Question

Are there more children diagnosed with Autism Spectrum Disorder (ASD) in states that have larger urban areas over states that are mostly rural? In the state of Pennsylvania, a fairly urban state, there are 245 eight year olds diagnosed with ASD out of 18,440 eight year olds evaluated. In the state of Utah, a fairly rural state, there are 45 eight year olds diagnosed with ASD out of 2,123 eight year olds evaluated ("Autism and developmental," 2008). Is there enough evidence to show that the proportion of children diagnosed with ASD in Pennsylvania is more than the proportion in Utah? Test at the 1% level.

Answer 1

Answer:

$z=\frac{0.0211-0.0132}{\sqrt{0.0141(1-0.0141)(\frac{1}{18440}+\frac{1}{2123})}}=24.827$  

$p_v =P(Z>24.827)\approx 0$  

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of children with ASD in Utah is significantly higher than the proportion of children with ASD in Pennsylvania

Step-by-step explanation:

Data given and notation  

$X_{P}=245$ represent the number of children with ASD in Pennsylvania

$X_{U}=45$ represent the number of children with ASD in Utah

$n_{P}=18440$ sample for Pennsylvania

$n_{U}=2123$ sample for Utah

$p_{P}=\frac{245}{18440}=0.0133$ represent the proportion of children with ASD in Pennsylvania

$p_{U}=\frac{45}{2123}=0.0212$ represent the proportion of children with ASD in Utah

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 the proportion for Utah is higher than the proportion for Penssylvania  , the system of hypothesis would be:  

Null hypothesis:$p_{U} \leq p_{P}$  

Alternative hypothesis:$p_{U} > \mu_{p}$  

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

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

Where $\hat p=\frac{X_{U}+X_{P}}{n_{U}+n_{P}}=\frac{245+45}{18440+2123}=0.0141$

Calculate the statistic

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

$z=\frac{0.0211-0.0132}{\sqrt{0.0141(1-0.0141)(\frac{1}{18440}+\frac{1}{2123})}}=24.827$  

Statistical decision

For this case we don't have a significance level provided $\alpha$, but we can calculate the p value for this test.  

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

$p_v =P(Z>24.827)\approx 0$  

So the p value is a very low value and using any significance level for example $\alpha=0.05, 0,1,0.15$ always $p_v$ so we can conclude that we have enough evidence to reject the null hypothesis, and we can say the the proportion of children with ASD in Utah is significantly higher than the proportion of children with ASD in Pennsylvania