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Transcript
```Wed Oct 6

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
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Confidence interval for  versus point estimate for 
What is Z2
5.1 Introduction

–
Objective of statistics is to make inferences about a population based on
information contained in a sample. Parameters are certain descriptive
measure of the population, e.g. , .
E.g. Population (1,2,3,4,5)  = 3.

Sample of size 2
1,1 y =1
1,2 y =1.5
…
3,3 y =3
5,5 y =5

Sample of size 3
1,1,1 y =1
…
…
…
5,5,5 y =5
One can use Sample mean to estimate population mean. Such is
an action of inference.
Two types of inferences:



1. Estimation : to estimate the value of the parameter
2. Hypothesis testing : to decide whether the parameter is a certain
specific value
e.g. 5.1 A researcher is interested in estimating the percentage of
registered voters in her state who have voted in at least one election
over the past 2 years.
–
–
–
a. Population of interest : registered voters
b. How to select a sample : simple random sample from a list of registered
voters.
c. Is it a problem of estimation or testing?

Estimating, parameter is “percentage of ……”
5.2 Estimation of 
Simplest: use a number to estimate .
E.g, y , such approach is called point estimate.



How good is the estimate?? Don’t know just by looking at the answer.
A better answer : give a confidence interval for .
E.g. A random sample of 36 measurements is drawn from a population
with a mean of 50 and a standard deviation of 3.
a. Describe the sampling distribution ofy .
b. Within what interval would you expect
y to lie approximately 90% of the time?
a. n = 36 > 30, by CLT,y has a normal distribution
with mean 50, standard deviation n = 336 = 0.5
b. P(? < y < ? ) = 0.9
P(c < y < d ) = 0.9
P( c 0.550  Z  d 0.550 ) = 0.9
P(-1.65 < Z < 1.65 ) = 0.9
.45 .45
-1.65
0
1.65
Thus
c  50
0.5
= -1.65
c = 50 – 1.65 * 0.5
d  50
= 1.65
0.5
d = 50 + 1.65 * 0.5
In general
P ( 1.65y <y < 1.65y ) = 0.9
Rearranging the terms inside the probability:
P(y 1.65y < <y 1.65y ) = 0.9
Even though we don’t know what is, we know that
it falls within (y 1.65y ,y 1.65y ) 90% of the
time, where  y = 
n
```
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