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)

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step-by-step explanation:

because i said it is

24

step-by-step explanation:

the graph shows 30% of students walk. 80 students were questioned, so the number that said they walk is

30% × 80 = 0.30 × 80 = 24