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.