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Suppose a psychologist specializing in learning disorders wanted to estimate the mean IQ score for children with a particular type of learning disorder. She obtained a random sample of 11 children with this learning disorder and recorded the following IQ scores.

82, 123, 102, 108, 113, 137, 114, 106, 92, 91, 98

IQ scores in the general population are normally distributed with a mean of 100.0 points and a standard deviation of 15.0 points. The psychologist was willing to assume that the distribution of IQ scores for all children with this particular type of learning disorder is approximately normal with a standard deviation equal to the standard deviation in the general population, 15.0 points.

Use the empirical rule, also known as the 68-95-99.7 percent rule, to estimate the margin of error of a 95% confidence interval for the average IQ score for all children with this learning disorder. Enter your answer with two decimal places of precision. _____ IQ points

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Answer:

Margin of error in 95% confidence level is 9,05 IQ points.

Step-by-step explanation:

margin of error (ME) around the mean using the formula

ME=[tex]\frac{z*s}{\sqrt{N} }[/tex] where

  • z is the corresponding statistic in the given confidence level(z-score or t-score)
  • s is the standard deviation of the sample(or of the population if it is known)
  • N is the sample size

Population standard deviation of IQ scores for all children with this particular type of learning disorder is assumed 15.

If the empirical rule is used, 95% confidence interval is 2 standard deviations around the mean. If we use the formula:

ME=[tex]\frac{2*15}{\sqrt{11} }[/tex] ≈ 9,05

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