Transcript Sampling Distributions for Proportions

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 of the
r
Proportion pˆ  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
Sampling Distribution of the
Proportion pˆ  r
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
The Standard Error for
p̂
The standard deviation of
the p̂ sampling distributi on 
 p̂ 
pq
n
Continuity Correction
• When using the normal distribution (which
is continuous) to approximate p-hat, 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.
Continuity Correction
If n = 20, convert a
p-hat interval from
5/8 to 6/8 to a
normal interval.
Note: 5/8 = 0.625
6/8 = 0.75
So p-hat interval is
0.625 to 0.75.
• Since n = 20,
.5/n = 0.025
• 5/8 - 0.025 = 0.6
• 6/8 + 0.025 = 0.775
• Required x interval is
0.6 to 0.775
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?
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.
Find the mean and the standard
deviation
 pˆ  p  0.12
 pˆ 
pq
.12(.88)

 0.023
n
200
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.
Calculate z-score for x = 0.0975
0.0975  0.12
z
 0.98
0.023
Calculate z-score for x =
0.1525
0.1525  0.12
z
 1.41
0.023
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.
Control Chart for Proportions
P-Chart
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.
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.
Finding the mean and the
standard deviation
Suppose  pˆ  p  0.12
then  pˆ 
pq
.12(.88)

 0.046
n
50
Is it appropriate to use the
normal distribution?
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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.
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
Control Limits
The center line is at 0.12.
Control limits are placed at -0.018, 0.028,
0.212, and 0.258.
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
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