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Dr Z decided to investigate the productivity of her plutonium mines. The standard deviation for the...

Dr Z decided to investigate the productivity of her plutonium mines. The standard deviation for the output is known (fat hint: use z-tables) and is 5 tons a day. The average output in a sample of a 100 mines was 66 tons. At 95 % confidence level, what is the margin of error? (Two decimals)

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Answer #1

solutionPopulation standard deviation =    = 5

Sample size n =100

At 95% confidence level the z is ,

= 1 - 95% = 1 - 0.95 = 0.05

/ 2 = 0.05 / 2 = 0.025

Z/2 = Z0.025 = 1.96   ( Using z table )

Margin of error = E =   Z/2    * ( /n)
= 1.96 * (5 / 100)

=0.98

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