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 _____. Group of answer choices 31.36 or less 1,936 or less 344.96 or less 7.84 or less

Answer 1

Answer:

Option d (7.84 or less) is the right alternative.

Step-by-step explanation:

Given:

$\sigma^2=1936$

$\sigma = \sqrt{1936}$

   $=44$

Random sample,

$n = 121$

The level of significance,

= 0.95

or,

$(1-\alpha) = 0.95$

        $\alpha = 1-0.95$

$Z_{\frac{\alpha}{2} } = 1.96$

hence,

The margin of error will be:

⇒ $E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )$

By putting the values, we get

        $=1.96(\frac{44}{\sqrt{121} } )$

        $=1.96(\frac{44}{11} )$

        $=1.96\times 4$

        $=7.84$