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Module 6: Hypothesis Testing with Sample Means

Jon Starkweather, PhD

Jon Starkweather, PhD
Research and Statistical Support

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The Research and Statistical Support (RSS) office at the University of North Texas hosts a number of ``Short Courses''.

A list of them is available at:

1. Sampling Distributions

From a score to a distribution of scores

Distribution of Means

3 rules for the Distribution of Means

  1. The mean of the distribution of means is the same as the mean of the population.
  2. The variability of the distribution of means is less than the variability of the population.
  3. The shape of the distribution of means is approximately normal if (a) each sample is of 30 or greater individuals, or (b) the distribution of the population is normal.

More on Rule 1

Sampling Error

Central Limit Theorem

More on Rule 2 (Part A)

More on Rule 2 (Part B)

More on Rule 3

Z-score for a mean?

2. Z-test with Means Example

Remember Scooby?

In Module 5, we had the example research question: Is Scooby's IQ greater than that of dogs not in cartoons?

We found that Scooby's Z-score was not more extreme than 1.64. Meaning, Scooby's IQ was not significantly greater than dogs not in cartoons.

Now, we have a sample of 3 cartoon dogs (Scooby, Pluto, & Goofy). We are expanding our research to include a sample greater than one individual, but we are pursuing the same research question. Do cartoon dogs have significantly higher IQ than other dogs (i.e., those not on cartoons)?

The Sample Data

Table 1: Raw Data
X Dog IQ
1 Scooby 123
2 Pluto 145
3 Goofy 133

\(\frac{\sum X}{n} = \frac{401}{3} = 133.67\)

Step 1

Define the populations and restate the research question as null and alternative hypotheses.

Population 1: Dogs on cartoons.
Population 2: Dogs not on cartoons.

Null Hypothesis: \(H_0: \mu_1 = \mu_2\)
Alternative Hypothesis: \(H_0: \mu_1 > \mu_2\)

Step 2

Determine the characteristics of the comparison distribution.

We then know: \(\mu_M = 100\)
Variance of the distribution of means:
\(\sigma_{M}^2 = \frac{\sigma^2}{n} = \frac{15^2}{3} = \frac{225}{3} = 75\)
Standard deviation of the distribution of means:
\(\sigma_{M} = \sqrt{\sigma_{M}^2} = \sqrt{75} = 8.67\)

Step 3

Determine the cutoff sample score on the comparison distribution at which the null should be rejected.

We know from previous experience our cutoff score or critical value is a Z-score of 1.64

Step 4

Compute or calculate your sample statistic.

\(Z = \frac{\overline{X} - \mu_M}{\sigma_M} = \frac{133.67 - 100}{8.67} = \frac{33.67}{8.67} = 3.88\)
Clearly 3.88 is a very extreme Z-score.

Step 5

Compare and make a decision about whether to reject the null hypothesis or fail to reject the null hypothesis.

\(Z_{calc} = 3.88 > 1.64 = Z_{crit}\)
We have demonstrated that our sample is more extreme than our cutoff of 5% (\(p = .05\)) and conclude that; based on our (very small) sample, dogs on cartoons have a higher avergae IQ than dogs not on cartoons.

3. Confidence Intervals

Confidence Intervals

Calculating an interval estimate

Confidence Limits

Calculating the Limits

Interpretation of Confidence Intervals

Controversy of Confidence Intervals

4. Summary of Module 6

Summary of Module 6

Module 6 covered the following topics:

Many of these topics will be revisited consistently in future modules.

This concludes Module 6

Next time Module 7.

This page was last updated on: October 8, 2010

This page was created using LATEX. This document was created in LATEX and converted to HTML using LATEX2HTML.

Return to the Short Course page by clicking the link below.

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Up: RSS Introduction to Statistics Home
jds0282 2010-10-08