Confidence Interval for the difference of two means would be used to compare two population means using samples from each. http://www.stat.yale.edu/Courses/1997-98/101/meancomp.htm

Confidence & Prediction Intervals

http://www.cliffsnotes.com/math/statistics/bivariate-relationships/simple-linear-regression (Find at the end)

## Confidence Interval Questions with Z Scores.

If z_{c} is the z value associated with c% then a c% confidence interval for the mean is:

(where X is the actual mean)

Z score = (X-XBar)/ (Standard deviation of sample / Sqrt(# samples))

1.96 = (X-4.32) / (0.17/ Sqrt(25)

Thus, X – 4.32 = 1.96*(0.17/ Sqrt(25) = 0.06664

We say that +/- 0.06664 is the margin of error.

The 95% confidence interval for the mean is 4.25 -> 4.39

### Example

http://www.ltcconline.net/greenl/courses/201/estimation/confintmean.htm

X# random samples.

XBar = Mean value.

S# Standard Deviation

95% confidence for the actual mean. AKA Confidence level.

### Description:

Usually, we do not know the population mean and standard deviation. Our goal is to estimate these numbers. The standard way to accomplish this is to use the sample mean and standard deviation as a best guess for the true population mean and standard deviation. We call this “best guess” a *point estimate*.

We are not only interested in finding the point estimate for the mean, but also determining how accurate the point estimate is. The Central Limit Theorem plays a key role here. We assume that the sample standard deviation is close to the population standard deviation (which will almost always be true for large samples).

### Solution:

We can use x to provide a point estimate for m. How accurate is x as a point estimate? We construct a *95% confidence interval for **m** *as follows. We draw the picture and realize that we need to use the table to find the z-score associated to the probability of .025 (there is .025 to the left and .025 to the right).

We arrive at z = -1.96. Now we solve for x:

Z score = (X-XBar)/ (Standard deviation of sample / Sqrt(# samples))

x – 14 x – 14

-1.96 = =

2/ 0.28

Hence

x – 14 = -0.55

We say that 0.55 is the *margin of error*.

We have that a 95% confidence interval for the mean clarity is

(13.45,14.55)

In other words there is a 95% chance that the mean clarity is between 13.45 and 14.55.

In general if z_{c} is the z value associated with c% then a c% confidence interval for the mean is

## Example 2

Try applying Normal Z transform.

Z 1= Xi – Xbar / Std Dev = 175 – 194 / 11.2 = -1.7 : Z1 score is 0.0446

Z 2= Xi – Xbar / Std Dev = 225 – 194 / 11.2 = 2.77 : Z2 score is 0.9972

Now subtract Z1 from Z2 : 0.9972 – 0.0446 = 0.9526 : This is 95.26%

## Also See:

How to Calculate a Sample Size Given Standard Deviation, Confidence Interval and Margin of Error.

## Other Confidence Interval Problems

http://www.ltcconline.net/greenl/courses/201/estimation/homeworkHandout/handout.htm

## ASQ Six Sigma Black Belt Exam Confidence Interval Questions

**Question: **Which of the following describes the 95% confidence interval of a 20% absentee rate in a department with 30 people?

(A) 6% to 34%

(B) 8% to 32%

(C) 13% to 27%

(D) 17% to 23%

**Answer:** A 6% to 34%.

This is a confidence intervals for proportion question. p + or – Z (α/2) * SQRT( (p*(1-p))/n )

p = 0.2

α = 5% (Use this to look up the Z Score on the Z table.)

n = 30

p + or – Z (α/2) * SQRT( (p*(1-p))/n )

0.2 + or – Z (5%) * SQRT( (0.2*(1-0.2))/30 )

0.2 + or -1.96 * SQRT( 0.0053)

0.2 + or – 0.1431

0.2 + 0.1431 = 0.3431 => 34.31% => Round down to 34%

0.2 – 0.1431 = 0.05686 => 5.68% => Round up to 5%

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