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=12 units bd= 10 units

step-by-step explanation:

answer is d-29

step-by-step explanation:

since you didn't include the picture, i can give you the description so you know how to pick the correct answer. the third quartile of a data is always shown in a box plot. in the middle of your graph should be a box. the third quartile is the value that goes with the far right edge of the box.