Question

in the state of texas, there are 205 eight - year - olds diagnosed with asd out of 18,781 eight - year - olds evaluated. in the state of kansas, there are 43 eight - year - olds diagnosed with asd out of 2,084 eight - year - olds evaluated. estimate the difference in proportion of children diagnosed with asd between texas and kansas. 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 texas and kansas ($p_1 - p_2$) is between and > next question

Explanation:

Step1: Calculate sample proportions

Let $n_1 = 18781$ be the number of eight - year - olds evaluated in Texas and $x_1=205$ be the number diagnosed with ASD in Texas. Let $n_2 = 2084$ be the number of eight - year - olds evaluated in Kansas and $x_2 = 43$ be the number diagnosed with ASD in Kansas.
The sample proportion for Texas $p_1=\frac{x_1}{n_1}=\frac{205}{18781}\approx0.01091$
The sample proportion for Kansas $p_2=\frac{x_2}{n_2}=\frac{43}{2084}\approx0.02063$

Step2: Calculate the difference in sample proportions

$\hat{p}_1-\hat{p}_2=0.01091 - 0.02063=- 0.00972$

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 $\hat{p}_2$ for $p_1$ and $p_2$.
$SE=\sqrt{\frac{0.01091\times(1 - 0.01091)}{18781}+\frac{0.02063\times(1 - 0.02063)}{2084}}$
$=\sqrt{\frac{0.01091\times0.98909}{18781}+\frac{0.02063\times0.97937}{2084}}$
$=\sqrt{\frac{0.010791}{18781}+\frac{0.020194}{2084}}$
$=\sqrt{0.0000005756+0.000009689}$
$=\sqrt{0.0000102646}\approx0.00320$

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

For a 99% confidence level, the significance level $\alpha=1 - 0.99 = 0.01$, and $\alpha/2=0.005$. The z - value $z_{\alpha/2}=z_{0.005}=2.576$

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 $=(\hat{p}_1-\hat{p}_2)-z_{\alpha/2}\times SE=-0.00972-2.576\times0.00320=-0.00972 - 0.00824=-0.01796$
Upper limit $=(\hat{p}_1-\hat{p}_2)+z_{\alpha/2}\times SE=-0.00972 + 2.576\times0.00320=-0.00972+0.00824=-0.00148$

Answer:

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