Question

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?

Answer 1

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