Question

in the state of hawaii, there are 273 eight - year - olds diagnosed with asd out of 18,667 eight - year - olds evaluated. in the state of new york, there are 50 eight - year - olds diagnosed with asd out of 2,371 eight - year - olds evaluated. estimate the difference in proportion of children diagnosed with asd between hawaii and new york. 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 hawaii and new york ($p_1 - p_2$) is between and

Explanation:

Step1: Calculate sample proportions

For Hawaii, $n_1 = 18667$, $x_1=273$, so $\hat{p}_1=\frac{x_1}{n_1}=\frac{273}{18667}\approx0.0146$. For New - York, $n_2 = 2371$, $x_2 = 50$, so $\hat{p}_2=\frac{x_2}{n_2}=\frac{50}{2371}\approx0.0211$.

Step2: Find critical value

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

Step3: Calculate the standard error

$SE=\sqrt{\frac{\hat{p}_1(1 - \hat{p}_1)}{n_1}+\frac{\hat{p}_2(1 - \hat{p}_2)}{n_2}}=\sqrt{\frac{0.0146\times(1 - 0.0146)}{18667}+\frac{0.0211\times(1 - 0.0211)}{2371}}\approx\sqrt{\frac{0.0146\times0.9854}{18667}+\frac{0.0211\times0.9789}{2371}}\approx\sqrt{\frac{0.014387}{18667}+\frac{0.020655}{2371}}\approx\sqrt{7.71\times10^{-7}+8.63\times10^{-5}}\approx\sqrt{8.71\times10^{-5}}\approx0.0093$.

Step4: Calculate the confidence interval

The difference in sample proportions is $\hat{p}_1-\hat{p}_2=0.0146 - 0.0211=- 0.0065$.
The margin of error $ME = z_{\alpha/2}\times SE=2.576\times0.0093\approx0.0239$.
The confidence interval is $(\hat{p}_1-\hat{p}_2)-ME$-0.0065-0.0239 < p_1 - p_2<-0.0065 + 0.0239$.
$-0.030 < p_1 - p_2<0.017$.

Answer:

With 99% confidence, it can be concluded that the difference in proportion of children diagnosed with ASD between Hawaii and New York ($p_1 - p_2$) is between - 0.030 and 0.017.