Question

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:

Answer 1

Answer: (16.47, 17.49)

Step-by-step explanation:

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

$\overline{x}\pm t_{\alpha/2}\dfrac{s}{\sqrt{n}}$

, where $\overline{x}$ = sample mean

n= sample size

s= sample standard deviation

$t_{\alpha/2}$ = 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

$\overline{x}$ = 16.98

s= 0.31

significance level$(\alpha)$ =1-0.99=0.01

Two tailed t- value for degree of freedom of 5 and significance level of 0.01 = $t_{\alpha/2}=4.0317$    [by student's t-table]

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

$16.98\pm 4.0317(\dfrac{0.31}{\sqrt{6}})\\\\=16.98\pm 4.0317(0.126557)\\\\=16.98\pm 0.51024\\\\=(16.98-0.51023,\ 16.98+0.51023)\\\\=(16.46977,\ 17.49023)\approx (16.47,\ 17.49)$

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