Question

A population has a known standard deviation of 1.27 and a sample space contains 85 values find the margin of error needed to create a 99% confidence interval estimate of the mean of the population

Answer 1

To create a 99% confidence interval estimate for a population mean with a known standard deviation, we need the sample mean and the margin of error. The formula for the margin of error is EBM = Z * (standard deviation / sqrt(sample size)), where Z is the z-score corresponding to the desired confidence level. For a 99% confidence interval, Z is approximately 2.576.

Step-by-step explanation:

To construct a confidence interval for a population mean with a known standard deviation, we need the sample mean, denoted as x, and the margin of error, denoted as EBM. The margin of error is dependent on the chosen confidence level. In this case, we require a 99% confidence interval. The formula for the margin of error is given as:

EBM = Z * (standard deviation / sqrt(sample size))

Here, Z represents the z-score corresponding to the chosen confidence level. For a 99% confidence interval, Z is approximately 2.576. Given the information provided, we can calculate the margin of error as:

EBM = 2.576 * (1.27 / sqrt(85)) = 0.300 (rounded to three decimal places)

Therefore, the margin of error needed to create a 99% confidence interval estimate of the population mean is 0.300.

Answer 2

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547

Step-by-step explanation:

We have that to find our
`\alpha` level, that is the subtraction of 1 by the confidence interval divided by 2. So:

`\alpha = (1-0.99)/(2) = 0.005`

Now, we have to find z in the Ztable as such z has a pvalue of
`1-\alpha`.

So it is z with a pvalue of
`1-0.005 = 0.995`, so
`z = 2.575`

Now, find the margin of error M as such

`M = z*(\sigma)/(√(n))`

In which
`\sigma` is the standard deviation of the population and n is the size of the sample.

In this question:

`\sigma = 1.27, n = 85`

So

`M = z*(\sigma)/(√(n))`

`M = 2.575*(1.27)/(√(85))`

`M = 0.3547`

The margin of error needed to create a 99% confidence interval estimate of the mean of the population is of 0.3547