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Sliva [168]
3 years ago
8

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.
Mathematics
2 answers:
aliina [53]3 years ago
4 0

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%.

Gnoma [55]3 years ago
4 0

Answer:

<h2>Sample size of at least 139 children is required.</h2>

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%.

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