Sampling Distribution of the Proportion
In this section, we’ll be talking about finding the probability of a sample proportion. You may remember that a sampling distribution is a distribution not of scores, but all possible sample outcomes which can be drawn given that we’re working with a specific n. In this case, the following equations can help us figure out, based on a population proportion, how likely it is that we’ll draw a sample that has a chosen proportion. A question which would require these equations may sound like the following:
“A nation-wide survey was conducted about the perception of a brand. People were asked whether they liked the brand or disliked the brand and the results showed that 65% of people liked the brand. If we were to draw a random sample of 200 people, what is the probability that 80% of people within that sample will say they like the brand?”
The population proportion, denoted as ?, is the proportion of items in the entire population with the particular characteristic that we’re interested in investigating. The sample proportion, denoted by p, is the proportion of items in the sample with the characteristic that we’re interested in investigating. The sample proportion, which by definition is a statistic, is used to estimate the population proportion, which by definition is a parameter.
The formula to calculate the Sample Proportion, p is:
p = X/n = Number of items having the characteristic of interest/Sample size
Likewise, the formula to calculate the Standard Error of the Proportion, ?p is:
?p= ?(1-?)/n
The formula to calculate Z for the Sampling Distribution of the Proportion is:
Z = p – ? / ? (1-?)/n
If these formulas are difficult to read, please make sure to go check out the Statistics Formula Glossary in a different post.
Let’s consider an example:
Data collected from customers at a local restaurant indicated that 58% of customers thought that great customer service was an important attribute in determining which restaurant they would patronage. A random sample of 1000 customers at the local restaurant was polled and the goal is to determine the probability that more than 62% of these customers indicated that customer service was an important attribute in deciding which restaurant to visit.
Recalling that the normal distribution is a useful estimation for the binomial distribution in instances where n? and n (1 – ?) are greater than or equal to 5.
n? = 1000 X 0.58 = 580 > 5
n (1 – ?) = 1000 (1 – 0.58) = 420 > 5
The calculation of the above two terms illustrates that the sample size is sufficiently large to make the assumption that the Sampling Distribution of the Proportion can be estimated by the normal distribution.
Since Z = p – ? / ? (1-?)/n, p = 0.62, ? = 0.58
Z = (0.62 – 0.58) / 0.58(1-0.58)/1000 = 2.5625
Therefore, the Probability = 1 – 0.9948 = 0.0052 = 0.52%