STAT 200 Week 5 Homework SOLUTION|GRADED A+

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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. ... 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 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) This study source was downloaded by 100000871878191230141379 from CourseHero.com on 096-1232-2021 14:1314:2520 GMT -05:00 = (0.2242 - 0.23) / sqrt(0.23*1-0.23/1432)= -0.52 Here - Z stat = -0.52 > -1.645 Fail to reject Ho, therefore, there is no sufficient evidence of identity theft lower than 23%. 7.2.6 In 2008, there were 507 children in Arizona out of 32,601 who were diagnosed with Autism Spectrum Disorder (ASD) ("Autism and developmental," 2008). Nationally 1 in 88 children are diagnosed with ASD ("CDC features -," 2013). Is there sufficient data to show that the incident of ASD is more in Arizona than nationally? Test at the 1% level. n=3260 1 x=507 Sample Proportion = 507/32601=0.0156 α=0.01 Ho:p=0.0114 Ha:p>0.0114 This corresponds to a right-tailed test, for which a z-test for one population proportion needs to be used. z=(0.0156-0.0114)/sqrt{(0.0114*[1-0.0114])/32601}= 7.061 Since the P-value is p=0, and p=0<0.01, it is concluded that the null hypothesis is rejected. therefore, there is enough evidence to claim that the population proportion p id greater than po, at the α=0.01 significance level. 7.3.6 The economic dynamism, which is the index of productive growth in dollars for countries that are designated by the World Bank as middle-income are in table #7.3.8 ("SOCR data 2008," 2013). Countries that are considered high- income have a mean economic dynamism of 60.29. Do the data show that the mean economic dynamism of middle-income countries is less than the mean for highincome countries? Test at the 5% level. Table #7.3.8: Economic Dynamism of Middle Income Countries 25.805 7 37.451 1 51.91 5 43.695 2 47.850 6 43.717 8 58.076 7 41.164 8 38.079 3 37.725 1 39.655 3 42.026 5 48.615 9 43.855 5 49.136 1 61.928 1 41.954 3 44.934 6 46.052 1 48.365 2 43.625 2 50.986 59.172 39.628 33.607 21.664 6 4 2 4 3 Sample mean = 43.87 =AVERAGE(C2:I4,G5,F5,E5,D5,C5) [Show More]

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