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It is known that the population variance equals 484. With a 0.95 probability, calculate the sample size that needs to be taken to estimate the population mean if the desired margin of error is 5 or less?

What is Pbar and how do we determine it's value?
Is the variance information we don't need?

User Stoefln
by
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1 Answer

6 votes

Answer:

The sample size is 74

Pbar = 0.05

The variance information is needed

Explanation:

The population variance ,
\sigma^(2) = 484


\sigma = 22

Confidence Interval level = 95% = 0.95

Significance Interval = 1 - CI

Significance Interval = 1 - 0.95 = 0.05

Error margin = 5

The critical value =
Z_{(\alpha)/(2) } = Z_(0.025) = 1.96 (From the z table)

The sample size is given by:


n \geq ((Z_(\alpha /2) * \sigma)/(E) )^(2)


n \geq ((1.96 * 22)/(5)) ^(2)


n \geq 74.373\\n \geq 75


\bar{P} = 1 - P

Since P = 0.95


\bar{P} = 1 - 0.95\\\bar{P} = 0.05

The variance information is needed in this question when calculating the sample size

User Fawwaz Yusran
by
8.2k points
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