Answer:
The margin of error is 6.45.
Step-by-step explanation:
The complete question is:
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 36 students with the disorder.
Determine the margin of error, 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.
The (1 - <em>α</em>)% confidence interval for population mean <em>μ</em> is:

The margin of error for this interval is:

Given:
<em>n</em> = 36
σ = 15
(1 - <em>α</em>)% = 99%
Compute the critical value of <em>z</em> for 99% confidence level as follows:

*Use a <em>z</em>-table.
Compute the value of MOE as follows:



Thus, the margin of error is 6.45.
Answer:
95
Step-by-step explanation:
add they're scores ( 79 ) that is brains score then at 16 which is colins. that brings us to the conclusion of 95
assuming the scale is by 1,
a) the slope is rise/run so it rises 3 units and moves 4 units right therefore the slope is 3/4
b) the y intercept is where the line crosses the y axis so by looking at the graph you can tell that the y int is 1
c) and the equation is made up of the slope and the y intercept: y= 3/4x + 1
0.1/0.4 converted to a decimal is 0.25 :) hope this helps