Question

in the state of alabama, there are 219 eight year olds diagnosed with asd out of 18,810 eight year olds evaluated. in the state of rhode island, there are 50 eight year olds diagnosed with asd out of 2,031 eight year olds evaluated. estimate the difference in proportion of children diagnosed with asd between alabama and rhode island. 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 alabama and rhode island ($p_1 - p_2$) is between and > next question

Explanation:

Step1: Calculate sample proportions

Let $n_1 = 18810$ (number of eight - year - olds evaluated in Alabama), $x_1=219$ (number of ASD diagnosed in Alabama), $n_2 = 2031$ (number of eight - year - olds evaluated in Rhode Island), $x_2 = 50$ (number of ASD diagnosed in Rhode Island).
The sample proportion for Alabama $p_1=\frac{x_1}{n_1}=\frac{219}{18810}\approx0.0116$
The sample proportion for Rhode Island $p_2=\frac{x_2}{n_2}=\frac{50}{2031}\approx0.0246$

Step2: Calculate the difference in sample proportions

$\hat{p}_1-\hat{p}_2=0.0116 - 0.0246=- 0.013$

Step3: Calculate the standard error

$SE=\sqrt{\frac{p_1(1 - p_1)}{n_1}+\frac{p_2(1 - p_2)}{n_2}}$
Since we are using sample proportions to estimate, we substitute $\hat{p}_1$ and $\hat{p}_2$
$SE=\sqrt{\frac{0.0116\times(1 - 0.0116)}{18810}+\frac{0.0246\times(1 - 0.0246)}{2031}}$
$=\sqrt{\frac{0.0116\times0.9884}{18810}+\frac{0.0246\times0.9754}{2031}}$
$=\sqrt{\frac{0.01146544}{18810}+\frac{0.02409484}{2031}}$
$=\sqrt{6.1\times10^{-7}+1.19\times10^{-4}}$
$=\sqrt{0.00011961}\approx0.0109$

Step4: Find the z - value for 99% confidence level

The z - value $z_{\alpha/2}$ for a 99% confidence level is 2.576

Step5: Calculate the confidence interval

The confidence interval for $p_1 - p_2$ is given by $(\hat{p}_1-\hat{p}_2)\pm z_{\alpha/2}\times SE$
Lower limit: $(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\times SE=-0.013-2.576\times0.0109=-0.013 - 0.0281=-0.041$
Upper limit: $(\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\times SE=-0.013 + 2.576\times0.0109=-0.013+0.0281 = 0.015$

Answer:

With 99% confidence, it can be concluded that the difference in proportion of children diagnosed with ASD between Alabama and Rhode Island $(p_1 - p_2)$ is between - 0.041 and 0.015.