TC3 → Stan Brown → Statistics → Ch 8 Guide
revised 5 Nov 2011 (What’s New?)

Guide to Chapter 8

Copyright © 2008–2012 by Stan Brown, Oak Road Systems

This is your guide to what’s important in the chapter, with comments on some things that the chapter leaves out or doesn’t explain well. Page numbers refer to Sullivan, Michael, Fundamentals of Statistics 3/e (Pearson Prentice Hall, 2011), which is equivalent to the “second custom edition” for TC3.

Always check Corrections to Sullivan’s Fundamentals of Statistics, 3rd Edition for known mistakes in the textbook. Caution: your calculator is more accurate than the book because the book rounds values and uses tables. When the book disagrees with your answer, check the errata.

Overview:  Last week we looked at probabilities of finding individuals in various regions within a ND. This week we shift focus to samples and ask how likely it is that we would find a sample mean or proportion within a given region in the ND. We know (Chapter 1) that samples vary; now we get a handle on just how much they vary.

Why do this? Because if we know how sample means (or proportions) behave for a known population, we can use the mean (or proportion) of one sample to make estimates about an unknown population. And when our sample differs from a baseline, we can say whether that represents a real effect or could be completely explained by chance.

For chapter summary, see Sampling Distributions — Summary.

Handout:  Instead of all the calculations and table lookups in your book, use section 2 of Normal Calculations on TI-83/84 or TI-89.

8.1  Distribution of the Sample Mean

377The definitions are pretty abstract, but you can understand them. Don’t just blow past them — plan to read this page more than once till you really understand a sampling distribution of the mean.

The mean of one sample is different from the mean of another, so is a random variable.

LAB (for larger classes): Central Limit Theorem Lab II (Roulette)

378Describe the distribution” in all problems means to give its shape, center, and spread.

378–9This example asks, if I took a lot of simple random samples of size n = 9 from a ND with mean μ = 100 and standard deviation σ = 15, what would the distribution of those sample means look like? The results are at the bottom of page 379: shape still ND, mean of sample means about the same, but s.d. of sample means less than s.d. of population. In other words, sample means stick closer to the mean than individual members of the population do.

380 Spend some time with the top two paragraphs. They help you understand why sample means behave the way they do.

380 Look at the histogram in Example 2 (n=25) compared to Example 1 (n=9). You have to look carefully at the scale of the x axis to see the difference.

381 The box at the top has a lot of information. You need it all, including symbols, formulas, and vocabulary.

Notice that it says nothing about the shape of the original distribution. The stuff in the box is true for samples drawn from any population.

382 Figure 4 plots the original population and sampling distribution of more or less on the same scale. Notice that the sampling distribution has a higher peak and skinnier tails. That means you’ll see most of the sample means quite close to the center, and a sample mean far from the population mean is rare. By contrast, an individual the same distance from the population mean is not rare at all.

382 Do example 3 on calculator, using section 2 of Normal Calculations on TI-83/84 or TI-89. The general procedure is below.

Pay attention to notation:  axis versus x axis, P(>110) versus P(x>110).

How to work Chapter 8 problems:

  1. Determine shape, center, and spread of the sampling distribution.

    (Ex. 3: σ = σ/√n = 15/√10 = 4.7)

  2. Sketch the curve. (See below.)
  3. TI-83 screen showing wrong and correct computations from the text Compute the probability using section 2 of Normal Calculations on TI-83/84 or TI-89. Use the unrounded SEM to compute probability to four decimal places:

    WRONG (shaded):

    normalcdf(110,10^99,100,4.7) = 0.0167

    CORRECT:

    normalcdf(110,10^99,100,15/√(10)) = 0.0175

  4. Show the numerical probability on the sketch.

sketched normal distribution for page 382 Example 3 Important points for your sketches:

  1. Draw the axis line.
  2. Label the axis, or as appropriate.
  3. Draw a vertical line in the middle of the distribution and show the numerical value of the mean.
  4. Draw a horizontal line at about the right spot and show the numerical value of the SEM or SEP, not σ of the original population.
  5. Draw a line and show the value for each boundary.
  6. Shade the area you’re trying to find.
  7. (optional) After you find the area, show its value.

TI-83 screen showing variable A as described in the text optional: To avoid keying in the formula for standard error twice: The first time you key in the standard error in a given problem, save it in a variable, for instance A. Then, each time you need that same value, use A instead.

To store a calculation into variable A, after the calculation press [STO→] and then [ALPHA MATH makes A]. To use the variable in a formula, press [ALPHA MATH makes A].

383–4 Now ask, what if the underlying distribution isn’t normal? Look at the skewed population on page 383 and then at the sampling distributions for various sample sizes on page 384.

385 Know the Central Limit Theorem.

385–6 Spend time with the examples and their graphs. Remember, this stuff is abstract and just glancing at it is nowhere near enough.

387 The CLT kicks in at different sample sizes depending on the shape of the original population. How do you know when a sample is big enough? The first paragraph has an important rule of thumb: if n≥30 it’s nearly always good enough. We say this is a conservative rule, meaning that we can count on it even if in some situations we could get away with lower n.

387–388 Work Example 5 with calculator. Note correct probability in “Using Technology” sidebar here and elsewhere.

Pay attention to the interpretation!

388 Review #1–10.

Extra problem: Suppose hotel guests who take elevators weigh on average 150 pounds with standard deviation of 35 pounds. An engineer is designing a large elevator, to lift 50 people. If she designs it to lift 4 tons (8000 pounds), what is the chance a random group of 50 people will overload it?   Answer: 0.0217

Need a hint? This is a problem in sample total. We haven’t studied that kind of problem, but we have studied problems in sample means. In math, when you have an unfamiliar type of problem, it’s always good to ask: Can I change this into some type of problem I do know how to solve? In this case, how do you change a problem about the total number of pounds in a sample (∑x) into a problem about the average number of pounds per person ()?

MOVIE: See Viewer’s Guide to Against All Odds Part 18, “The Sample Mean and Control Charts”. (Based on student feedback, this won’t be shown in class. But if you’ve missed class you may want to print the guide, then go to the media center and view the movie.)

8.2  Distribution of the Sample Proportion

392 definition: sample proportion  = x/n

is a random variable for samples from binomial data, just as the sample mean is a random variable for samples from numeric data.

394–5 sampling distribution of the proportion for p=0.82, n=various

396 characteristics of sampling distribution of

Note similarity to characteristics of sampling distribution of the mean.

For binomial data, the rule of thumb is  np(1−p)≥10, not n≥30.

The s.d. of the sampling distribution of the proportion is called the standard error of the proportion or SEP; symbol σ. Know symbol and formula.

396Note additional requirement in the text below the purple box: sample size must be ≤ 5% of population size: n≤0.05N. This is problematic if you don’t know the size of the population. So multiply both sides by 20 and get 20n ≤ N or N ≥ 20n. Even when you don’t know N, you can usually know if it’s greater than 20 times sample size.

Interesting factoid (which you don’t have to know): the standard deviation of a binomial population is σ = √(p(1−p)). Therefore the SEP is σ/√n, just like the SEM.

396 TI-83/84 calculations as described in text Work Example 3.

Don’t just assert that 20n is ≤ N; compute and show the value of 20n. Don’t just assert that np(1−p) is ≤ 10; compute and show the value.

The SEP is an ugly calculation. If you like, you can use the “save in a variable technique” so that you have to type it only once.

397–8 Example 4 — work on calculator.

Always use unrounded numbers. 
WRONG: normalcdf(.217 ... 
CORRECT: normalcdf(26/120 ...

398 Review #1–6

What’s New

5 Nov 2011: Show an example of sketching the sampling distribution, with commentary.

24–25 Oct 2011: Add to the Why do this? paragraph; add clarifications here, here, and here.

29–30 Mar 2011: Add a note on page 377 that is a random variable, a hint to the elevator problem, and a disclaimer about the video.

20 Mar 2011:

10 Jan 2011: Start over for the third edition of the textbook.

(intervening changes suppressed)

12 Mar 2008: new document (MS-Word format)


This page is used in instruction at Tompkins Cortland Community College in Dryden, New York; it’s not an official statement of the College. Please visit www.tc3.edu/instruct/sbrown/ to report errors or ask to copy it.

For updates and new info, go to http://www.tc3.edu/instruct/sbrown/stat/