AP Stats ~ Lesson 9B: 1-proportion z-tests and power OBJECTIVES:

9STATE and CHECK the Random, 10%, and Large Counts conditions for performing a significance test about a population proportion. 9PERFORM a significance test about a population proportion. 9INTERPRET the power of a test and DESCRIBE what factors affect the power of a test. 9DESCRIBE the relationship among the probability of a Type I error (significance level), the probability of a Type II error, and the power of a test.

Recall our basketball player who claimed to be an 80% free-throw shooter. In an SRS of 50 free-throws, he made 32. His sample proportion of made shots, 32/50 = 0.64, is much lower than what he claimed. Does it provide convincing evidence against his claim?

To find out, we must perform a significance test of

H0: p = 0.80 Ha: p < 0.80

where p = the actual proportion of free throws the shooter makes in the long run.

In Chapter 8, we introduced three conditions that should be met before we construct a confidence interval for an unknown population proportion. These same three conditions must be verified before carrying out a significance test.

Look familiar? Be a little bit careful, here. The Large Counts condition is slightly different. With confidence intervals, we used our estimated proportion, but with Significance Tests, we will always use the population proportion (mostly because we know it). Use the same value you use in the hypothesis statements.

If the null hypothesis H0 : p = 0.80 is true, then the player’s sample proportion of made free throws in an SRS of 50 shots would vary according to an approximately Normal sampling distribution with mean

ˆ p = p = 0.80 and standard deviation ˆ p =p(1 p)

n=(0.8)(0.2)50

= 0.0566

A significance test uses sample data to measure the strength of evidence against H0. Here are some principles that apply to most tests: • The test compares a statistic calculated from sample data with the value of the parameter stated by the null hypothesis. • Values of the statistic far from the null parameter value in the direction specified by the alternative hypothesis give evidence against H0.

A test statistic measures how far a sample statistic diverges from what we would expect if the null hypothesis H0 were true, in standardized units. That is,

test statistic = statistic - parameterstandard deviation of statistic

The test statistic says how far the sample result is from the null parameter value, and in what direction, on a standardized scale. You can use the test statistic to find the P-value of the test. In our free-throw shooter example, the sample proportion 0.64 is pretty far below the hypothesized value H0: p = 0.80. Standardizing, we get

test statistic = statistic - parameterstandard deviation of statistic

z =0.64 0.800.0566

= 2.83

The shaded area under the curve in (a) shows the P-value. (b) shows the corresponding area on the standard Normal curve, which displays the distribution of the z test statistic. Using Table A, we find that the P-value is P(z ≤ – 2.83) = 0.0023.

So if H0 is true, and the player makes 80% of his free throws in the long run, there’s only about a 2-in-1000 chance that the player would make as few as 32 of 50 shots.

To perform a significance test, we state hypotheses, check conditions, calculate a test statistic and P-value, and draw a conclusion in the context of the problem. The four-step process is ideal for organizing our work. It also correlates directly to the grading rubric on the AP Test free–response questions.

When the conditions are met—Random, 10%, and Large Counts,

the sampling distribution of ˆ p is approximately Normal with mean

ˆ p = p and standard deviation ˆ p =p(1 p)

n.

The z statistic has approximately the standard Normal distribution when H0 is true. P-values therefore come from the standard Normal distribution.

Example: On TV shows like American Idol, contestants often wonder if there is an advantage to performing last. To investigate, researchers selected a random sample of 600 college students and showed each student the audition video of 12 different singers. For each student, the videos were shown in random order. So we would expect approximately 1/12 of the students to prefer the last singer they view, assuming the order doesn’t matter. In this study, 59 of the 600 students preferred the last singer they viewed. Do these data provide convincing evidence at the 5% significance level that there is an advantage to going last?

Work for our example: State: Plan: DO: Conclude:

Something to think about. Sometimes the data doesn’t support our alternate hypothesis. For example, we just tested the hypothesis that more than 1/12 or .0833 of the subjects choose the last performer viewed. What if our data gave us a proportion of .07? It isn’t greater than .0833, so there obviously won’t be evidence showing that there is an advantage to going last. We could stop before doing any other work.

TWO-SIDED TESTS: The P-value in a one-sided test is the area in one tail of a standard Normal distribution—the tail specified by Ha. In a two-sided test, the alternative hypothesis has the form

Ha : p ≠p0.

The P-value in such a test is the probability of getting a sample proportion as far as or farther from p0 in either direction than the observed value of p-hat. As a result, you have to find the area in both tails of a standard Normal distribution to get the P-value.

Example: When the accounting firm AJL and Associates audits a company’s financial records for fraud, it often uses a test based on Benford’s Law. Benford’s law states that the distribution of first digits in many real-life sources of data is not uniform. In fact, when there is no fraud, about 30.1% of the numbers in financial records begin with the digit 1. If the proportion of first digits that are 1 is significantly different from .301 in a random sample of records, AJL and Associates does a much more thorough investigation of the company. Suppose that a random sample of 300 expenses from a company’s financial records results in only 68 expenses that begin with the digit 1. Should AJL and Associates do a more thorough investigation of the company? Justify your answer.

Work for our example: State: Plan: DO: Conclude:

Isn’t this similar to what we did with confidence intervals? In some ways, yes. A confidence interval gives us an interval where the true proportion likely is. If we did a confidence interval and the null hypothesis value is not in that interval, we consider that statistically significant. This means that you have the option of using a confidence interval rather than a significance test whenever you are looking at a two-sided test. You need to approach the problem a little bit differently, but it is an acceptable alternative. However, if you are looking at a one-sided test, you must do a significance test.

Type II Errors and the POWER of a Test A significance test makes a Type II error when it fails to reject a null

hypothesis H0 that really is false.

There are many values of the parameter that make the alternative hypothesis Ha true, so we concentrate on one value.

The probability of making a Type II error depends on several factors, including the actual value of the parameter. A high probability of Type II error for a specific alternative parameter value means that the test is not sensitive enough to usually detect that alternative.

The significance level of a test is the probability of reaching the wrong conclusion when the null hypothesis is true. The power of a test to detect a specific alternative is the probability of reaching the right conclusion when that alternative is true. We can just as easily describe the test by giving the probability of making a Type II error (sometimes called β).

Here’s the basic idea for finding the power of a test.

How can we increase the power of a significance test? 9Increase the sample size. Increasing the sample size reduces the variability of both the null and the alternate hypothesis distribution, so the overlap is minimal.

9Increase the significance level α. Using a larger significance level will reduce the probability of a type II error, and that will, in turn, increase the power of the test.

9Increase the difference between the null and alternative parameter values. It is easier to detect large differences between the parameters than it is to detect small differences, so it is easier to avoid making a mistake.

Homework:

Page 570: #31-35, 37-57 odds, 59-62 Read pp. 574-595