Question

A researcher is going to estimate the average typing speed of students of a college. He selects a random sample of 20 students and finds the average typing speed of 70 wpm and the standard deviation of 5 wpm. Estimate the margin of error of the 90% confidence interval of a student's average typing speed. Express your answer using THREE decimal places.

Answer 1

Answer:

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.

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 = 20 - 1 = 19

90% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 19 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.9}{2} = 0.95$. So we have T = 1.7291

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. For this question, we have $s = 5, n = 20$. So

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

$M = 1.7291\frac{5}{\sqrt{20}}$

$M = 1.933$

The margin of error of the 90% confidence interval of a student's average typing speed is of 1.933 wpm.