A particular brand of diet margarine was analyzed to determine the level of poly

Question

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.

1) Calculate a 99% two-sided confidence interval for the mean level of polyunsaturated fatty acid for this particularbrand of diet margarine. Is your method used in part b justified? Why or why not?

2) Calculate a 99% lower confidence bound for the mean level of polyunsaturated fatty acid for this particular brandof diet margarine. Interpret.

3) Calculate a 95% two-sided confidence interval for the standard deviation of the level of polyunsaturated fatty acid for this particular brand of diet margarine.

Explanation / Answer

Sample of six packages: 16.8, 17.2, 17.4, 16.9, 16.5, 17.1.

Sample mean, X_bar = 101.9/6 = 16.98333 and sample standard deviation, sd = 0.31885.

1. 99% two sided CI for mean level = [ X_bar - t0.005,5*sd/sqrt(6), X_bar + t0.005,5*sd/sqrt(6) ] = [16.45847, 17.5082].

2. 99% lower confidence bound for mean level = X_bar - t0.01,5*sd/sqrt(6) = 16.54532. That means out of 100 occasions, the mean level is more than 16.54532 for 99 occasions on an average.

3. 95% two sided CI for sd of the level = [ sd*sqrt(5/chi-sq0.025) , sd*sqrt(5/chi-sq0.975) ] = [ 0.05026, 0.1975].

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