Z-scores with sample mean
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Scenario: Given the distribution of all IQ scores has a mean (µ) of 100 and standard deviation (σ) of 15, what's the probability of finding a sample of 30 people with a sample mean IQ score less than 105?
| Clue | Insight |
|---|---|
| We've got a mean (µ) of 100. | We're working with means here. |
| We're being asked to find a mean from a sample of 30 people. | We'll be working with a sampling distribution, not a population distribution. |
| The population distribution for IQ scores is normal, so our sampling distribution will be too. | We can utilize z-scores, not t-scores. |
What we'll do:
- Compute z-score, given our x-value of 105
- 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!
The Standard Error (SE) formula occurs in the denominator!
x = 105
µ = 100
σ = 15
n = 30
z = (105 - 100) / [15 / √(30)]
z = (5) / [15 / (5.48)]
z = (5) / [2.74]
z = 1.83
p-value = 0.9664 or 96.64%
Answer: 96.64% is the probability of finding a sample of 30 people with a sample mean IQ score less than 105.