Module 10: Inference for Means
Distribution of Sample Means (4 of 4)
Distribution of Sample Means (4 of 4)
Learning outcomes
- Estimate the probability of an event using a normal model of the sampling distribution.
Let’s compare what we have learned about sampling distributions for proportions and for means.
| Sampling Distribution | |||||
|---|---|---|---|---|---|
| Variable | Parameter | Statistic | Center | Spread | Shape |
| Categorical (example: left-handed or not) | p = population proportion | [latex]\hat{p}[/latex]= sample proportion | p | [latex]\sqrt{\frac{p(1-p)}{n}}[/latex] | Normal when np ≥ 10 and n(1 – p) ≥ 10 |
| Quantitative (example: age) | μ = population mean, σ = population standard deviation | [latex]\bar{x}[/latex]= sample mean | μ | [latex]\frac{\sigma}{\sqrt{n}}[/latex] | When will the distribution of sample means be approximately normal? |
Now we know the conditions that allow us to use a normal model for the sampling distribution of means. As we have done before, we now convert sample means to z-scores and use a standard normal curve to find probabilities and identify unusual sample means.
Normal Model Simulation Useful Again
Recall the standard normal model simulation we first used in Probability and Probability Distribution. It was our tool for converting between intervals of z-scores and probabilities.
Click here to open this simulation in its own window.
Example
Surprising Heights for Individual Basketball Players
Suppose we have a population of adult male basketball players and we know their heights: the mean height is μ = 190 cm and the standard deviation of their heights is σ = 7.2 cm. The heights are normally distributed, which is often the case with body measurements.
Would it be surprising to find a randomly chosen player from this population with a height of 195 cm?
We can answer this question by computing the probability that a randomly chosen player from this population has height greater than 195 cm. To carry out the analysis, let’s use X to denote the height of a randomly chosen individual from this population. Since heights are normally distributed, we can convert heights to z-scores and use our simulation to find the probability P(X > 195).
- Convert the interval X > 195 to an interval of z-scores.Recall that the z-score of an X-value is the number of standard deviations that value is away from the mean. The formula is
[latex]Z = \frac{X-\mu}{\sigma}[/latex]
So the z-score of X = 195 is
[latex]\frac{X-\mu}{\sigma}= \frac{195-190}{7.2}=\frac{5}{7.2}= 0.69[/latex]
That means that the interval of X-values “X > 195” corresponds to the interval of Z-values “Z > 0.69.”
2. Convert the interval Z > 0.69 to a probability statement. We use the simulation (or some sort of technology) for this step. Below is a picture of the simulation with the settings for this problem. We moved one flag out of the way and the other flag to the position Z = 0.69. For a “greater than” probability, we want the area to the right of Z =