Sampling Distributions for Proportions
Allow us to work with the proportion of successes rather
than the actual number of successes in binomial
experiments.
Sampling Distribution of the Proportion
• 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
hat"-p" read is ˆn
rp
Sampling Distribution of the Proportion
If np > 5 and nq > 5 then p-hat = r/n can be approximated by a normal random variable (x) with:
n
pqp
p
p̂ˆ and
The Standard Error for p̂
n
pq
p̂
ondistributi sampling p̂ the
of deviation standard The
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 between10 % and 15% of them will be in favor of the new park?
• 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.
Is it appropriate to the normal distribution?
Find the mean and the standard deviation
023.0200
)88(.12.
12.0
ˆ
ˆ
n
pq
p
p
p
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
98.0023.0
12.00975.0
z
Calculate z-score for x = 0.1525
41.1023.0
12.01525.0
z
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
046.050
)88(.12.
12.0
ˆ
ˆ
n
pqthen
pSuppose
p
p
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.
Control Limits
138.012.050
)88.0(12.0312.03
092.012.050
)88.0(12.0212.02
n
qpp
n
qpp
Control limits are placed at two and three standard deviations above and below the
mean.
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