Z-scores with sample proportion

Scenario: At Crammer Nation University, 60% of students join Greek Life. What's the probability of finding a sample of 30 people with a sample proportion of less than 65%?

60% of students join Greek Life is a proportion of students who joined Greek Life.We're working with proportions here, not means.
We're being asked to consider a sample of 30 people.We'll be working with a sampling distributions, not the population distribution.
Given our sample abides by the assumptions for sample proportions, and therefore the Central Limit Theorem applies, our sampling distribution is normal.We can use z-scores instead of t-scores.

What we'll do:

  • Compute z-score, given our p-hat value of 0.65
  • Find the area under the z-distribution to the left of our z-score (seen in red below)...
  • ...which will be equal to the p-value corresponding to our z-score!
Created with Statistics Kingdom

p-hat is the sample proportion we're testing on.
p is the population proportion.
q is the "proportion of failure", a.k.a. the opposite of the population proportion
is the sample size.

Standard Error (SE) looks a little different with proportions. That's because standard deviation (σ) doesn't occur with proportions. It gets replaced by √(pq)!

We can see Standard Error (SE) still occurs in the denominator of our z-score calculation.

p-hat = 0.65
p = 0.60
q = 1 - p = 1 - 0.60 = 0.40
n = 30

z = (0.65 - 0.60) / √([0.60 * 0.40] / 30)
z = (0.05) / √([0.24] / 30)
z = (0.05) / √(0.008)
z = (0.05) / (0.089)
z = 0.56

p-value = 0.7123 or 71.23%

Answer: 71.23% is the probability of finding a sample of 30 people with a sample proportion less than 0.65.

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