#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). Find the proportion of children with ASD in Arizona with a confidence level of 99%.
Blank #1: Report the bounds of the 99% confidence interval as lower bound,upper bound with commas in between and no additional spaces. Report bounds to three decimal places. So, in context, 99% confident that the true proportion of children with ASD in Arizona is between ________________ and ____________________.
#12. 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 incidence of ASD is more in Arizona than nationally? Test at the 1% level.
Use the framework below to guide your work.
Hypotheses: : Ho: p1 = 1/88; Ha: p > 1/88
Blank #2: Test statistic = _________ (round to two decimal places)
Blank #3: p-value = __________ (round to four decimal places)
Blank #4: Test decision: We decide to ___________ Ho (reject or do not reject)
Blank #5: Conclusion back into the words of problem: The evidence __________(favors or does not favor) that the proportion of children with ASD in Arizona is more than the national proportion of 1/88.
Solution:
#6:
Given:
x = 507
n = 32601
Therefore,
Sample propotion is,
Confidence level = 0.99
So,level of significance = = 0.01
Therefore, Zc = 2.576 ...Using standard normal table.
99% confidence interval for population proportion is,
#12.
Null and alternative hypotheses:
Ho: p1 = 1/88 = 0.011; Ha: p > 1/88=0.011
Test statistic:
P-value:
P(Z>7.879) = 1-P(Z<7.879) = 1-1 = 0
P-value =0
Decision: since p=0<0.01, it is concluded that the null hypothesis is rejected.
Conclusion:
The evidence favors that the proportion of children with ASD in Arizona is more than the national proportion of 1/88.
Done
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