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)

sss (side, side, side)

sas (side, angle, side)

asa (angle, side, angle)

aas (angle, angle, side)

hl (hypotenuse, leg)

you would have to sell 24 bracelets.

step-by-step explanation:

84/3.5 = 24