STAT 200 Week 5 Homework Problems

.docx

School

University of Maryland Global Campus (UMGC) *

*We aren’t endorsed by this school

Course

200

Subject

Statistics

Date

Apr 3, 2024

Type

docx

Pages

9

Uploaded by JudgeSummer12858

Report
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. 1. X= identity theft µ=mean identify theft 2. Population parameter: Alaska population complaints 3. Hypothesis: H o : µ=0.23 complaints H A : µ<0.23 complaints α: 0.05 There is not enough information provided. 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. Type I: Rejecting the problem of identify theft when there is one. By saying the proportion of complaints of identity theft in Alaska is less than 23% when it’s 23%, is rejecting a problem when there is a problem. Type I Consequences: The consequences for this error could be giving people in Alaska a false sense of security for fraud. Type II: Saying the proportion of complaints of Alaska are 23% when they are less than 23%. This can make consumers in Alaska believe they are at higher probability of identity theft than they are. Type II Consequences: The consequences for this error could be people in Alaska panicking about identity theft, when there isn’t a greater chance than 23%. Significance level α: α=0.05 The alpha level is appropriate because both errors are equally as bad. 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.
1. X= number of Alaska identity theft p=proportion of Alaska identify theft 2. Hypothesis: H o : p=0.23 complaints H A : p<0.23 complaints α: 0.05 3. A. A sample random sample of 1,432 consumer complaints in Alaska for identity theft was taken. B. There are 1,432 consumers in this case. The consumers are all from Alaska, so you are not mixing people from Alaska and people not from Alaska. There are only two outcomes, either there is identity theft or there is not. The conditions for the binomial distribution are satisfied. c. In this case p=0.23, n=1432, np=329.36≥5, and nq=1102.64≥5. So, the sampling distribution for p^ is a normal distribution. 4. The z in the results is the test statistics. The p=0.2998057839 is the p-value, and the p^=0.224 is the sample proportion. The p-value is approximately 0.2998. 5. Conclusion: Since the p-value>0.05 then reject H o Since .2998 > .05, fail to reject the null hypothesis. 6. Interpretation: There is not evidence to show that the proportion of Alaska provides enough evidence to show that Alaska had a lower proportion of identity theft 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. 1. x=number of children diagnosed with ASD in Arizona p=proportion of children diagnosed with ASD in Arizona 2.H o : p=0.011 H A : p>0.011 α=0.01 3.A. a simple random sample of 32,601 children in Arizona who were diagnosed with ASD was taken. The data is actually all of children who were diagnosed with ASD in Arizona. However, the information is still sample information since it is only for the one year that the data was collected. B. There were 32,601 Arizona children were diagnosed. There are only two outcomes, either diagnosed or not diagnosed. The conditions for the binomial distribution are satisfied. C. In this case, n=32,601, p=0.011, np=32,601*0.011=358.611≥1, nq=32,601(1- 0.011)=32242.389≥1 4.
The z in the results is the test statistic. The p=4.899440939 E -13 is the p value which is actually 4.8990.x13 -13 , and the p^=0.015556702 is the sample proportion. The p-value is approximately 4.899. You can't have a p-value of 4.899 because a p-value is a probability. You ignored the E-13 in the display of the p-value, which means the p-value is 4.889 x 10-13 = .0000000000004889; e Since the p-value is < .01 we will reject the null hypothesis; f There is sufficient evidence to show that the proportion of Arizona children in 2008 with ASD is greater than the national proportion 4. Conclusion: Since the p-value is more than 0.01, you fail to reject the H o . 5.Interpretation: There is not enough evidence to show that the proportion of children in Arizona who are diagnosed with ASD is sufficient data to show that the incident of ASD is more in Arizona than nationally. 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 high-income countries? Test at the 5% level. Table #7.3.8: Economic Dynamism of Middle Income Countries 25.8057 37.4511 51.915 43.6952 47.8506 43.7178 58.0767 41.1648 38.0793 37.7251 39.6553 42.0265 48.6159 43.8555 49.1361 61.9281 41.9543 44.9346 46.0521 48.3652 43.6252 50.9866 59.1724 39.6282 33.6074 21.6643 1.x= economic dynamism of middle-income countries µ=mean of dynamism of middle-income countries 2. H o : µ =$60.29 H A : µ < $60.29 α= 0.05 3.A. A random sample 26 economic dynamism of middle- income countries in 2013 was taken. The data is actually all of the economic dynamism of middle-income countries designed by the World Bank. However, the information is still sample information since it is only for one year that is collected. B. The distribution of economic dynamism of middle-income countries in 2013 is normally distribution. To see if this condition has been met, we look at the histogram.
Your preview ends here
Eager to read complete document? Join bartleby learn and gain access to the full version
  • Access to all documents
  • Unlimited textbook solutions
  • 24/7 expert homework help
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
SEE MORE TEXTBOOKS
Recommended textbooks for you
Text book image
Glencoe Algebra 1, Student Edition, 9780079039897...
Algebra
ISBN:9780079039897
Author:Carter
Publisher:McGraw Hill
Text book image
College Algebra (MindTap Course List)
Algebra
ISBN:9781305652231
Author:R. David Gustafson, Jeff Hughes
Publisher:Cengage Learning
SEE MORE TEXTBOOKS