1 Chapter Seven Introduction to Sampling Distributions Section 3 Sampling Distributions for Proportions.

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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 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 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

n

rp ˆ

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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 = 0.6256/8 = 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

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

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

n

qpp

n

qpp

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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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