Question

in the state of missouri, there are 273 eight year olds diagnosed with asd out of 18,422 eight year olds evaluated. in the state of arizona, there are 47 eight year olds diagnosed with asd out of 2,013 eight year olds evaluated. estimate the difference in proportion of children diagnosed with asd between missouri and arizona. use a 99% confidence level. round to three decimal places. with % confidence, it can be concluded that the difference in proportion of children diagnosed with asd between missouri and arizona ($p_1 - p_2$) is between and

Explanation:

Step1: Calculate sample proportions

For Missouri, $n_1 = 18422$, $x_1=273$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{273}{18422}\approx0.0148$.
For Arizona, $n_2 = 2013$, $x_2 = 47$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{47}{2013}\approx0.0233$.

Step2: Find the z - value for 99% confidence level

The critical value $z_{\alpha/2}$ for a 99% confidence level. Since $\alpha=1 - 0.99 = 0.01$, then $\alpha/2=0.005$. Looking up in the standard - normal table, $z_{\alpha/2}=2.576$.

Step3: Calculate the standard error

The standard error $SE=\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}$.
$\hat{p}_1(1 - \hat{p}_1)=0.0148\times(1 - 0.0148)=0.0148\times0.9852\approx0.0146$.
$\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}=\frac{0.0146}{18422}\approx7.936\times10^{-7}$.
$\hat{p}_2(1 - \hat{p}_2)=0.0233\times(1 - 0.0233)=0.0233\times0.9767\approx0.0227$.
$\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}=\frac{0.0227}{2013}\approx1.128\times10^{-5}$.
$SE=\sqrt{7.936\times10^{-7}+1.128\times10^{-5}}\approx\sqrt{1.207\times10^{-5}}\approx0.0035$.

Step4: Calculate the margin of error

The margin of error $ME = z_{\alpha/2}\times SE=2.576\times0.0035\approx0.009$.

Step5: Calculate the confidence interval

The point - estimate of the difference in proportions is $\hat{p}_1-\hat{p}_2=0.0148 - 0.0233=- 0.0085$.
The confidence interval is $(\hat{p}_1-\hat{p}_2)-ME$-0.0085-0.009 < p_1 - p_2<-0.0085 + 0.009$.
$-0.018 < p_1 - p_2<0.001$.

Answer:

With 99% confidence, it can be concluded that the difference in proportion of children diagnosed with ASD between Missouri and Arizona $(p_1 - p_2)$ is between - 0.018 and 0.001.