Chapter Seven Introduction to Sampling Distributions Section 3
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Transcript of Chapter Seven Introduction to Sampling Distributions Section 3
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Chapter SevenIntroduction to Sampling
DistributionsSection 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 of the Proportion n
rp ˆ
• 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 ˆnrp
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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:
npqpp p̂ˆ and
nrp ˆ
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The Standard Error for p̂
npq
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 = 0.6256/8 = 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 between 10 % 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
npq
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
npqthen
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
Control limits are placed at two and three standard deviations above and
below the mean.
138.012.050
)88.0(12.0312.03
092.012.050
)88.0(12.0212.02
nqpp
nqpp
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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 ProportionsEmployee 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
• We use the z formula:
• A Calculator is used to compute
• The result rounds to z= -3.16
x
xxz
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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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