# Explain the concept of a confidence interval estimate. A) Discuss what impact changing the sample size has on the calculation of the CIE holding all other variables constant. (Discuss how the CIE...

Explain the concept of a confidence interval estimate.

A) Discuss what impact changing the sample size has on the calculation of the CIE holding all other variables constant. (Discuss how the CIE will change as the sample size "n" changes.)

B) Discuss what impact changing the standard deviation has on the calculation of the CIE holding all other variables constant. (Discuss how the CIE will change as the standard deviation changes.)

C) Discuss how the confidence interval estimate will change as the level of confidence changes holding all other variables constant. (Discuss how the CIE will change as the confidence level changes).

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A.) **As the sample size increases the width of the confidence interval decreases**.

For example:

`sigma = 1.4`

` ` `bar x = 15.7`

`n = 30`

with a` 95%` confidence level

Finding the confidence interval:

`barx +- z_(alpha/2) * (sigma)/(sqrt(n))`

`15.7 +- 1.96*(1.4)/(sqrt(30))`

`15.7 +- .5009835659`

`15.20, 16.20`

For this one the width of the interval is `16.20 - 15.20 = 1` .

If we change the sample size to` 60` , we will have:

`15.7 +- 1.96*(1.4)/(sqrt(60))`

`15.7 +- .3542488767`

`15.35, 16.05`

For this interval the width of the interval is `16.05 - 15.35 = 0.70` .

B.) If we increase the standard deviation to `2.4` , we will have:

`15.7 +- 1.96*(2.4)/(sqrt(30))`

`15.7 +- 0.8588289702`

`14.84, 16.56`

For this confidence interval the width is `1.72` . That is larger compare to the width of `1` for a standard deviation of ` 1.4 ` from part a.

**Hence, as we increase the standard deviation, the width of the confidence interval also increases**.

C.)** As the confidence level increases the width of the interval also increases.**

If we change the 95% to 99% we will have:

`15.7 +- 2.575*(1.4)/(sqrt(30))`

`15.7 +- .6581799399`

`15.04, 16.36`

The width is `16.36 - 15.04 = 1.32` , which is larger than the width of `1`

for a confidence level of `95%` from part a.