STAT 200 Week 6 Homework Problems, full solution guide, 100% correct.

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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. Difference in the proportion of female students taking the Biology exam and female students taking the calculus AB exam using a 90% CI: Here, we have it that: n1 = 144,796 n2 = 211,693 x1 = 84,199 x2 = 102,598 C = 1 - α = 1 - 0.9 = 0.1 For this problem, we are to estimate a 90% confidence interval on the difference in proportion of female students taking the biology and calculus AB exam. This difference is given by: p1 – p2 Where; x1 = number of female students taking the biology exam x2 = number of female students taking the calculus AB exam p1 = proportion of female students taking the biology exam p2 = proportion of female students taking the calculus AB exam x1 = 84,199 n1 - x1 = 144,796 - 84,199 = 60597 x2 = 102,598 n2 - x2 = 211,693 - 102,598 = 109095 are all greater than or equal to 5, we assume both sampling distributions of �̂1 and �̂2 can be approximated with a normal distribution. �̂1 = x1/n1 = 84199/144796 = 0.5815, �̂2 = x2/n2 = 102598/211693 = 0.4847 �̂1 = 1 − �̂1 = 1 − 84199/144796 = 0.4185, �̂2 = 1 − �̂2 = 1 − 102598/211693 = 0.5153 To calculate the z-value in excel, =NORM.S.INV (0.95) = 1.645, and use this to find the margin error: E = 1.645*(sqrt (((0.5815*0.4185)/144796) + ((0.4847*0.5153)/211693))) = 0.0028 Then use it to compute the confidence interval on p1 - p2: (�1 − �̂2) − � < (�1 − �2) < (�̂1 − �̂2) + � (0.5815 − 0.4847) − 0.0028 < (�1 − �2) < (0.5815 − 0.4847) + 0.0028 �. �??� < (?� − ?�) < �. �??� Statistical Interpretation: There is a 90% chance that �. �??� < (?� − ?�) < �. �??� contains the true difference in proportions. Conclusion: The proportion of female students taking the biology exam is anywhere from 9.41% to 9.96% higher than the proportion of female students taking the calculus AB exam. 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. Testing at 1% level of significance: Here, we have it that; n1 = 18,440 n2 = 2,123 x1 = 245 x2 = 45 α = 0.01 For this problem, we are testing whether the proportion of children diagnosed with ADS in Pennsylvania is more than that of Utah. Where; x1 = number of children diagnosed with Autism Spectrum Disorder (ASD) in Pennsylvania x2 = number of children diagnosed with ASD in Utah p1 = proportion of children diagnosed with ASD in Pennsylvania p2 = proportion of children diagnosed with ASD in Utah The hypothesis will be: H0 : p1 = p2 Vs. H1 : p1 > p2 x1 = 245 n1 - x1 = 18440 -245 = 18195 x2 = 45 n2 - x2 = 2123 - 45 = 2078 are all greater than or equal to 5, we assume both sampling distributions of �̂1 and �̂2 can be approximated with a normal distribution. �̂1 = �1/�1 = 245/18144 = 0.0133, �̂2 = �2/�2 = 45/2123 = 0.0212 �̂1 = 1 − �̂1 = 1 − 245/18144 = 0.9867, �̂2 = 1 − �̂2 = 1 − 45/2123 = 0.9788 Then compute the pooled proportions, �̅ = (245 + 45) / (18144 + 2123) = 0.0141 and �̅ = 1 – 0.0141 = 0.9859 the test statistic z = -2.927 Conclusion Since the p-value is greater than the level of significance (i.e. [p-value = 0.9983] > [α = 0.01]), we fail to reject ?�. This is not enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion of children diagnosed with ASD in Utah. 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 Testing at 5% level of significance: We have that; n = 9 α = 0.05 For this problem, we are testing whether a west coast fish wholesaler is more expensive than an east coast wholesaler. Where; x1 = wholesale prices from west coast fishery x2 = wholesale prices from east coast fishery μ1 = mean wholesale prices from west coast fishery μ2 = mean wholesale prices from east coast fishery the hypothesis will be; H0: �1 = �2 Vs. H1: �1 > �2 at � = 0.05 Thus �̅= $2.446 and s� = $7.399, where �̅ is the sum of the difference between x1 and x2 divided by n and sd is the standard deviation of the difference. Thus t = (2.446 – 0) / (7.399/ sqrt(9)) = 0.9915 We can get the p- value in excel by: =1 - T.DIST( [Show More]

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