1. Interval Estimation Download this presentation.
Chapter 8
Interval Estimation
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2. I. Introduction
The previous chapter found a point estimate of the population parameter . But we know that for every sample, we will never have a point estimate that is exactly equal to the true population parameter.
Thus this chapter shows us how interval estimates of  and p can be developed to provide information about the precision of that estimate.

3. II. Interval Estimation of 
In a large sample case, n>=30.
We will show how a sampling distribution of can be used to develop an interval estimate of .

4. A. An example CJW is a mail-order sporting equipment firm that
conducts a monthly customer service survey. Their
scale is where 100 is “Excellent” service.
They find that the population =20, changes every month and  is unknown.
Most recent survey: =82, =20, n=100

5. B. Sampling Error
Question: How good is the sample estimate of the population parameter?
Sampling Error =
We don’t precisely know the sampling error because we don’t know  . But we can draw some probability conclusions about the size of the sampling error.

6. How?
The Central Limit Theorem allows us to conclude that the sampling distribution of can be approximated by a normal probability distribution.
n=100, =20 

7. C. Probability Statements
From the standard normal probability table (z-table) we know that 95% of all values are within z=1.96* of the mean.
95% of all sample x-bar values fall here.
Precision statement: “There is a 95% probability that will provide a sampling error of 3.92 or less.” 

8. D. General Lingo
: probability that the sampling error is larger than the sampling error in the precision statement.
/2: probability in each tail of the distribution
(1-): probability that a sample mean will provide a sampling error less than or equal to the sampling error in the precision statement.
Z/2 is the value of the standard normal random variable corresponding to an area of /2 in the upper tail.

9. A picture should help... In our example (1-) =.95, =.05, /2=.025.
(1- ) of all x-bar values.
/2
/2 
Precision statement: There is a (1-) probability that the value of a sample mean will provide a sampling error of z/2* or less.

10. E. Interval Estimate: large sample,  known.
z/2 is the z-value providing an area of /2 in the upper tail of the standard normal probability distribution.

11. F. Interval Estimate: large sample,  unknown.
If the population standard deviation is unknown, use the sample value.
Example: Suppose a 50-point exam is given to 36 statistics students. The mean is 39.5 and the standard deviation is 7.77.
Construct a 95% confidence interval.

12. Solution
With 95% confidence, we find a z.025= Using the rest of our information, we construct:
39.5 ± 1.96(7.77/6)
or: 39.5 ± 2.54
What does this interval tell us?