Transcript Understanding Basic Statistics By Brase and Brase

Chapter Seven
Introduction to Sampling
Distributions
Section 3
Sampling Distributions for
Proportions
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Key Points 7.3
• Compute the mean and standard deviation
for the proportion p hat = r/n
• Use the normal approximation to compute
probabilities for proportions p hat = r/n
• Construct P-charts and interpret what they
tell you
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Sampling Distributions for
Proportions
Allow us to work with the
proportion of successes rather than
the actual number of successes in
binomial experiments.
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Sampling Distribution ofr the
ˆ 
p
Proportion
n
•
•
•
•
n= number of binomial trials
r = number of successes
p = probability of success on each trial
q = 1 - p = probability of failure on each
trial
r
ˆ 
 p
is read " p - hat"
n
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Sampling Distribution ofr the
pˆ 
Proportion
n
If np > 5 and nq > 5 then p-hat = r/n can be
approximated by a normal random variable (x)
with:
 pˆ  p and  p̂ 
pq
n
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The Standard Error for
p̂
The standard deviation of
the p̂ sampling distributi on 
 p̂ 
pq
n
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Continuity Correction
• When using the normal distribution
(which is continuous) to approximate phat, a discrete distribution, always use
the continuity correction.
• Add or subtract 0.5/n to the endpoints of
a (discrete) p-hat interval to convert it to
a (continuous) normal interval.
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Continuity Correction
If n = 20, convert a p- • Since n = 20,
hat interval from 5/8
.5/n = 0.025
to 6/8 to a normal
interval.
• 5/8 - 0.025 = 0.6
• 6/8 + 0.025 = 0.775
Note: 5/8 = 0.625
6/8 = 0.75
• Required x interval is
So p-hat interval is 0.625
0.6 to 0.775
to 0.75.
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Suppose 12% of the population
is in favor of a new park.
• Two hundred citizen are surveyed.
• What is the probability that between
10 % and 15% of them will be in favor of
the new park?
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Is it appropriate to the normal
distribution?
• 12% of the population is in favor of a
new park.
p = 0.12, q= 0.88
• Two hundred citizen are surveyed.
n = 200
• Both np and nq are greater than five.
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Find the mean and the
standard deviation
 pˆ  p  0.12
 pˆ 
pq
.12(.88)

 0.023
n
200
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What is the probability that
between 10 % and 15%of them
will be in favor of the new park?
• Use the continuity correction
• Since n = 200, .5/n = .0025
• The interval for p-hat (0.10 to 0.15)
converts to 0.0975 to 0.1525.
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Calculate z-score for x = 0.0975
0.0975  0.12
z
 0.98
0.023
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Calculate z-score for x = 0.1525
0.1525  0.12
z
 1.41
0.023
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P(-0.98 < z < 1.41)
0.9207 -- 0.1635 = 0.7572
There is about a 75.7% chance that
between 10% and 15% of the
citizens surveyed will be in favor of
the park.
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Control Chart for Proportions
P-Chart
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Constructing a P-Chart
• Select samples of fixed size n at regular
intervals.
• Count the number of successes r from the
n trials.
• Use the normal approximation for r/n to
plot control limits.
• Interpret results.
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Determining Control Limits for
a P-Chart
• Suppose employee absences are to be
plotted.
• In a daily sample of 50 employees, the
number of employees absent is recorded.
• p/n for each day = number absent/50.For
the random variable p-hat = p/n, we can
find the mean and the standard
deviation.
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Finding the mean and the
standard deviation
Suppose  pˆ  p  0.12
then  pˆ 
pq
.12(.88)

 0.046
n
50
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Is it appropriate to use the
normal distribution?
•
•
•
•
•
The mean of p-hat = p = 0.12
The value of n = 50.
The value of q = 1 - p = 0.88.
Both np and nq are greater than five.
The normal distribution will be a good
approximation of the p-hat distribution.
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Control Limits
Control limits are placed at two and three
standard deviations above and below the
mean.
pq
0.12(0.88)
p2
 0.12  2
 0.12  0.092
n
50
pq
0.12(0.88)
p3
 0.12  3
 0.12  0.138
n
50
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Control Limits
The center line is at 0.12.
Control limits are placed at -0.018, 0.028,
0.212, and 0.258.
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Control Chart for Proportions
Employee Absences
0.3
+3s = 0.258
0.2
+2s = 0.212
0.1
mean = 0.12
0.0
-2s = 0.028
-0.1
-3s = -0.018
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Daily absences can now be
plotted and evaluated.
Employee Absences
0.3
+3s = 0.258
0.2
+2s = 0.212
0.1
mean = 0.12
0.0
-2s = 0.028
-0.1
-3s = -0.018
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Calculator – Chapter 7
• In this chapter use the TI-83 or TI-84
Plus graphing calculator to do any
computations with formulas from the
chapter. For example, computing the z
score corresponding to a raw score from
an x bar distribution.
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Calculator – Chapter 7
•
•
Example: If a random sample of size 40 is taken from a distribution with
mean = 10 and standard deviation = 2, find the z score corresponding to
x=9
x  x
z
We use the z formula:
•
A Calculator is used to compute
•
The result rounds to z= -3.16
x
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• Statistics are like bikinis. What they reveal
is suggestive, but what they conceal is
vital. ~Aaron Levenstein
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