Question

A professor at a university wants to estimate the average number of hours of sleep students get during exam week. On the first day of exams, she asked 20 students how many hours they had slept the night before. The average of the sample was 5.88 with a standard deviation of 1.942. When estimating the average amount of sleep with a 95% confidence interval, what is the margin of error?a) 0.3463b) 0.4893c) 0.6344d) 0.6336e) 0.3725

Answer 1

Answer:

The margin of error is of 0.9089 hours.

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

95% 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 $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.093

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.

20 students:

This means that $n = 20$

Standard deviation of 1.942.

This means that $s = 1.942$

When estimating the average amount of sleep with a 95% confidence interval, what is the margin of error?

$M = T\frac{s}{\sqrt{n}} = 2.093\frac{1.942}{\sqrt{20}} = 0.9089$

The margin of error is of 0.9089 hours.