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

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

Explanation:

Given:

`\sigma^2=1936`

`\sigma = √(1936)`

`=44`

Random sample,

`n = 121`

The level of significance,

= 0.95

or,

`(1-\alpha) = 0.95`

`\alpha = 1-0.95`

`Z_{(\alpha)/(2) } = 1.96`

hence,

The margin of error will be:

⇒
`E = Z_{(\alpha)/(2) }((\sigma)/(√(n) ) )`

By putting the values, we get

`=1.96((44)/(√(121) ) )`

`=1.96((44)/(11) )`

`=1.96* 4`

`=7.84`