PLEASE HELP A study by the department of education of a certain state was trying to determine the mean SAT scores of the graduating high school seniors. The study examined the scores of a random sample of 224 graduating seniors and found the mean score to be 492 with a standard deviation of 90. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth. (Do not write
To find the margin of error for the mean at a 95% confidence level, you can use the formula:
Margin of Error = (Critical Value) * (Standard Deviation / Square Root of Sample Size)
First, we need to find the critical value associated with a 95% confidence level. For a z-distribution, which is commonly used for large sample sizes, the critical value for a 95% confidence level is approximately 1.96.
Next, we need to determine the standard deviation and the sample size. In this case, the standard deviation is given as 90, and the sample size is 224.
Using these values, we can calculate the margin of error:
Margin of Error = (1.96) * (90 / √224)
Computing this expression, we get:
Margin of Error ≈ 13.3
Therefore, the margin of error for the mean at the 95% confidence level is approximately 13.3.