- Slides 1-6
- Introduction
- Effect of Sample Size on Hypothesis Testing

- Slides 7 – 11
- Statistical Significance vs. Practical Importance

- Slides 12 – 17
- Using Confidence Intervals to Conduct Hypothesis Tests

- Slides 18 – 21
- What Confidence Intervals ADD to our analyses
- Summary

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]]>- Slides 1-7: Finding P-Values

**There is an error in the transcript for SLIDE 13: It says****“We can enter 2.5 in for “x” and select P(X > x) from the list to calculate a probability of 0.01044.”**

**It should read****“We can enter 2.31 in for “x” and select P(X > x) from the list to calculate a probability of 0.01044.”**

- Examples and Summary

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]]>- Slides 1-4: Introduction to Steps and Motivating Examples

- Slides 5-12: Steps for Motivating Examples

- Slides 13-18: Final Comments

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]]>To test the first figure, let p be the proportion of email users who feel that spam has increased in their personal email. The first set of hypotheses that the student wants to test is then:

Ho: p = 0.37

Ha: p ≠ 0.37

Based upon the data collected by this student, a 95% confidence interval for p was found to be:

(0.25, 0.32).

Based on the collected data, a 95% confidence interval for p was found to be (0.08, 0.14).

For testing the second figure in the report, let p be the proportion of email users who feel that spam has increased in their work email. The second set of hypotheses that the students wants to test, is then:

Ho: p = 0.29

Ha: p ≠ 0.29

Based upon the data collected by this student, a 95% confidence interval for p was found to be:

(0.273, 0.304).

According to a study completed in 2006 by Pew Internet, 42% of all Americans had a broadband Internet connection at home. This same statistics student wanted to see if this percentage is different for students at his university.

Ho: p = 0.42

Ha: p ≠ 0.42

Based upon the data the student collected, a 95% confidence interval for p was found to be:

(0.439, 0.457).

According to the same Pew Internet study, 8% of those with broadband connections are using fixed wireless. let p be the proportion of broadband users who use fixed wireless, and consider the hypotheses:

Ho: p = 0.08

Ha: p ≠ 0.08

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]]>Ho: p = 0.087

Ha: p ≠ 0.087

Based on the collected data, a 95% confidence interval for p was found to be (0.08, 0.14).

The UCLA Internet Report (February 2003) estimated that roughly 60.5% of U.S. adults use the Internet at work for personal use. A follow-up study was conducted in order to explore whether that figure has changed since. Let p be the proportion of U.S. adults who use the Internet at work for personal use.

Ho: p = 0.605

Ha: p ≠ 0.605

Based on the collected data, the p-value of the test was found to be 0.001.

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]]>Background:

For this activity, we will use example 1. Here is a summary of what we have found:

The results of this study—64 defective products out of 400—were statistically significant in the sense that they provided enough evidence to conclude that the repair indeed reduced the proportion of defective products from 0.20 (the proportion prior to the repair).

Even though the results—a sample proportion of defective products of 0.16—are statistically significant, it is not clear whether the results indicate that the repair was effective enough to meet the company’s needs, or, in other words, whether these results have a practical importance.

If the company expected the repair to eliminate defective products almost entirely, then even though statistically, the results indicate a significant reduction in the proportion of defective products, this reduction has very little practical importance, because the repair was not effective in achieving what it was supposed to.

To make sure you understand this important distinction between statistical significance and practical importance, we will push this a bit further.

Consider the same example, but suppose that when the company examined the 400 randomly selected products, they found that 78 of them were defective (instead of 64 in the original problem):

Consider now another variation on the same problem. Assume now that over a period of a month following the repair, the company randomly selected 20,000 products, and found that 3,900 of them were defective.

Note that the sample proportion of defective products is the same as before , 0.195, which as we established before, does not indicate any practically important reduction in the proportion of defective products.

**Summary:** This is perhaps an “extreme” example, yet it is effective in illustrating the important distinction between practical importance and statistical significance. A reduction of 0.005 (or 0.5%) in the proportion of defective products probably does not carry any practical importance, however, because of the large sample size, this reduction is statistically significant. In general, with a sufficiently large sample size you can make any result that has very little practical importance statistically significant. This suggests that when interpreting the results of a test, you should always think not only about the statistical significance of the results but also about their practical importance.

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]]>Ho: p = .40

Ha: p > .40

The results are reported to be not statistically significant, with a p-value of 0.214.

Decide whether each of the following statements is a valid conclusion or an invalid conclusion, based on the study:

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]]>**Background: **This activity is based on the results of a recent study on the safety of airplane drinking water that was conducted by the U.S. Environmental Protection Agency (EPA). A study found that out of a random sample of 316 airplanes tested, 40 had coliform bacteria in the drinking water drawn from restrooms and kitchens. As a benchmark comparison, in 2003 the EPA found that about 3.5% of the U.S. population have coliform bacteria-infected drinking water. The question of interest is whether, based on the results of this study, we can conclude that drinking water on airplanes is more contaminated than drinking water in general.

Here is an example of possible output from software for this problem. We will verify the results ourselves.

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]]>- In test I, a significance level of .05 was used, and the p-value was calculated to be 0.025.

- In test II, a significance level of .01 was used, and the p-value was calculated to be 0.025.

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