Question

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 81 graduating seniors and found the mean score to be 506 with a standard deviation of 83. At the 95% confidence level, find the margin of error for the mean, rounding to the nearest tenth. (Do not write ±).

Answer 1

Answer:

18.1

Step-by-step explanation:

Margin of error = critical value × standard error

At 95% confidence, the critical value is z = 1.960.

The standard error is σ/√n = 83/√81 = 9.222.

So the margin of error is:

1.960 × 9.222 = 18.1