A particular brand of dietmargarine was analyzed to determine the level of polyu

Question

A particular brand of dietmargarine was analyzed to determine the level of polyunsaturatedfatty acid (in percentages). A sample of six packages resulted inthe following data: 16.8, 17.2, 17.4, 16.9, 16.5, and 17.1.

(a) Check the assumption that the levels of polyunsaturated fattyacid is normally distributed.
(b) Calculate a 99% confidence interval on the mean. Provide a practical interpretation of thisinterval.
(c) Calculate a 99% lower confidence bound on the mean. Compare this bound with the lower bound of the two-sided confidenceinterval and discuss why they are different.


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Explanation / Answer

first arrange the data in ascending order as follows: 16.5, 16.8, 16.9, 17.1, 17.2, 17.4 mean of the above data, =(16.5+16.8+16.9+17.1+17.2+17.4)/6= 16.983 median = (16.9+17.1)/2 = 17 standard deviation of the above data, = 0.319 a) since the mean and median are not equal, hence theassumption that the levels of polyunsaturated fatty acid isnormally distributed is wrong. b) at 99% confidence for two tail distribution, z =2.58 99% confidence inteval on the mean = ± z =16.983 ± (2.58 * 0.319) = (16.16, 17.8) this means 99% of the data lies within the range 16.16 to17.8 c) at 99% confidence, for one tails distribution, z=2.33 99% lower confidence bound to mean = - z= 16.983-(2.33*0.319) = 16.24 this is different from the above because this is one taildistribution and the above question is two tail distribution

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