 -4 0.7346938776 3 0.7346938776 -3 26.4489795918 27 _=(_ )^2 =42.8571428571 _= (_ ) _= 29 =_/(_ _ )= 0.8371578903 81.0000 29.0000 1.035714286 3.107142857 0.75 i 1 2 3 4 5 6 7 8 9 10 xi yi 39793 99 39391 88 39239 98 38495 86 38064 87 37899 83 37838 87 37486 95 37258 95 37138 89 _^2 _ = _ = n=10 382601 907 xi2 yi2 xiyi 1583482849 9801 3939507 1551650881 7744 3466408 1539699121 9604 3845422 1481865025 7396 3310570 1448868096 7569 3311568 1436334201 6889 3145617 1431714244 7569 3291906 1405200196 9025 3561170 1388158564 9025 3539510 1379231044 7921 3305282 = _^2 = _ _ = 14646204221 82543 34716960 38,260.1 90.7 146,383,525,201 14,638,352,520.1 =(_ )/=382,601/10 =38,260.1 7,851,700.9 =(_ )/=907/10 =90.7 _=_^2 (_ ) ^2/=14,646,204,221(382,601)^2/10=14,646,204,221146,383,525,201/10=14,646,204,2 =7,851,700.9 _= _ _ (_ )(_ )/=34,716,960(382,601) (907)/10=34,716,960347,019,107/10=34,716,96034,701,910.7_=15,049.3 _1=_/_ =15,049.3/7,851,700.9_1=0.001916693_1=0.001 9167 _0= _1 =90.7(0.0019167)(38,260.1)=90.773.3328651_0=17.367135 = _0+_1 =0.751.0357 "Escriba aqu la ecuacin." (_ )^2 (_ )_ 2349782.41 68.89 151757.1 1278934.81 7.29 99519.2 958245.21 53.29 95932.2 55178.01 22.09 20201.4 38455.21 13.69 -17060.7 130393.21 59.29 -29971.3 178168.41 13.69 -36722.7 599230.81 18.49 -73539.5 1004204.41 18.49 -95199.5 1259108.41 2.89 -99866.9 _=(_ )^2 =7851700.90 _=(_ )^2 = 278.10 _= (_ ) _= 15049.3 (_ )^2 =_/(_ _ )= 0.3220578813 347,019,107.0 25,201/10=14,646,204,22114,638,352,520.1_ _=15,049.3 17.367135 34,701,910.7000 15,049.3000 0.001916693 73.332865103 17.36713490 Run i 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 Message rate xi Revenue yi 1367.9 149 1215.6 80 678.8 67 581.4 33 459.8 31 410.4 29 310.8 24 292.5 23 254.8 23 160.1 22 150.7 22 146.2 21 144.8 21 125.9 21 119.4 20 108.6 20 104.5 19 98.3 17 85.6 14 73.8 7 55.7 5 38.1 2 2.8 1 _^2 _ = _ = n=23 6986.5 671 xi2 yi2 x iyi 1871150.41 22201 203,817.1 1477683.36 6400 97,248.0 460769.44 4489 45,479.6 338025.96 1089 19,186.2 211416.04 961 14,253.8 168428.16 841 11,901.6 96596.64 576 7,459.2 85556.25 529 6,727.5 64923.04 529 5,860.4 25632.01 484 3,522.2 22710.49 484 3,315.4 21374.44 441 3,070.2 20967.04 441 3,040.8 15850.81 441 2,643.9 14256.36 400 2,388.0 11793.96 400 2,172.0 10920.25 361 1,985.5 9662.89 289 1,671.1 7327.36 196 1,198.4 5446.44 49 516.6 3102.49 25 278.5 1451.61 4 76.2 7.84 1 2.8 = _^2 = _ _ = 4945053.29 41631 437,815.0 303.760870 29.1739130 48811182.25 2122225.3152174 =(_ )/=6,986.5/23 =303.7608695 2822827.9748 =(_ )/=671/23 =29.17391304 _=_^2 (_ ) ^2/=4,945,053.29(6,986,5)^2/23=4,945,053.2948,811,182,25/23=4,945,053.292,122 74 _= _ _ (_ )(_ )/=437,815.0(6,986.5) (671)/23=437,815.04,687,941.5/23=437,815.0203,823.5434_=233,991.4565 _1=_/_ =233,991,4565/2,822,827.974_1=0.082892567_10. 083 _0= _1 =29.17391304(0.082892567)(303.7608695)=29.1739130425.17951817 _0=3 = _0+_1 =3.994+0.083 (_ )^2 1,132,392 14,358.2911153119 831,451 2,583.2911153119 140,654 1,430.8128544424 77,083 14.6389413989 24,348 3.3345935728 11,372 0.0302457467 50 26.7693761815 127 38.1172022684 2,397 38.1172022684 20,638 51.4650283554 23,428 51.4650283554 24,825 66.8128544423 25,269 66.8128544423 31,634 66.8128544423 33,989 84.1606805293 38,088 84.1606805293 39,705 103.5085066163 42,214 148.2041587902 47,594 230.2476370510 52,882 491.6824196597 61,534 584.3780718336 70,576 738.4215500945 90,577 793.7693761815 _=(_ )^2 = 2,822,828 _=(_ )^2 = 22,055.3043478261 _= (_ ) _= (_ )^2 4,687,941.5 3=4,945,053.292,122,225.315 _=2,822,827.9 =233,991.4565 425.17951817 _0=3.994394861_03.994 203,823.5435 (_ )^2 (_ )_ 158556.7304 72947.1304 25127.6217 9162.0913 4837.2130 3092.5348 168.9391 -259.0000 -1126.1000 -3160.5391 -3367.3391 -3308.7783 -3338.1783 -3735.0783 -3687.2174 -3903.2174 -3785.9565 -3492.8348 -3054.2522 -1609.7261 -1240.3043 -531.3217 -300.9609 _= (_ ) _= 233991.4565 =_/(_ _ )= 0.9377806489 233,991.4565 0.082892567 25.179518178 3.994394866 Run 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 Sweetness_Index Pectin_(ppm) 5.1 218 5.4 226 5.9 262 5.8 211 5.8 224 6.1 213 5.8 229 5.5 270 5.7 240 5.9 212 5.3 412 5.5 258 5.8 304 5.5 258 5.3 286 5.3 386 5.8 269 5.6 264 5.8 227 5.3 265 5.9 231 5.8 218 5.8 245 5.9 243 x y 5 6 3 4 2 0 1 3 5 3 6 1 1 2 School 1 2 3 4 5 6 7 8 9 10 Annual_tuition ($) %_with_Job_Offer 39789 94 39417 91 39146 99 38871 94 38019 96 37892 84 37544 99 37231 89 37043 96 36375 91 1. Find the Test Statistic 2. Find the P value Construct a scattergram for each data set. Then calculate r and r squared for each data set. Interpret their values. Complete part A x y -5 -4 -3 -2 -1 1. Calculate r 2. Calculate r^2 -3 0 1 4 6 To study how social media may influence the products consumers buy, researchers collected the opening weekend box office revenue (in millions of dollars) for 23 recent movies and the social media message rate (average number of messages referring to the movie per hour). The data are available below. Conduct a complete simple linear regression analysis of the relationship between revenue (y) and message rate (x). Message_Rate Revenue_($millions) 1368.1 143 1213.4 78 683.5 66 577.8 35 457.3 34 411.4 33 308.3 24 286.7 22 252.1 21 164.2 21 148.5 20 148.1 20 144.5 20 121.9 20 116.3 20 113.4 20 100.2 19 98.1 17 86.9 13 79.1 11 57.9 6 35.8 3 8.4 2 The least squares regression equation is y^= Just need part B