Question

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.

Answer 1

Answer:

$z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318$    

$p_v =P(Z>0.318)=0.375$  

So the p value is a very high value and using the significance level provided $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona is not significantly higher than the national incident rate at 1% of significance.

Step-by-step explanation:

Data given and notation  

$X_{1}=507$ represent the number of children diagnosed with Autism Spectrum Disorder (ASD) in Arizona

$n_{1}=32601$ sample elected from arizona

$\hat p_{1}=\frac{507}{32601} =0.0156$ represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona

$p_{2}=\frac{1}{88}= 0.0114$ represent the proportion of children diagnosed with Autism Spectrum Disorder (ASD)  in the nation

z would represent the statistic (variable of interest)  

$p_v$ represent the value for the test (variable of interest)  

Concepts and formulas to use  

We need to conduct a hypothesis in order to check if that the incident of ASD is more in Arizona than nationally, the system of hypothesis are:

Null hypothesis:$p_{1} \leq p_{2}$  

Alternative hypothesis:$p_{1} > p_{2}$  

We need to apply a z test to compare proportions, and the statistic is given by:  

$z=\frac{p_{1}-p_{2}}{\sqrt{\hat p (1-\hat p)(\frac{1}{n_{1}}+\frac{1}{n_{2}})}}$   (1)  

Where $\hat p=\frac{X_{1}+X_{2}}{n_{1}+n_{2}}=\frac{507+1}{32601 +88}=0.01554$  

Calculate the statistic  

Replacing in formula (1) the values obtained we got this:  

$z=\frac{0.0156-0.0114}{\sqrt{0.01554(1-0.01554)(\frac{1}{32601}+\frac{1}{88})}}=0.318$    

Statistical decision

The significance level provided is $\alpha=0.01$, and we can calculate the p value for this test.    

Since is a one right sided test the p value would be:  

$p_v =P(Z>0.318)=0.375$  

So the p value is a very high value and using the significance level provided $\alpha=0.01$ we see that $p_v>\alpha$ so we can conclude that we have enough evidence to FAIL to reject the null hypothesis, and we can say the the proportion of children diagnosed with Autism Spectrum Disorder (ASD) in arizona is not significantly higher than the national incident rate at 1% of significance.