A simple random sample of 20 third-grade children from a certain school district is selected, and each is given a test to measure his/her reading ability. You are interested in calculating a 95% confidence interval for the population mean score. In the sample, the mean score is 64 points, and the standard deviation is 12 points. What is the margin of error associated with the confidence interval

Margin of Error = ME =± 5.2592

Step-by-step explanation:

In the given question n= 20 < 30

Then according to the central limit theorem z test will be applied in which the standard error will be  σ/√n.

Sample Mean = μ = 64

Standard Deviation= S= σ = 12

Confidence Interval = 95 %

α= 0.05

Critical Value for two tailed test for ∝= 0.05 = ±1.96

Margin of Error = ME = Standard Error *Critical Value

ME = 12/√20( ±1.96)=

ME    = 2.6833*( ±1.96)= ± 5.2592

The standard error for this test is σ/√n

=12/√20

=2.6833