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Lesson 8 - 2
Distribution of the Sample
Proportion
Objectives
• Describe the sampling distribution of a sample
proportion
• Compute probabilities of a sample proportion
Vocabulary
• Sample proportion – p-hat is x / n ; where x is the
number of individuals in the sample with the
specified characteristic (x can be thought of as the
number of successes in n trials of a binomial
experiment). The sample proportion is a statistic
that estimates the population portion, p.
Conclusions regarding the
distribution of the sample proportion
• Shape: as the size of the sample, n, increases, the
shape of the distribution of the sample proportion
becomes approximately normal
• Center: the mean of the distribution of the sample
proportion equals the population proportion, p.
• Spread: standard deviation of the distribution of the
sample proportion decreases as the sample size, n,
increases
Sampling Distribution of p-hat
• For a simple random sample of size n such that n ≤
0.05N (sample size is ≤ 5% of the population size)
• The shape of the sampling distribution of p-hat is
approximately normal provided np(1 – p) ≥ 10
• The mean of the sampling distribution of p-hat is
μ p-hat = p
• The standard deviation of the sampling distribution
of p-hat is σ = √(p(1 – p)/n)
Summary of Distribution of p
x, number of individuals with specified characteristic
p-hat or p = --n, number of sample size
For a simple random sample of size n such that n ≤ 0.05N
(that is less than 5% of the population size)
• Shape of sampling distribution of p is approximately
normal,
provided np(1 – p) ≥ 10
• Mean of the sampling distribution of p is μp = p
• Standard Deviation of the sampling distribution of p is
σp =
p(1 – p)
-----------n
Example 1
Assume that 80% of the people taking aerobics classes
are female and a simple random sample of n = 100
students is taken What is the probability that at most
75% of the sample students are female?
P(p < 75%)
μp = 0.80
n = 100
σp = (0.8)(0.2)/100 = 0.04
p - μp
-0.05
0.75 – 0.8
Z = ------------- = ----------------- = ----------------σx
0.04
0.04
normalcdf(-E99,-1.25) = 0.1056
normalcdf(-E99,0.75,0.8,0.04) = 0.1056
= -1.25
a
Example 2
Assume that 80% of the people taking aerobics classes
are female and a simple random sample of n = 100
students is taken If the sample had exactly 90 female
students, would that be unusual?
P(p > 90%)
μp = 0.80
n = 100
σp = (0.8)(0.2)/100 = 0.04
p - μp
0.1
0.90 – 0.8
Z = ------------- = ----------------- = ----------------σx
0.04
0.04
normalcdf(2.5,E99) = 0.0062
a
= 2.5
less than 5% so it is unusual
normalcdf(0.9,e99,0.8,0.04) = 0.0062
Example 3
According to the National Center for Health Statistics,
15% of all Americans have hearing trouble. In a
random sample of 120 Americans, what is the
probability at least 18% have hearing trouble?
P(p > 18%)
a
μp = 0.15
n = 120
σp = (0.15)(0.85)/120 = 0.0326
p - μp
0.03
0.18 – 0.15
Z = ------------- = ----------------- = ----------------σx
0.0326
0.0326
normalcdf(0.92,E99) = 0.1788
normalcdf(0.18,E99,0.15,0.0326) = 0.1787
= 0.92
Example 4
According to the National Center for Health Statistics,
15% of all Americans have hearing trouble. Would it be
unusual if the sample above had exactly 10 having
hearing trouble?
P(x < 10)
μp = 0.15
n = 120
a
p = 10/120 = 0.083
σp = (0.15)(0.85)/120 = 0.0326
p - μp
-0.067
0.083 – 0.15
Z = ------------- = ----------------- = ----------------σx
0.0326
0.0326
normalcdf(-E99,-2.06) = 0.0197
= -2.06
which is < 5% so unusual
normalcdf(-E99,0.083,0.15,0.0326) = 0.01993
Summary and Homework
• Summary: The sample proportion, like the
sample mean, is a random variable
– If the sample size n is sufficiently large and the
population proportion p isn’t close to either 0 or 1,
then this distribution is approximately normal
– The mean of the sampling distribution is equal to
the population proportion p
– The standard deviation of the sampling
distribution is equal to p(1-p)/n
• Homework
– pg 439 – 441; 1, 2, 9, 13, 17, 19
Homework
1: p = 0.44
2: sample proportion = x / n
9: apx normal μ = 0.103 σ ≈ 0.010
13: a. apx normal μ = 0.103 σ ≈ 0.015
b. P(x ≥ 390) = 0.00383
normalcdf(0.39,E99,0.35,0.015)
c. P(x ≤ 320) = 0.02275
normalcdf(-E99,0.32,0.35,0.015)
17: a. apx normal μ = 0.26 σ ≈ 0.0196
b. P(p-hat < 0.24) = 0.15377
c. P(x ≥ 150) = 0.02063, Yes
19: a. P(p-hat ≤ 0.40) = 0.04324
b. P(p-hat ≥ 0.45) = 0.1265, No
normalcdf(-E99,0.24,0.26,0.0196)
normalcdf(0.3,E99,0.26,0.0196)
normalcdf(-E99,0.4,0.43,0.0175)
normalcdf(0.45,E99,0.43,0.0175)