Question

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 16 students with the disorder. Determine the margin of error, m, of a 99% 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, sigma = 15 points.

Answer 1

Answer:

The  margin of error is $MOE = 9.68$

Step-by-step explanation:

From the question we are told that

    The sample size is $n= 16$

     The  standard deviation is $\sigma = 15$

     The  confidence level is  $C = 99$%

Generally the level of significance is mathematically evaluated as

     $\alpha = 100 - C$

     $\alpha = 100 - 99$

     $\alpha = 1%$%

   $\alpha = 0.01$

The  critical value of  $\frac{\alpha }{2}$ obtained from the normal distribution table is  

     $Z_{\frac{\alpha }{2} } = 2.58$

The  reason obtaining the critical value of  $\frac{\alpha }{2}$ instead of  $\alpha$ is because we are  considering the two tails of the area normal distribution curve which is not inside the 99% confidence interval

   Now the margin of error is evaluated as

              $MOE = Z_{\frac{\alpha }{2} } * \frac{\sigma}{\sqrt{n} }$

substituting values

            $MOE = 2.58 * \frac{15}{\sqrt{16} }$

           $MOE = 9.68$