Question

A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percentages). A sample of six packages resulted in the following data: 16.8, 17.2, 17.4, 16.9, 16.6, and 17.2. Find a 99% two-sided confidence interval on the true mean yield. Assume population is approximately normally distributed. (a) Calculate the sample mean and standard deviation. Round the mean to 2 decimal places, and round the standard deviation to 3 decimal places.

Answer 1

(16.528, 17.512) is a 99% two-sided confidence interval for the true mean yield. The sample mean is 17.02 and the sample standard deviation is 0.299.

Explanation:

We have a small sample of size n = 6, where the sample mean and standard deviation are
`\bar{x} =  17.02` and s = 0.299. The confidence interval is given by
`\bar{x}\pm t_(\alpha/2)((s)/(√(n)))` where
`t_(\alpha/2)` is the
`\alpha/2`th quantile of the t distribution with n-1=5 degrees of freedom, this because we are dealing with a small sample which comes from the normal distribution (the pivotal quantity is
`T=\frac{\bar{X}-\mu}{S/√(n)}`). As we want the 99% confidence interval, we have that
`\alpha = 0.01` and the confidence interval is
`17.02\pm t_(0.005)((0.299)/(√(6)))` where
`t_(0.005)` is the 0.5th quantile of the t distribution with 5 df, i.e.,
`t_(0.005) =  -4.0321`. Then, we have
`17.02\pm (-4.0321)((0.299)/(√(6)))` and the 99% confidence interval is given by (16.528, 17.512)