Question

An electronics company just finished designing a new tablet computer and is interested in estimating its battery-life. A random sample of 20 laptops with a full charge was tested and the battery-life was found to be approximately normal with a mean of 6 hours and a sample standard deviation of 1.5 hours. Which of the following is the correct form for a 99% confidence interval?
a) .99( ) 6 2.576(0.3354) CI
b) .99( ) 6 2.576(1.5) CI
c) .99( ) 6 2.861(0.3354)CI
d) .99( ) 6 2.861(1.5) CI

Answer 1

Answer:

$6 \pm 2.861(0.3354)$

Step-by-step explanation:

We have the standard deviation for 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 = 20 - 1 = 19

99% 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 $1 - \frac{1 - 0.99}{2} = 0.995$. So we have T = 2.861.

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.861\frac{1.5}{\sqrt{20}} = 2.861(0.3354)$

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

The format of the confidence interval is:

$S_{M} \pm M$

In which $S_{M}$ is the sample mean

So

$6 \pm 2.861(0.3354)$