Section 8.3 ~ Estimating Population Proportions

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Transcript Section 8.3 ~ Estimating Population Proportions

Section 8.3 ~
Estimating Population Proportions
Introduction to Probability and Statistics
Ms. Young
Sec. 8.3
Objective

After this section you will learn how to
estimate population proportions and compute
the associated margins of error and
confidence intervals.
Sec. 8.3
The Basics of Estimating a Population Proportion

The process for estimating a population proportion, p, with a 95%
confidence level using a sample proportion, p̂ , is the same as the
process of estimating a population mean using a sample mean (section
8.2)

The only difference is the way that the margin of error is defined:
 ˆ (1  ˆ ) 
E  2 

n



The confidence interval is written as:
ˆ  E    ˆ  E
or
ˆ  E
Sec. 8.3
Example 1
The Nielsen ratings for television use a random sample of households. A
Nielsen survey results in an estimate that a women’s World Cup soccer
game had 72.3% of the entire viewing audience. Assuming that the
sample consists of n = 5,000 randomly selected households, find the
margin of error and the 95% confidence interval for this estimate.
 ˆ (1  ˆ ) 
E  2 

n



 0.723(1  0.723) 
E  2 
 
5000


E  0.013
The 95% confidence interval is:
0.723 – 0.013 < p < 0.723 + 0.013
or
0.710 < p < 0.736
With 95% confidence, we can conclude that between 71% and 73.6% of the
entire viewing audience watched the women’s World Cup soccer game.
Sec. 8.3
Choosing Sample Size

Choosing a sample size appropriate to satisfy a desired margin of error
is found by manipulating this APPROXIMATE formula for margin of
error:
1
E
n
 E n 1

1
 n 
E
1
n
E
2
 n
1
E2
Used to approximate appropriate sample size

Note: any value equal to or larger than the value found using the
formula would be sufficient
1
n 2
E
Sec. 8.3
Example 2
You plan a survey to estimate the proportion of students on your campus
who carry a cell phone regularly. How many students should be in the
sample if you want (with 95% confidence) a margin of error of no more
than 4 percentage points?
1
n 2
E
1
 n
0.042
1
 n
 625
.0016
You should survey at least 625 students.