Statistics
Estimates and Sample Sizes
Chapter 7, Part 2
Example Problems
Finding Margin of Error
• Question: Assume that a random sample is
used to estimate a population proportion p.
Find the margin of error E that corresponds to
the given statistics and confidence level.
n = 600, x = 120, 95% confidence
• Answer: The margin of error is found with
the formula
^ ^
E=z /2
pq
n
Finding Margin of Error
• Question: n = 600, x = 120, 95% confidence
• Formula:
^ ^
E=z /2
pq
n
x 120
 0.2
• We first need to find p = 
n 600
^
^
• therefore q = 1 – 0.2 = 0.8
Finding Margin of Error
• The z is found by 1 – 0.95 (confidence) = 0.05
and divided in half 0.05/2 = 0.025
– Therefore in the z table you look up
1 – 0.025 = 0.975 and find the z value of 1.96.
Finding Margin of Error
• Plugging everything into the formula
^ ^
E=z /2
pq
0.2  0.8
 1.96
 0.032
n
600
Finding Margin of Error with TI83/84
• In your TI-83/84 calculator:
– STAT
– TESTS
– 1-PropZInt
• Enter n 600, x 120 and confidence 0.95 as decimal
– Calculate
• You should get the interval (.16799, .23201).
– Subtract these .23201-.16799 = .06402.
– Divide this by 2 to get .06402/2 = .03201 or rounded
to the same answer as previously 0.032
Confidence Intervals
• Question: In the week before and the week after
a holiday, there were 12,000 total deaths, and
5988 of them occurred in the week before the
holiday.
– Construct a 99% confidence interval estimate of the
proportion of deaths in the week before the holiday to
the total deaths in the week before and the week
after the holiday.
– Based on the result, does there appear to be any
indication that people can temporarily postpone their
death to survive the holiday?
Confidence Intervals
• Use the same technique of finding the margin of
error and then the confidence interval is the
proportion +/- the error
𝑥
5988
𝑝= =
= 0.499
𝑛 12,000
• Then 𝑞 = 1 − 𝑝= 1 – 0.499 = 0.501
𝐸𝑟𝑟𝑜𝑟 = 1.96
0.499 ∗ 0.501
= 0.009
12,000
• Thus, the interval is
– Low: 0.499 – 0.009 = 0.490
– High: 0.499 + 0.009 = 0.508
Confidence Intervals TI83/84
• In your TI-83 calculator:
– STAT
– TESTS
– 1-PropZInt
• Enter x=5988, n=12000 and confidence as decimal 0.95
• Calculate
– You should get (.490, .508) for the interval.
Confidence Intervals
• Based on the result, does there appear to be any
indication that people can temporarily postpone
their death to survive the holiday?
– This would be no because the interval is not less
than 0.5 (might close to it) and thus there is no
proof that the number of deaths is different the
week before and the week after a holiday
– If the confidence interval only contains
proportions below 0.5 then there is evidence that
the number of deaths the week before the holiday
is different than the week after the holiday
Finding Sample Size
• Question: Use the margin of error, confidence
level, and population standard deviation to
find the minimum sample size required to
estimate an unknown population mean.
Margin of error = 0.9 inches
Confidence Level = 90%
σ = 2.8 inches
Finding Sample Size
 z  
n    /2

 E 
2
• We need the sample size formula
• We know σ = 2.8 inches and E = 0.9 inches,
so we just need zα/2 which can be found in the
lower right corner of a Positive Z values for
0.90 confidence level is 1.645.
• We plug everything into the formula.
 z  
 1.645 2.8 
2
n    /2
 
  5.118  26.19
0.9 

 E 
2
2
and we always round up one so n = 27.