Respuesta :
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 [tex]\pi[/tex], and a confidence interval [tex]1-\alpha[/tex], we have the following confidence interval of proportions.
[tex]\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
In which
z is the zscore that has a pvalue of [tex]1 - \frac{\alpha}{2}[/tex].
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex] is the margin of error.
95% confidence interval
So [tex]\alpha = 0.05[/tex], z is the value of Z that has a pvalue of [tex]1 - \frac{0.05}{2} = 0.975[/tex], so [tex]z = 1.96[/tex].
We have that:
[tex]M = 0.02, \pi = 0.0147[/tex]
So
[tex]M = z\sqrt{\frac{\pi(1-\pi)}{n}}[/tex]
[tex]0.02 = 1.96\sqrt{\frac{0.0147*(1-0.0147)}{n}}[/tex]
[tex]0.02\sqrt{n} = 0.2359[/tex]
[tex]\sqrt{n} = 11.79[/tex]
[tex]n = 139.1[/tex]
The researcher should use a sample size of at least 140 children to get a margin of error to be within 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 = [tex]Z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} }[/tex]
where, [tex]Z_\frac{\alpha}{2}[/tex] = At 5% level of significance z score has value of 1.96
[tex]\sigma[/tex] = [tex]\sqrt{p(1-p)}[/tex] = [tex]\sqrt{0.0147(1-0.0147)}[/tex] = 0.1203
So, Margin of error = [tex]1.96 *\frac{\sqrt{0.0147*(1-0.0147)} }{\sqrt{n} }[/tex]
0.02 = [tex]1.96 *\frac{0.1203 }{\sqrt{n} }[/tex]
n = [tex](\frac{1.96*0.1203}{0.02})^{2}[/tex] = 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%.
