sampling distributions for proportions
DESCRIPTION
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 - PowerPoint PPT PresentationTRANSCRIPT
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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.
![Page 2: Sampling Distributions for Proportions](https://reader035.vdocument.in/reader035/viewer/2022070404/56813bd6550346895da4fd90/html5/thumbnails/2.jpg)
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
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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
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The Standard Error for p̂
n
pq
p̂
ondistributi sampling p̂ the
of deviation standard The
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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.
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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
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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 between10 % and 15% of them will be in favor of the new park?
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• 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?
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Find the mean and the standard deviation
023.0200
)88(.12.
12.0
ˆ
ˆ
n
pq
p
p
p
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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
98.0023.0
12.00975.0
z
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Calculate z-score for x = 0.1525
41.1023.0
12.01525.0
z
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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
046.050
)88(.12.
12.0
ˆ
ˆ
n
pqthen
pSuppose
p
p
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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
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.
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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