A small amount of the trace element selenium, 50–200 micrograms (µg) per day, is

Question

A small amount of the trace element selenium, 50–200 micrograms (µg) per day, is considered essential to good health. Suppose that random samples of

n1 = n2 = 36

adults were selected from two regions of the United States and that a day's intake of selenium was recorded for each person. The mean and standard deviation of the selenium daily intakes for the 36 adults from region 1 were

x1 = 167.1

and

s1 = 25.8 µg,

respectively. The corresponding statistics for the 36 adults from region 2 were

x2 = 140.9

and

s2 = 17.7 µg.

Find a 95% confidence interval for the difference

(1 2)

in the mean selenium intakes for the two regions. (Round your answers to three decimal places.)

Explanation / Answer

The difference of the two samples follows a standard normal distribution as the variances of the sample are known.

Mean = X1 - X2 = 167.1 - 140.9 = 26.2

Standard error^2 = S1^2/n1 + S2^2/n2 = 25.8^2/36 + 17.7^2/36 = 27.19

Standard error = sqrt(27.19) = 5.21

At alpha = 0.05, Zalpha/2 = 1.96

So, 95% CI of difference of means is : 26.2 - 1.96*5.21, 26.2+1.96*5.21 = (15.988, 36.412)

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