Question

Match the confidence level with the confidence interval for the population mean. type the correct letter in each box. equation editorequation editor 1. x¯¯¯±1.96(σn√) equation editorequation editor 2. x¯¯¯±1.645(σn√) equation editorequation editor 3. x¯¯¯±2.575(σn√)

Answer 1

Answer:

$\bar X \pm 1.96 \frac{\sigma}{\sqrt{n}}$

"=1-2*(1-NORM.DIST(1.96,0,1,TRUE))"

And we got 95%

$\bar X \pm 1.645 \frac{\sigma}{\sqrt{n}}$

"=1-2*(1-NORM.DIST(1.645,0,1,TRUE))"

And we got 90%

$\bar X \pm 2.575 \frac{\sigma}{\sqrt{n}}$

"=1-2*(1-NORM.DIST(2.575,0,1,TRUE))"

And we got 99%

Step-by-step explanation:

Previous concepts  

A confidence interval is "a range of values that’s likely to include a population value with a certain degree of confidence. It is often expressed a % whereby a population means lies between an upper and lower interval".  

The margin of error is the range of values below and above the sample statistic in a confidence interval.  

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".  

Solution to the problem

The confidence interval for the mean is given by the following formula:  

$\bar X \pm z_{\alpha/2}\frac{\sigma}{\sqrt{n}}$ (1)  

For this case we have the following cases:

$\bar X \pm 1.96 \frac{\sigma}{\sqrt{n}}$

And for this case we can find the % of confidence with the following excel code:

"=1-2*(1-NORM.DIST(1.96,0,1,TRUE))"

And we got 95%

For the other case:

$\bar X \pm 1.645 \frac{\sigma}{\sqrt{n}}$

And for this case we can find the % of confidence with the following excel code:

"=1-2*(1-NORM.DIST(1.645,0,1,TRUE))"

And we got 90%

For the other case:

$\bar X \pm 2.575 \frac{\sigma}{\sqrt{n}}$

And for this case we can find the % of confidence with the following excel code:

"=1-2*(1-NORM.DIST(2.575,0,1,TRUE))"

And we got 99%