Question

A large soda producer is interested in determining the average number of ounces of soda in their bottles labeled 16 ounces. A random sample of 600 bottles is selected from that day's total production. The average number of ounces of soda is found to be 16.18 with a standard deviation of 0.32 ounces. Find a 95% confidence interval for the mean number of ounces of soda in such packaging.

a) is this a "proportion of success" or a "means" problem?

b) what calculator function are you using?

c) what confidence interval did you get?

d) interpret the confidence interval in context. Be sure to use the words of the problem.

Answer 1

a) This problem is a means problem, as we are measuring a physical variable, not a proportion.

b) The t-students distribution.

c) The 95% confidence interval for the mean is (16.154, 16.206).

d) We are 95% confident that the true average number of ounces of soda in the bottles is between 16.154 and 16.206 ounces.

Explanation:

We have to calculate a 95% confidence interval for the mean.

The population standard deviation is not known, so we have to estimate it from the sample standard deviation and use a t-students distribution to calculate the critical value.

The sample mean is M=16.18.

The sample size is N=600.

When σ is not known, s divided by the square root of N is used as an estimate of σM:

`s_M=(s)/(√(N))=(0.32)/(√(600))=(0.32)/(24.4949)=0.0131`

The t-value for a 95% confidence interval is t=1.964.

The margin of error (MOE) can be calculated as:

`MOE=t\cdot s_M=1.964 \cdot 0.0131=0.026`

Then, the lower and upper bounds of the confidence interval are:

`LL=M-t \cdot s_M = 16.18-0.026=16.154\\\\UL=M+t \cdot s_M = 16.18+0.026=16.206`

The 95% confidence interval for the mean is (16.154, 16.206).