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.

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.

Step-by-step 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:

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

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

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

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

your answer would be 10x - 1

72 is the answer to this problem