Statistics > QUESTIONS & ANSWERS > STAT 200 Week 6 Homework Problems, full solution guide, 100% correct (All)

STAT 200 Week 6 Homework Problems 9.1.2 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)-0.0028<p1-p2<(0.582-0.485)+0.0028 0.0942<p1-p2<0.998 There is a 90% chance that 0.0942<p1-p2<0.998 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-value>0.05, accept Ho. 9.2.6 The British Department of Transportation studied to see if people avoid driving on Friday the 13 th . They did a traffic count on a Friday and then again on a Friday the 13 th 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 6 th and the 13 th 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 1835.8-679.962<ud<1835.8+679.962 There is a 90% chance that 2515.762<ud<2515.762 contains true mean difference betwen the 6 th and the 13 th . 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: 2520976841.08 s21 = 2520976841.08/(52-1) = 49430918.45 Female Earnings N2: 52 df2 = N - 1 = 52 - 1 = 51 M2: $35,376.38 SS2: 4047089204.31 s22 = 4047089204.31/(52-1) = 79354690.28 T-value Calculation s2p = ((51/102) * 49430918.45) + ((51/102) * 79354690.28) = 64392804.37 s2M1 = 64392804.37/52 = 1238323.16 s2M2 = 64392804.37/52 = 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. [Show More]

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