A particular brand of diet margarine was analyzed to determi

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, 16.9, and 17.2. Find a 99% two-sided confidence interval on the true mean yield.

(a) Calculate the sample mean and standard deviation. Round the mean to 2 decimal places, and round the standard deviation to 3 decimal places.

(b) Calculate the 99% two-sided confidence interval on the true mean yield. Round your answers to 2 decimal places.

Solution

a)
Mean(x)=17.066
Standard deviation( sd )=0.2338
b)
Confidence Interval
CI = x ± t a/2 * (sd/ Sqrt(n))
Where,
x = Mean
sd = Standard Deviation
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval

Sample Size(n)=6
Confidence Interval = [ 17.066 ± t a/2 ( 0.2338/ Sqrt ( 6) ) ]
= [ 17.066 - 4.032 * (0.095) , 17.066 + 4.032 * (0.095) ]
= [ 16.68,17.45]