Question

A sociologist develops a test to measure attitudes towards public transportation, and 27 randomly selected subjects are given the test. Their mean score is 76.2 and their standard deviation is 21.4. Find the Margin of Error (E) for the 95% confidence interval for the mean score of all such subjects.

Answer 1

Answer:

The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.

Step-by-step explanation:

We have the standard deviation for the sample, which means that the t-distribution is used 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 = 27 - 1 = 26

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 26 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.0518

The margin of error is:

$M = T\frac{s}{\sqrt{n}}$

In which s is the standard deviation of the sample and n is the size of the sample.

In this question:

$n = 27, s = 21.4$. So

$M = 2.0518\frac{21.4}{\sqrt{27}}$

$M = 8.45$

The margin of error for the 95% confidence interval for the mean score of all such subjects is of 8.45.