The Stanford-Binet Intelligence Scale is an intelligence test, which, like many other IQ tests, is standardized in order to have a normal distribution with a mean of 100 and a standard deviation of 15 points.

As an early intervention effort, a school psychologist wants to estimate the average score on the Stanford-Binet Intelligence Scale for all students with a specific type of learning disorder using a simple random sample of 36 students with the disorder.

Determine the margin of error,
, of a 90% confidence interval for the mean IQ score of all students with the disorder. Assume that the standard deviation IQ score among the population of all students with the disorder is the same as the standard deviation of IQ score for the general population, =15

points.

Give your answer precise to at least two decimal places.

The margin of error of a 90% confidence interval for the mean IQ score of all students with the disorder is 4.1125 and this can be determined by using the formula of margin of error.

Given :

Standardized in order to have a normal distribution with a mean of 100 and a standard deviation of 15 points.
A school psychologist wants to estimate the average score on the Stanford-Binet Intelligence Scale for all students with a specific type of learning disorder using a simple random sample of 36 students with the disorder.
90% confidence interval.

According to the given data, the confidence is 90% and n > 30, so the critical value is z = 1.645.

Now, to determine the standard error using the below formula:

[tex]\rm SE = \dfrac{\sigma}{\sqrt{n} }[/tex]

[tex]\rm SE = \dfrac{15}{\sqrt{36} }=\dfrac{15}{6}[/tex]

SE = 2.5

Now, the formula of margin of error is given by:

[tex]\rm MOE = z\times SE[/tex]

MOE = (1.645) [tex]\times[/tex] (2.5)

MOE = 4.1125

So, the margin of error is 4.1125.

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https://brainly.com/question/22771970