Question

A survey was initiated and intended to capture the prevalence of specific learning disorder (SLD) among school-aged children with autism spectrum disorder (ASD). Out of a sample of 1,483 participants, a total of 241 were found to have SLD. Calculate 95% confidence interval for the proportion of participants who have SLD among the children with ASD.

Answer 1

The 95% confidence interval for the proportion of participants who have SLD among the children with ASD is (0.1437, 0.1813).

Explanation:

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

`\pi \pm z\sqrt{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\alpha)/(2)`.

Out of a sample of 1,483 participants, a total of 241 were found to have SLD.

This means that
`n = 1483, \pi = (241)/(1483) = 0.1625`

95% confidence level

So
`\alpha = 0.05`, z is the value of Z that has a pvalue of
`1 - (0.05)/(2) = 0.975`, so
`Z = 1.96`.

The lower limit of this interval is:

`\pi - z\sqrt{(\pi(1-\pi))/(n)} = 0.1625 - 1.96\sqrt{(0.1625*0.8375)/(1483)} = 0.1437`

The upper limit of this interval is:

`\pi + z\sqrt{(\pi(1-\pi))/(n)} = 0.1625 + 1.96\sqrt{(0.1625*0.8375)/(1483)} = 0.1813`

The 95% confidence interval for the proportion of participants who have SLD among the children with ASD is (0.1437, 0.1813).