Question

P286 8-38 (Modified) (28pt)

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, 16.5, 17.1.

a.Is the level of polyunsaturated fatty acid normally distributed? Why or why not?

b. Calculate a 99% two-sided confidence interval for the mean level of polyunsaturated fatty acid for this particular brand of diet margarine.

c. Is your method used in part (b) justified? Why or why not?

d. Repeat part (b) using R. Attached the R codes and output.

Answer 1

a)

> shapiro.test(x)

        Shapiro-Wilk normality test

data: x
W = 0.98779, p-value = 0.9831

The data is normally distributed by Shapiro test since p-vale > 0.05.

b)

c) Yes, the method used in part (b) is justified, since all the assumptions for t-test are valid here.

d)

> t.test(x,conf.level=0.99)

        One Sample t-test

data: x
t = 130.47, df = 5, p-value = 5.017e-10
alternative hypothesis: true mean is not equal to 0
99 percent confidence interval:
16.45847 17.50820
sample estimates:
mean of x
16.98333