Question

It is known that the variance of a population equals 1,936. A random sample of 121 has been selected from the population. There is a .95 probability that the sample mean will provide a margin of error of

Answer 1

Given that C = 0.95, sample size (n) = 121, variance (σ²) = 1936

Standard deviation (σ) = √variance = √1936 = 44

α = 1 - C = 1 - 0.95 = 0.05

α/2 = 0.025

The z score of α/2 is the same as the z score of 0.475 (0.5 - 0.025) which is equal to 1.96.

The margin of error (E) is given by:

$E=Z_\frac{\alpha}{2} *\frac{\sigma}{\sqrt{n} } \\\\E=1.96*\frac{44}{\sqrt{121} } =7.84$

The 0.95 probability that the sample mean will provide a margin of error of 7.84

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