Assume that the distribution of home voltage amounts has an unknown mean µ and a

Question

Assume that the distribution of home voltage amounts has an unknown mean µ and a known standard deviation = 0.2404. Construct a 1-=97.5% confidence interval of µ manually by first finding the z-critical values using Rcmdr.

Data:

Day Home Generator UPS 1 123.8 124.8 123.1 2 123.9 124.3 123.1 3 123.9 125.2 123.6 4 123.3 124.5 123.6 5 123.4 125.1 123.6 6 123.3 124.8 123.7 7 123.3 125.1 123.7 8 123.6 125.0 123.6 9 123.5 124.8 123.6 10 123.5 124.7 123.8 11 123.5 124.5 123.7 12 123.7 125.2 123.8 13 123.6 124.4 123.5 14 123.7 124.7 123.7 15 123.9 124.9 123.0 16 124.0 124.5 123.8 17 124.2 124.8 123.8 18 123.9 124.8 123.1 19 123.8 124.5 123.7 20 123.8 124.6 123.7 21 124.0 125.0 123.8 22 123.9 124.7 123.8 23 123.6 124.9 123.7 24 123.5 124.9 123.8 25 123.4 124.7 123.7 26 123.4 124.2 123.8 27 123.4 124.7 123.8 28 123.4 124.8 123.8 29 123.3 124.4 123.9 30 123.3 124.6 123.8 31 123.5 124.4 123.9 32 123.6 124.0 123.9 33 123.8 124.7 123.9 34 123.9 124.4 123.9 35 123.9 124.6 123.6 36 123.8 124.6 123.2 37 123.9 124.6 123.1 38 123.7 124.8 123.0 39 123.8 124.3 122.9 40 123.8 124.0 123.0 The sample mean The two z-critical values are X and -z 0.0125 0.0125 Therefore, a level 100(1-a)% 97.5% confidence interval of u is given by X z a/2 vn

Explanation / Answer

Z critical = 1.96

sigma = .2404

N = 40

and X bar = 123.68

so confidence intervals 123.68 - (1.96*.2404/ sqrt(40)) to 123.68 + (1.96*.2404/ sqrt(40))

= 123.525 123.755

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