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 10 children with this learning disorder and recorded the following IQ scores.

105,96,102,95,91,98,109,122,88,129

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, m, 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.

Answer 1

Solution:

By using empirical rule, we know that about 95% of the area lies within two standard deviations from the mean. So, 95% confidence interval for the population mean based on the empirical rule is given as below:

Confidence interval = Xbar ± 2*S/sqrt(n)

Standard error = S/sqrt(n)

Margin of error = 2*S/sqrt(n)

From given data, we have

Xbar = 103.5

S = 13.26021619

n = 10

Standard error = S/sqrt(n) = 13.26021619/sqrt(10) = 4.193249

Margin of error = 2*S/sqrt(n) = 2*4.193249 = 8.386498

Confidence interval = Xbar ± 2*S/sqrt(n)

Confidence interval = 103.5 ± 8.386498

Lower limit = 103.5 - 8.386498 = 95.11

Upper limit = 103.5 + 8.386498 =111.89

Confidence interval = (95.11, 111.89)