Question

4. All the savvy business statisticians know that the variance of a population equals 1780. A random sample of 121 observations is going to be taken from the population. a. With a 0.95 probability, what statement can be made about the size of the margin of error? b. With a 0.95 probability, how large of a sample would have to be taken to provide a margin of error of 8 or less?

Answer 1

Solution:-

4)

Variance = 1780

a) The size of the margin of error is 7.5175.

$M.E = z_{\alpha /2}\times{\frac{\sigma }{\sqrt{n}}}$

$M.E = 1.96\times{\frac{42.19005 }{\sqrt{121}}}$

M.E = 7.5175

b) The required sample size is 107.

$n = \left (\frac{z_{\alpha /2}\times \sigma }{M.E} \right )^{2}$

$n = \left (\frac{1.96\times 42.19005 }{8} \right )^{2}$

n = 106.8445

n = 107