Stat 200 Week 6 Homework Problems

Many high school students take the AP tests in different subject areas. In 2007, of the 144,796 students who took the biology exam 84,199 of them were female. In that same year, of the 211,693 students who took the calculus AB exam 102,598 of them were female ("AP exam scores," 2013). Estimate the difference in the proportion of female students taking the biology exam and female students taking the calculus AB exam using a 90% confidence level. n1=144,796 p1=84199/144796 = 0.582 q1=1-0.582 = 0.418 n2=211,693 p2=102598/211693 = 0.485 q1=1-0.485 = 0.515 zc = 1.645 E=1.645*sqrt[{(0.582*0.418)/144796}+{90.485*0.515)/211693}] E= 0.0028 (0.582-0.485)-0.0028p1-p2(0.582-0.485)+0.0028 0.0942p1-p20.998 There is a 90% chance that 0.0942p1-p20.998 contains true difference in proportions. 9.1.5 Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated ("Autism and developmental," 2008). Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah? Test at the 1% level. n1=18,440 x1=245 p1=245/18440 = 0.013 q1=1-0.013 = 0.987 n2=2,123 x2=45 p2=45/2123 = 0.021 q1=1-0.021 = 0.979 p-= (245+45)/(18440+2123) = 0.014 q-=1-0.014 = 0.986 z={(0.013+0.021)-0}/sqrt[{(0.014*0.986)/18440}+{(0.014*0.986)/2123}] z=12.760 9.2.3 All Fresh Seafood is a wholesale fish company based on the east coast of the U.S. Catalina Offshore Products is a wholesale fish company based on the west coast of the U.S. Table #9.2.5 contains prices from both companies for specific fish types ("Seafood online," 2013) ("Buy sushi grade," 2013). Do the data provide enough evidence to show that a west coast fish wholesaler is more expensive than an east coast wholesaler? Test at the 5% level. Table #9.2.5: Wholesale Prices of Fish in Dollars Fish All Fresh Seafood Prices Catalina Offshore Products Prices Cod 19.99 17.99 Tilapi 6.00 13.99 Farmed Salmon 19.99 22.99 Organic Salmon 24.99 24.99 Grouper Fillet 29.99 19.99 Tuna 28.99 31.99 Swordfish 23.99 23.99 Sea Bass 32.99 23.99 Striped Bass 29.99 14.99 All Fresh Seafood Prices Catalina Offshore Products Prices Fish d=x1-x2 Cod 19.99 17.99 2 Tilapi 6 13.99 -7.99 Farmed Salmon 19.99 22.99 -3 Organic Salmon 24.99 24.99 0 Grouper Fillet 29.99 19.99 10 Tuna 28.99 31.99 -3 Swordfish 23.99 23.99 0 Sea Bass 32.99 23.99 9 Striped Bass 29.99 14.99 15 Mean= 2.446 Standard Dev= 7.399 Test Stat= 0.110 df= 8 p-value= 0.542 Since p-value0.05, accept Ho. 9.2.6 The British Department of Transportation studied to see if people avoid driving on Friday the 13th. They did a traffic count on a Friday and then again on a Friday the 13th at the same two locations ("Friday the 13th," 2013). The data for each location on the two different dates is in table #9.2.6. Estimate the mean difference in traffic count between the 6th and the 13th using a 90% level. Table #9.2.6: Traffic Count Dates 6th 13th 1990, July 139246 138548 1990, July 134012 132908 1991, September 137055 136018 1991, September 133732 131843 1991, December 123552 121641 1991, December 121139 118723 1992, March 128293 125532 1992, March 124631 120249 1992, November 124609 122770 1992, November 117584 117263 Dates 6th 13th d=x1-x2 1990, July 139246 138548 698 1990, July 134012 132908 1104 1991, September 137055 136018 1037 1991, September 133732 131843 1889 1991, December 123552 121641 1911 1991, December 121139 118723 2416 1992, March 128293 125532 2761 1992, March 124631 120249 4382 1992, November 124609 122770 1839 1992, November 117584 117263 321 Mean= 1835.8 Standard Dev= 1173.065 df= 9 tc= 1.833 E= 679.962 1835.8-679.962ud1835.8+679.962 There is a 90% chance that 2515.762ud2515.762 contains true mean difference betwen the 6th and the 13th. 9.3.1 The income of males in each state of the United States, including the District of Columbia and Puerto Rico, are given in table #9.3.3, and the income of females is given in table #9.3.4 ("Median income of," 2013). Is there enough evidence to show that the mean income of males is more than of females? Test at the 1% level. Table #9.3.3: Data of Income for Males $42,951 $52,379 $42,544 $37,488 $49,281 $50,987 $60,705 $50,411 $66,760 $40,951 $43,902 $45,494 $41,528 $50,746 $45,183 $43,624 $43,993 $41,612 $46,313 $43,944 $56,708 $60,264 $50,053 $50,580 $40,202 $43,146 $41,635 $42,182 $41,803 $53,033 $60,568 $41,037 $50,388 $41,950 $44,660 $46,176 $41,420 $45,976 $47,956 $22,529 $48,842 $41,464 $40,285 $41,309 $43,160 $47,573 $44,057 $52,805 $53,046 $42,125 $46,214 $51,630 Table #9.3.4: Data of Income for Females $31,862 $40,550 $36,048 $30,752 $41,817 $40,236 $47,476 $40,500 $60,332 $33,823 $35,438 $37,242 $31,238 $39,150 $34,023 $33,745 $33,269 $32,684 $31,844 $34,599 $48,748 $46,185 $36,931 $40,416 $29,548 $33,865 $31,067 $33,424 $35,484 $41,021 $47,155 $32,316 $42,113 $33,459 $32,462 $35,746 $31,274 $36,027 $37,089 $22,117 $41,412 $31,330 $31,329 $33,184 $35,301 $32,843 $38,177 $40,969 $40,993 $29,688 $35,890 $34,381 Male Earnings N1: 52 df1 = N - 1 = 52 - 1 = 51 M1: $46,453.31 SS1: .08 s21 = .08/(52-1) = .45 Female Earnings N2: 52 df2 = N - 1 = 52 - 1 = 51 M2: $35,376.38 SS2: .31 s22 = .31/(52-1) = .28 T-value Calculation s2p = ((51/102) * .45) + ((51/102) * .28) = .37 s2M1 = .37/52 = 1238323.16 s2M2 = .37/52 = 1238323.16 t = 11076.92/sqrt2476646.32 = 7.04 The t-value is 7.04. The p-value is .00001. The result is therefore significant at p 0.01. Therefore there is enough evidence to show that the mean income of males is more than females at a 1% level 9.3.3 A study was conducted that measured the total brain volume (TBV) (in ) of patients that had schizophrenia and patients that are considered normal. Table #9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Is there enough evidence to show that the patients with schizophrenia have less TBV on average than a patient that is considered normal? Test at the 10% level. Table #9.3.5: Total Brain Volume (in ) of Normal Patients 1663407 1583940 1299470 1535137 1431890 1578698 1453510 1650348 1288971 1366346 1326402 1503005 1474790 1317156 1441045 1463498 1650207 1523045 1441636 1432033 1420416 1480171 1360810 1410213 1574808 1502702 1203344 1319737 1688990 1292641 1512571 1635918 Table #9.3.6: Total Brain Volume (in ) of Schizophrenia Patients 1331777 1487886 1066075 1297327 1499983 1861991 1368378 1476891 1443775 1337827 1658258 1588132 1690182 1569413 1177002 1387893 1483763 1688950 1563593 1317885 1420249 1363859 1238979 1286638 1325525 1588573 1476254 1648209 1354054 1354649 1636119 Mean1 1463339.219 Mean 2 1451293.194 Standard Dev1 125458.276 Standard Dev2 171932.2264 DF= 61 t=(1451293..219)/sqrt({171932.226^2/31}+{125458.276^2/32}) = -0.32 t critical value = 1.296 Since the absolute value of the t statistic (0.32) is less than the t critical value, plus the p value associated with t = -0.32 at 61 degrees freedom is 0.7501. Since the p-value is greater than the level of significance (0.1), we fail to reject the null hypothesis. This means there isn't enough evidence to suggest that patients with schizophrenia have less TBV on average than a patient that is considered normal. P-value computed from online calculator. 9.3.4 A study was conducted that measured the total brain volume (TBV) (in ) of patients that had schizophrenia and patients that are considered normal. Table #9.3.5 contains the TBV of the normal patients and table #9.3.6 contains the TBV of schizophrenia patients ("SOCR data oct2009," 2013). Compute a 90% confidence interval for the difference in TBV of normal patients and patients with Schizophrenia. 9.3.8 The number of cell phones per 100 residents in countries in Europe is given in table #9.3.9 for the year 2010. The number of cell phones per 100 residents in countries of the Americas is given in table #9.3.10 also for the year 2010 ("Population reference bureau," 2013). Find the 98% confidence interval for the different in mean number of cell phones per 100 residents in Europe and the Americas. Table #9.3.9: Number of Cell Phones per 100 Residents in Europe 100 76 100 130 75 84 112 84 138 133 118 134 126 188 129 93 64 128 124 122 109 121 127 152 96 63 99 95 151 147 123 95 67 67 118 125 110 115 140 115 141 77 98 102 102 112 118 118 54 23 121 126 47 Table #9.3.10: Number of Cell Phones per 100 Residents in the Americas 158 117 106 159 53 50 78 66 88 92 42 3 150 72 86 113 50 58 70 109 37 32 85 101 75 69 55 115 95 73 86 157 100 119 81 113 87 105 96 Mean1 = 108.15 Standard Dev1 = 897.90 n1 = 53 Mean2 = 87.205 Standard Dev2 = 1235.904 n2 = 39 df= 39+53-2 = 90 t=2.37 CI= (108.15-87.205)+-2.37*sqrt(897.90/53+1235.904/39) = 20.9450+-16.5275 = (4.4175;37.4725) Since 0 is not within the interval, there is a significant difference in mean number of cell phones per 100 residents in Europe and the Americas 11.3.2 Levi-Strauss Co manufactures clothing. The quality control department measures weekly values of different suppliers for the percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up). The data is in table #11.3.3, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer ("Waste run up," 2013). Do the data show that there is a difference between some of the suppliers? Test at the 1% level. Table #11.3.3: Run-ups for Different Plants Making Levi Strauss Clothing Plant 1 Plant 2 Plant 3 Plant 4 Plant 5 1.2 16.4 12.1 11.5 24 10.1 -6 9.7 10.2 -3.7 -2 -11.6 7.4 3.8 8.2 1.5 -1.3 -2.1 8.3 9.2 -3 4 10.1 6.6 -9.3 -0.7 17 4.7 10.2 8 3.2 3.8 4.6 8.8 15.8 2.7 4.3 3.9 2.7 22.3 -3.2 10.4 3.6 5.1 3.1 -1.7 4.2 9.6 11.2 16.8 2.4 8.5 9.8 5.9 11.3 0.3 6.3 6.5 13 12.3 3.5 9 5.7 6.8 16.9 -0.8 7.1 5.1 14.5 19.4 4.3 3.4 5.2 2.8 19.7 -0.8 7.3 13 3 -3.9 7.1 42.7 7.6 0.9 3.4 1.4 70.2 1.5 0.7 3 8.5 2.4 6 1.3 2.9 Mean 4.5227 8.8318 4.8316 7.4895 10.3769 Standard Dev 10.0320 15.3535 4.4032 3.6571 9.5550 N 22 22 19 19 13 Anova: Single Factor SUMMARY Groups Count Sum Average Variance 1.2 21 98.3 4. 1 105.095619 16.4 21 177.9 8. 1 244.515142 9 12.1 18 79.7 4. 8 17.2480065 4 11.5 18 130.8 7. 7 13.1623529 4 24 12 110.9 9. 7 81.3208333 3 ANOVA urce of Variat SS df MS F P-value F crit Between Groups 327.415484 1 4 81.8538710 3 0. 92 0. 15 3. 6 Within Groups 8403.72051 6 85 98.8673001 9 Total 8731.136 89 Since p-value is more than 0.01, we accept Ho, therefore there is no evidence between the suppliers.