Aparticular 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.

(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.

Step-by-step explanation:

We have a small sample of size n = 6, where the sample mean and standard deviation are and s = 0.299. The confidence interval is given by where is the 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 ). As we want the 99% confidence interval, we have that and the confidence interval is where is the 0.5th quantile of the t distribution with 5 df, i.e., . Then, we have and the 99% confidence interval is given by (16.528, 17.512)