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rusak2 [61]
3 years ago
9

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
Mathematics
1 answer:
kumpel [21]3 years ago
4 0

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    

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