Statistics > QUESTIONS & ANSWERS > STAT 200 Week 5 HW Sean Aldridge_With Complete Solutions 2021 (All)

STAT 200 Week 5 HW Sean Aldridge_With Complete Solutions 2021

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STAT 200 Week 5 Homework Problems 7.1.2 According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. I n ... that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? State the random variable, population parameter, and hypotheses. x = number of consumer complaints from identity theft in Alaska Population parameter = proportion of consumer complaints from identity theft in Alaska Hypothesis = HO : P=0.23 HA : P<0.23 7.1.6 According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? State the type I and type II errors in this case, consequences of each error type for this situation, and the appropriate alpha level to use. Skipped problem per direction by professor. 7.2.4 According to the February 2008 Federal Trade Commission report on consumer fraud and identity theft, 23% of all complaints in 2007 were for identity theft. In that year, Alaska had 321 complaints of identity theft out of 1,432 consumer complaints ("Consumer fraud and," 2008). Does this data provide enough evidence to show that Alaska had a lower proportion of identity theft than 23%? Test at the 5% level. x = 321 n = 1432 HO : p=0.23 HA : p<0.23 α = 0.05 q = 1 – 0.23 = 0.77 p-hat = 321 / 1432 = 0.224162011 z = (0.224162011 – 0.591) / √((0.591 * 0.77)/321) = -0.36683798 / 0.03765187314599194 = -9.74 p-value = normalcdf(-1e99,-9.74,0,1) [Show More]

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