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.

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