Question

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, σ = 15 points.

Answer 1

Answer:

E= 6.45

standard deviation is σ = 15,

The critical value is z(α/2) = 2.58.

Step-by-step explanation:

Margin error is the value that is lie above and below the sample.It gives percentage of numbers.Its is the product of critical value standard deviation and standard error of statistic.

General formula for the margin of error is

Margin of error = critical value  ×   standard error of statistic

                        = $z \alpha /2$  ×   σ √n

z-value from two tailed is listed below:

From the table of standard normal distribution, probability value of 0.10.

row and column values gives the area to the two tail of z.

The positive z value is 2.58.

standard deviation is σ = 15,

The critical value is z(α/2) = 2.58.

after putting these vales we obtain the margin of error value that is

E= 6.45