Section 9.2 February 13th, 2018

9.2 Learning Objectives �  Check the Random, Large Counts, & 10% Conditions

to perform a significance test for a population proportion

�  Perform a one-sided and two-sided significance tests for Population Proportions

�  Explain why confidence intervals provide more information than significance tests

�  Interpret the power of a test

�  Describe the relationship between the probability of a Type I error, probability of Type II error, and the power of a test

Conditions for performing a Significance Test about a Proportion

Example 1: I’m a Great Free Throw Shooter! Pg. 555

In Section 9.1, we met a virtual basketball player who claimed to make 80% of free throws. We suspected that the player might be exaggerating. In an SRS of 50 shots, the player made only 32. This player’s sample proportion of made free throws was therefore p-hat = 32/50 = 0.64. To find out if it provides convincing evidence against the player’s claim, we need to perform a significance test of:

Ho:p=0.80

Ha:p<0.80

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

EXAMPLE 1: Check the conditions for performing a significance test of the virtual basketball player’s claim.

EXAMPLE 1: Check the conditions for performing a significance test of the virtual basketball player’s claim.

�  Random: the 50 computer-generated shots were a simple random sample from the population of all possible shots that the virtual player takes.

�  10% Condition: we don’t need to check the 10% Condition because we’re not sampling without replacement from a finite population since the applet can keep on shooting free-throws

�  Large Counts Condition: Assuming Ho is true, p = 0.80. Then npo=(50(0.80)=40≥10 and n(1-po)=(50)(0.20)=10≥10 so the Large Counts Condition has been met

Significance Tests A significance test uses sample data to measure the strength of evidence against the null hypothesis HO and in favor of the alternative hypothesis Ha.

To test how far the statistic is from the parameter, standardize the statistic.

This is called a test statistic.

Test Statistic

EXAMPLE 2: In an SRS of 50 free throws, the virtual player made 32.

(a)  Calculate the test statistic. (State pO and p).

The virtual player’s sample proportion of made shots is p-hat=0.64 and po=0.80.

z = 0.64− 0.80(0.80)(0.20)

50

=−0.160.0566

= −2.83

(b) Find the P-value using Table A or technology. Show this result as an area under a standard Normal curve.

Using Table A, we find that P(z ≤ -2.83) = 0.0023. The curve below shows the area of 0.0023 to the left of z = -2.83.

(c) Interpret the p-value.

If Ho is true and the player truly makes 80% of his or her free throw shots in the long run, there is only about a 0.0023 probability that the player would make 32 or fewer shots by chance alone.

Read pg. 3 in your notes

The Four Step Process for a Significance Test about a Population Proportion

Read EXAMPLE: “One Potato, Two Potato” on page 559 or watch video on TPS5e’s website.

Let’s Try it! “Better to be Last?” On 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?

STATE

PLAN

DO

CONCLUDE

Two-Sided Tests In a two-sided test, the alternative hypothesis has the form Ha: p ≠ pO which gives you a P-value that is the probability of getting a sample proportion as far or further from pO in either direction.

As a result, you must find the area in both tails by multiplying the p-value by two. Easy!

On Thursday 2/15/17:

� Complete 9.2 Notes

� Complete 9.3 Notes packet

Read Example “Nonsmokers” on pg. 562

HW: Benford’s Law and Fraud “Benford’s Law and Fraud”

When the accounting firm AJL and Associates audits a company’s financial records for fraud, they often use 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 0.301 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 this company? Justify your answer.

STATE

PLAN

DO

CONCLUDE

Benford’s law and fraud Part 2

Problem:

(a) Find and interpret a confidence interval for the proportion of all expenses that begin with the digit 1 for the company in the previous Alternate Example.

(b) Does your interval from part (a) lead to the same conclusion as the significance test? Explain.

(a) Find and interpret a confidence interval for the proportion of all expenses that begin with the digit 1 for the company.

(b) Does your interval from part (a) lead to the same conclusion as the significance test? Explain.