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Sagot :
Answer:
The 95% confidence interval for the proportion of participants who have SLD among the children with ASD is (0.1437, 0.1813).
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 level of [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].
Out of a sample of 1,483 participants, a total of 241 were found to have SLD.
This means that [tex]n = 1483, \pi = \frac{241}{1483} = 0.1625[/tex]
95% confidence level
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].
The lower limit of this interval is:
[tex]\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1625 - 1.96\sqrt{\frac{0.1625*0.8375}{1483}} = 0.1437[/tex]
The upper limit of this interval is:
[tex]\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.1625 + 1.96\sqrt{\frac{0.1625*0.8375}{1483}} = 0.1813[/tex]
The 95% confidence interval for the proportion of participants who have SLD among the children with ASD is (0.1437, 0.1813).
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