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STAT 200 Week 5 Homework SOLUTIONGRADED A+ LATEST 2022-2023

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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 i ... dentity 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.7.1.6 The population proportion=0.23 n=number of complaints=1432 Alaska had complaints=321=x So, sample proportion=nx=1432321 =0.2242 H0: p=0.23 against H1: p<0.23 Z statistic, Z=np^(1−p^)p^−0.23 ~N(0,1) Obs(Z)=(-)0.52 H0: p=0.23 against H1: p<0.23 p value=0.30 (Excel Formula: NORM.S.DIST(0.52,1)) As,p value> α (level of significance). Hence there is not enough evidence to support the claim that, Alaska had a lower proportion of identity theft than 23% . Appropriate level of significance = 0.05 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. H0 : p = 0.23 μ=299,710.5 km/sec μ<299,710.5 lm/sec H1 : p < 0.23 Z critical value = -1.645 at 5% level Sample proportion 'p' = 321/1432 = 0.2242 Z stat = (p - P) / sqrt(PQ/n) [Show More]

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