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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 :

Cricetus

Answer:

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

Step-by-step explanation:

Given:

[tex]\sigma^2=1936[/tex]

[tex]\sigma = \sqrt{1936}[/tex]

   [tex]=44[/tex]

Random sample,

[tex]n = 121[/tex]

The level of significance,

= 0.95

or,

[tex](1-\alpha) = 0.95[/tex]

        [tex]\alpha = 1-0.95[/tex]

[tex]Z_{\frac{\alpha}{2} } = 1.96[/tex]

hence,

The margin of error will be:

⇒ [tex]E = Z_{\frac{\alpha}{2} }(\frac{\sigma}{\sqrt{n} } )[/tex]

By putting the values, we get

        [tex]=1.96(\frac{44}{\sqrt{121} } )[/tex]

        [tex]=1.96(\frac{44}{11} )[/tex]

        [tex]=1.96\times 4[/tex]

        [tex]=7.84[/tex]    

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