Suppose a psychologist specializing in learning disorders wanted to estimate the

Question

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

Explanation / Answer

Solution

Back-up Theory

If a random variable X ~ N(µ, 2), i.e., X has Normal Distribution with mean µ and variance 2, then

Z = (X - µ)/ ~ N(0, 1), i.e., Standard Normal Distribution ………………………..(1)

P(X or t) = P[{(X - µ)/ } or {(t - µ)/ }] = P[Z or {(t - µ)/ }] .………(2)

X bar ~ N(µ, 2/n),…………………………………………………………….…….(3),

where X bar is average of a sample of size n from population of X.

So, P(X bar or t) = P[Z or {(n)(t - µ)/}] …………………………………(4)

Probability values for the Standard Normal Variable, Z, can be directly read off from

Standard Normal Tables or can be obtained using Excel Function…………………..(5)

Empirical rule, also known as 68 – 95 – 99.7 percent rule:

P{ µ - ) X (µ + )} = 0.68;

P{(µ - 2) X (µ + 2)} = 0.95;

P{(µ - 3) X (µ + 3)} = 0.997 ……………………………………………….(6)

Now, to work out the solution

Let X = IQ score. Then, we are given, X ~ N(100, 152).

And what we require is a lower bound, L say, and an upper bound, say U, such that

P(L µ U) = 0.95………………………………………………………………….. (7)

Now, [vide (3) under Back-up Theory], X bar ~ N(100, 225/11), [11 = sample size.]

[vide (1) under Back-up Theory] {(n)(Xbar - µ)/} ~ N(0, 1).

By Empirical Rule, [vide (6) under Back-up Theory],

P[ (0 – 2) {(n)(Xbar - µ)/} (0 + 2)] = 0.95. i.e., P[ - 2 {(n)(Xbar – µ)/} 2)] = 0.95   

Or, P[{Xbar – (2/n)} µ {Xbar + (2/n)}] = 0.95 …………………………….(8)

(7) and (8) => L = {Xbar – (2/n)} and U = {Xbar + (2/n)}

Now, from the given data, we compute, Xbar = 106.

So, L = 106 – 9.045 and U = 106 + 9.045 or 106 ± 9.05 ANSWER

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