Question

A simple random sample of size n is drawn from a normally distributed population, and the mean of the sample is (x with a line over it), while the standard deviation is s. What is the 99% confidence interval for the population mean? Use the table below to help you answer the question.
Confidence Level. 90%. 95%. 99%
z*-score. 1.645. 1.96. 2.58

Answer 1

Answer:

the answer is the (1.645 over root of n) on e 2020

Step-by-step explanation:

Answer 2

Answer:

Last option

${\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}$

Step-by-step explanation:

${\displaystyle {\overline {x}}}$

If we extract a sample of size n from a normal population and we want to estimate the population mean then the confidence interval for the mean will be:

${\displaystyle {\overline {x}}} \± Z_{\alpha/ 2}*\frac{s}{\sqrt{n}}$

Where

Where ${\displaystyle {\overline {x}}}$  is an estimator for the mean μ.

n is the sample size and s is its standard deviation.

$\alpha$ is the level of significance

$\alpha$= 1-Confidence Level

$\alpha= 1-0.99$

$\alpha = 0.01$

Then $Z_{\frac{\alpha}{2}} = Z_{\frac{0.01}{2}} = Z_{0.005}=2.58$

Then the 99% confidence interval for the population mean is

${\displaystyle {\overline {x}}} \± \frac{2.58*s}{\sqrt{n}}$