This document discusses sampling distributions of sample proportions. It provides examples of how to calculate the probability that a sample proportion will fall within a certain range of the true population proportion. In one example, a candidate claims 53% of students support her candidacy. The document calculates the probability that a random sample of 400 students will show less than 49% support as 5.48%. Another example calculates the probability that a candidate with actual 80% support will receive over 50% in a sample of 100 students as 77.34%, indicating a good chance of meeting the majority requirement.

1. Sampling Distribution of
Sample Proportion
Unit 5
This is the last part of Unit 5

3. Sampling Distribution of Sample Proportion
Assume that in an election race between
Candidate A and Candidate B, 0.60 of the voters
prefer Candidate A.
If a random sample of 10 voters were polled, it is
unlikely that exactly 60% of them, or 6, would
prefer Candidate A.
By chance the proportion in the sample preferring
Candidate A could easily be a little lower than 0.60
or a little higher than 0.60.

4. Sampling Distribution of Sample Proportion
The sampling distribution of p is the
distribution that would result if you
repeatedly sampled 10 voters and
determined the proportion p that favored
Candidate A. Of course it did not have to be
10! 10 was only the sample size which we
can call n and N is the total number in the
population.

5. Central Limit Theorem
According to the Central Limit
Theorem, the sampling distribution
of p-hat is approximately normal for
sufficiently large sample size, such
that np > 5 and nq > 5.

6. Calculations
The formulae for calculating these proportions are given as follows:
p = X/N and Ƹ
𝑝 = x/n
where:
◦ X is the number of elements in the population possessing the
characteristic of interest; N is the total number of elements in the
population;
◦ x is the number of elements in the sample possessing the characteristic
of interest; and n is the total number of elements in the sample.
The distribution of p is closely related to the Binomial
distribution.

7. Mean and Standard Deviation of p

8. Let us look at some examples…
On page 75 of your unit notes, there is Example 5.1:
We have five students A, B, C, D and E and we are interested
in knowing which of these took statistics in sixth form. It
turns out that A , D and E did. The others not.
Based on this information, we have the proportion of
students who took statistics in sixth form is
P = X/N = 3/5 = 0.60
If we select all possible samples of the students who took
statistics in sixth form we have:

9. Statistics in Sixth Form

10. Example 5.3: Ms. Smarthead
Imagine that Ms. Smarthead is running for
president of the Guild of Students at the St.
Augustine campus of UWI. She claims that
53% of students favour her as their choice
for the post of president. Assume that this
claim is true. What is the probability that in a
random sample of 400 students at the St.
Augustine campus, less than 49% will favour
her as their choice for president?

11. Imagine that Ms. Smarthead is running for president of the Guild of Students at
the St. Augustine campus of UWI. She claims that 53% of students favour her as
their choice for the post of president. Assume that this claim is true. What is
the probability that in a random sample of 400 students at the St. Augustine
campus, less than 49% will favour her as their choice for president?
From the information given, we know that the population proportion
p = 0.53 (53%) and therefore q = 0.47.
We shall use ෝ
𝑝 = 0.49

12. Imagine that Ms. Smarthead is running for president of the Guild of Students at
the St. Augustine campus of UWI. She claims that 53% of students favour her as
their choice for the post of president. Assume that this claim is true. What is
the probability that in a random sample of 400 students at the St. Augustine
campus, less than 49% will favour her as their choice for president?
We are looking for the probability that less than 49% will favour her as
their choice of president.
The probability we are interested in is finding P( Ƹ
𝑝 < 0.49). Therefore,
the corresponding z value is z = (0.49 – 0.53)/0.025 = -1.60.
Hence, P( ෝ
𝑝 < 0.49) = P( 𝑧 < -1.60) = 0.0548.
So there is only a small probability that less than 49% will favour her
as president.
We can conclude that the probability that in a random sample of 400
students at the St. Augustine campus, less than 49% will favour her
as their choice for president is 0.0548. So, she has a pretty good
chance of being elected president.

13. Looking at the problem another way…
Imagine that Ms. Smarthead is running for
president of the Guild of Students at the St.
Augustine campus of UWI. She claims that 80% of
students favour her as their choice for the post of
president. Assume that this claim is true. What is
the probability that in a random sample of 100
students at the St. Augustine campus, she will be
selected if the requirement is at least a simple
majority in favour will confirm her as their choice
for president?

14. Solution
This is another example of Sampling Distribution of Sample Proportion. 80% of
students favour her as their choice for the post of president so p = 0.80 and q =
0.20 with n = 100.
Std dev, σ Ƹ
𝑝
= √(0.80 * 0.20)/100 = 0.04
The z-value for (0.50 - 0.80)/0.04 = -0.75
We are looking for P( Ƹ
𝑝 > 0.50) = P(z > -0.75) = 77.34% so she has a good chance.

17. Questions