Question

A researcher would like to estimate the proportion of all children that have been diagnosed with Autism Spectrum Disorder (ASD) in their county. They are using 95% confidence level and the CDC national estimate that 1 in 68 ≈ 0.0147 children are diagnosed with ASD. What sample size should the researcher use to get a margin of error to be within 2%? Round up to the nearest integer.

Answer 1

Answer:

The researcher should use a sample size of at least 140 children to get a margin of error to be within 2%.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence interval $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$ is the margin of error.

95% confidence interval

So $\alpha = 0.05$, z is the value of Z that has a pvalue of $1 - \frac{0.05}{2} = 0.975$, so $z = 1.96$.

We have that:

$M = 0.02, \pi = 0.0147$

So

$M = z\sqrt{\frac{\pi(1-\pi)}{n}}$

$0.02 = 1.96\sqrt{\frac{0.0147*(1-0.0147)}{n}}$

$0.02\sqrt{n} = 0.2359$

$\sqrt{n} = 11.79$

$n = 139.1$

The researcher should use a sample size of at least 140 children to get a margin of error to be within 2%.

Answer 2

Answer:

Sample size of at least 139 children is required.

Step-by-step explanation:

We are given that the a researcher would like to estimate the proportion of all children that have been diagnosed with Autism Spectrum Disorder (ASD) in their country.

Let p = proportion of children diagnosed with ASD = 1/68 = 0.0147

Also, Margin of error = 3%

       Confidence level = 95%

Margin of error formula = $Z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} }$

where, $Z_\frac{\alpha}{2}$ = At 5% level of significance z score has value of 1.96

           $\sigma$  = $\sqrt{p(1-p)}$ = $\sqrt{0.0147(1-0.0147)}$ = 0.1203

So, Margin of error = $1.96 *\frac{\sqrt{0.0147*(1-0.0147)} }{\sqrt{n} }$

                     0.02   = $1.96 *\frac{0.1203 }{\sqrt{n} }$

                       n = $(\frac{1.96*0.1203}{0.02})^{2}$ = 139.10 ≈ 139.

Therefore, the researcher must use a sample size of at least 139 children to get a margin of error to be within 2%.