Question

A machine at a soft-drink bottling factory is calibrated to dispense 12 ounces of cola into cans. A simple random sample of 35 cans is pulled from the line after being filled and the contents are measured. The mean content of the 35 cans is 11.92 ounces with a standard deviation of 0.085 ounce.
Estimate the true mean contents of the cans being filled by this machine with 95% confidence.

Answer 1

Answer:

The true mean contents of the cans being filled by this machine with 95% confidence is between 11.891 ounces and 11.949 ounces.

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 = 35 - 1 = 34

95% confidence interval

Now, we have to find a value of T, which is found looking at the t table, with 34 degrees of freedom(y-axis) and a confidence level of $1 - \frac{1 - 0.95}{2} = 0.975$. So we have T = 2.032

The margin of error is:

$M = T\frac{s}{\sqrt{n}} = 2.032\frac{0.085}{\sqrt{35}} = 0.029$

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

The lower end of the interval is the sample mean subtracted by M. So it is 11.92 - 0.029 = 11.891 ounces.

The upper end of the interval is the sample mean added to M. So it is 11.92 + 0.029 = 11.949 ounces.

The true mean contents of the cans being filled by this machine with 95% confidence is between 11.891 ounces and 11.949 ounces.