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

Answer:

(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 $\bar{x} = 17.02$ and s = 0.299. The confidence interval is given by $\bar{x}\pm t_{\alpha/2}(\frac{s}{\sqrt{n}})$ 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 $T=\frac{\bar{X}-\mu}{S/\sqrt{n}}$). As we want the 99% confidence interval, we have that $\alpha = 0.01$ and the confidence interval is $17.02\pm t_{0.005}(\frac{0.299}{\sqrt{6}})$ 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 $17.02\pm (-4.0321)(\frac{0.299}{\sqrt{6}})$ and the 99% confidence interval is given by (16.528, 17.512)