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

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. X1 = number of female students who took biology exam X2 = number of female students who took AP exam P1 = proportion of female students who took biology exam P2 = proportion of female students who took AP exam Ho: P1 = P2 Ha: P1 P2 or Ho: P1 – P2 = 0 Ha: P1 – P2 0 a = 0.90 z = 56.86 p = 0 Therefore we can reject null hypothesis There is a 90% chance that 0.0941 P1 – P2 0.0996 contains the true difference in proportions The proportion of female students who took biology test is 9.41% to 9.96% higher than the proportion of female students who took AP 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. X1 = number of children diagnosed with ASD in Pennsylvania X2 = number of children diagnosed with ASD Utah P1 = proportion of children diagnosed with ASD in Pennsylvania P2 = proportion of children diagnosed with ASD in Utah Ho: P1 = P2 or Ha: P1 – P2 = 0 Ha: P1 P2 Ha: P1 – P2 0 a = .01 z = -2.93 p = 0.998 Since p-value is greater than 0.01 we fail to reject the null hypothesis There is not enough evidence to show that proportion of eight years old diagnosed with ASD in Pennsylvania is more than proportion 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 X1 = price of fish in All Fresh Seafood X2 = price of fish in Catalina Offshore Products U1 = mean price of fish in All Fresh Seafood U2 = mean price of fish in Catalina Offshore Products H0: Ud = 0 Ha: Ud 0 a = 0.05 t = 0.9915 p-value = 0.825 Since p-value is more than level of significance we reject null hypothesis and conclude that there is not sufficient evidence to support the claim that west coast fish wholesalers are more expensive than east cost wholesaler. 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 1990, July 1991, September 1991, September 1991, December 1991, December 1992, March 1992, March 1992, November 1992, November X1 = traffic count on 6th X2 = traffic count on 13th U1 = mean traffic count on 6th U2 = mean traffic count on 13th H0: Ud = 0 Ha: Ud ≠ 0 a = 0.90 The confidence interval is (1154.1, 2517.5). Therefore, the mean differences in traffic counts between 1154.1 and 2517.5. Therefore, there is 90% confidence that the true mean difference in traffic counts between Friday the 6th and Friday the 13th is between 1154.1 and 2517.5. 9.3.1