Question

You want to estimate the mean hourly yield for a process that manufactures an antibiotic. You observe the process for 100 hourly periods chosen at random, with the results of 32 ounces per hour and a standard deviation of 7 ounces. Calculate the 95% margin of error (in ounces per hour). (Round your answers to three decimal places.)

Answer 1

Answer:

The margin of error for the 95% confidence interval is of 1.389 ounces per hour.

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 = 100 - 1 = 99

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 99 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 1.9842

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.

Then

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

$M = 1.9842\frac{7}{\sqrt{100}}$

$M = 1.389$

The margin of error for the 95% confidence interval is of 1.389 ounces per hour.