11. A certain brand of margarine was analyzed to determine the level of polyunsaturated fatty acid (in percent). A sample of 6 packages had an average of 16.98 and sample standard deviation of 0.31. Assuming normality, a 99% confidence interval for the true mean of fatty acid level is:

(16.47, 17.49)

Step-by-step explanation:

Formula for confidence interval for the true mean if population stanmdard deviation is unknown:

, where = sample mean

n= sample size

s= sample standard deviation

= Two tailed critical value.

We assume that the level of polyunsaturated fatty acid is normally distributed.

Given,

n= 6

degree of freedom = n-1 =5

= 16.98

s= 0.31

significance level =1-0.99=0.01

Two tailed t- value for degree of freedom of 5 and significance level of 0.01 =    [by student's t-table]

Now , the 99% confidence interval for the true mean of fatty acid level is:

Hence, a 99% confidence interval for the true mean of fatty acid level is: (16.47, 17.49)

x -

median = 1.25

range = 12-14.5

interquartile = q12-q14.5

y -

median = 1

range = 15-17

interquartile = q15-q17

graph them on a graphing calculator, press 2nd trace. select intersect. hit enter until it says answer. ti 84 works best

step-by-step explanation: