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 \pm±).

Answer 1

Answer:

The margin of error for the mean is of 165.18 points.

Step-by-step explanation:

We have the standard deviation of the sample, so we use the t-distribution to solve this question.

The first step to solve this problem is finding how many degrees of freedom, we have. This is the sample size subtracted by 1. So

df = 81 - 1 = 80

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 80 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 1.9901

The margin of error is:

M = T*s = 1.9901*83 = 165.18.

In which s is the standard deviation of the sample.

The margin of error for the mean is of 165.18 points.