INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE COURSE: JUST 3900 INTRODUCTORY STATISTICS FOR CRIMINAL JUSTICE Chapter 8: Hypothesis Testing Peer Tutor Slides Instructor: Mr. Ethan W. Cooper, Lead Tutor © 2013 - - PLEASE DO NOT CITE, QUOTE, OR REPRODUCE WITHOUT THE WRITTEN PERMISSION OF THE AUTHOR. FOR PERMISSION OR QUESTIONS, PLEASE EMAIL MR. COOPER AT THE FOLLWING: coopere07@students.ecu.edu

Key Terms: Don’t Forget Notecards Hypothesis Test (p. 233) Null Hypothesis (p. 236) Alternative Hypothesis (p. 236) Alpha Level (level of significance) (pp. 238 & 245) Critical Region (p. 238) Type I Error (p. 244) Type II Error (p. 245) Statistically Significant (p. 251) Directional (one-tailed) Hypothesis Test (p. 256) Effect Size (p. 262) Power (p. 265)

Formulas Standard Error of M: 𝜎 𝑀 = 𝜎 𝑛 = 𝜎 2 𝑛 = 𝜎 2 𝑛 z-Score Formula: 𝑧= 𝑀−𝜇 𝜎 𝑀 Cohen’s d: 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝜇 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝜎 estimated Cohen’s d: 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑀 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝜎

Logic of Hypothesis Testing Question 1: The city school district is considering increasing class size in the elementary schools. However, some members of the school board are concerned that larger classes may have a negative effect on student learning. In words, what would the null hypothesis say about the effect of class size on student learning?

Logic of Hypothesis Testing Question 1 Answers: For a two-tailed test: The null hypothesis would say that class size has no effect on student learning. The alternative hypothesis would say that class size does have an effect on student learning. For a one-tailed test: The null hypothesis would say that class size does not have a negative effect on student learning. The alternative hypothesis would say that class size has a negative effect on student learning.

Alpha Level and the Critical Region Question 2: If the alpha level is decreased from α = 0.01 to α = 0.001, then the boundaries for the critical region move farther away from the center of the distribution. (True or False?)

Alpha Level and the Critical Region Question 2 Answer: True. A smaller alpha level means that the boundaries for the critical region move further away from the center of the distribution.

Possible Outcomes of a Hypothesis Test Question 3: Define Type 1 and Type II Error.

Possible Outcomes of a Hypothesis Test Question 3 Answer: Type I error is rejecting a true null hypothesis – that is, saying that treatment has an effect when, in fact, it doesn’t. Type I error = false (+) = Alpha (α) = level of significance Type II error is the failure to reject a null hypothesis. In terms of a research study, a Type II error occurs when a study fails to detect a treatment that really exists. Type II error = false (-) = beta error = (β) A Type II error is likely to occur when a treatment effect is very small.

Two-Tailed Hypothesis Test Question 4: After years of teaching driver’s education, an instructor knows that students hit an average of µ = 10.5 orange cones while driving the obstacle course in their final exam. The distribution of run-over cones is approximately normal with a standard deviation of σ = 4.8. To test a theory about text messaging and driving, the instructor recruits a sample of n = 16 student drivers to attempt the obstacle course while sending a text message. The individuals in this sample hit an average of M = 15.9 cones. Do the data indicate that texting has a significant effect on driving? Test with α = 0.01.

Two-Tailed Hypothesis Test Question 4 Answer: Step 1: State hypotheses H0: Texting has no effect on driving. (µ = 10.5) H1: Texting has an effect on driving. (µ ≠ 10.5) Step 2: Set Criteria for Decision (α = 0.01) z = ± 2.58 Reject H0 Reject H0 z = - 2.58 z = 2.58

Two-Tailed Hypothesis Test Question 4 Answer: Step 3: Compute sample statistic 𝜎 𝑀 = 𝜎 𝑛 = 4.8 16 = 4.8 4 =1.20 𝑧= 𝑀−𝜇 𝜎 𝑀 = 15.9−10.5 1.20 = 5.4 1.20 =4.50

Two-Tailed Hypothesis Test Question 4 Answer Step 4: Make a decision For a Two-tailed Test: zsample (4.50) > zcritical (2.58) Thus, we reject the null and note that texting has a significant effect on driving. If -2.58 < zsample < 2.58, fail to reject H0 If zsample ≤ -2.58 or zsample ≥ 2.58, reject H0

Factors that Influence a Hypothesis Test Question 5: If other factors are held constant, increasing the size of a sample increases the likelihood of rejecting the null hypothesis. (True or False?)

Factors that Influence a Hypothesis Test Question 5 Answer: True. A larger sample produces a smaller standard error, which leads to a larger z-score. For 𝑧= 𝑀−𝜇 𝜎 𝑀 , where 𝜎 𝑀 = 𝜎 𝑛 , as sample size (n) increases, standard error ( 𝜎 𝑀 ) decreases, which then increases z. Consequently, as z increases so does the probability of rejecting the null hypothesis.

Factors that Influence a Hypothesis Test Question 6: If other factors remain constant, are you more likely to reject the null hypothesis with a standard deviation of σ = 2 or σ = 10?

Factors that Influence a Hypothesis Test Question 6 answer: σ = 2. A smaller standard deviation produces a smaller standard error, which leads to a larger z-score. Thus, increasing the probability of rejecting the null hypothesis. 𝜎 𝑀 = 𝜎 𝑛 = 10 25 = 10 5 =2 𝜎 𝑀 = 𝜎 𝑛 = 20 25 = 20 5 =4

One-tailed Hypothesis Test Question 7: A researcher is testing the hypothesis that consuming a sports drink during exercise improves endurance. A sample of n = 50 male college students is obtained and each student is given a series of three endurance tasks and asked to consume 4 ounces of the drink during each break between tasks. The overall endurance score for this sample is M = 53. For the general population of male college students, without any sports drink, the scores average μ = 50 with a standard deviation of σ = 10. Can the researcher conclude that endurance scores with the sports drink are significantly higher than score without the drink? (Use a one-tailed test, α = 0.05)

One-tailed Hypothesis Test Question 7 Answer: Step 1: State hypotheses H0: Endurance scores are not significantly higher with the sports drink. (µ ≤ 50) H1: Endurance scores are significantly higher with the sports drink. (µ > 50) Step 2: Set Criteria for Decision (α = 0.05) z = 1.65 Reject H0 z = 1.65

One-tailed Hypothesis Test Question 7 Answer: Step 3: Compute sample statistic 𝜎 𝑀 = 𝜎 𝑛 = 10 50 = 10 7.07 =1.41 𝑧= 𝑀−𝜇 𝜎 𝑀 = 53−50 1.41 = 3 1.41 =2.13

One-tailed Hypothesis Test Question 7 Answer: Step 4: Make a decision For a One-tailed Test: zsample (2.13) > zcritical (1.65) Thus, we reject the null and note that the sports drink does raise endurance scores. If zsample ≤ 1.65, fail to reject H0 If zsample > 1.65, reject H0

Effect Size and Cohen’s d Question 8: A researcher selects a sample from a population with µ = 40 and σ = 8. A treatment is administered to the sample and, after treatment, the sample mean is found to be M = 47. Compute Cohen’s d to measure the size of the treatment effect.

Effect Size and Cohen’s d Question 8 Answer: estimated Cohen’s d: 𝑚𝑒𝑎𝑛 𝑑𝑖𝑓𝑓𝑒𝑟𝑒𝑛𝑐𝑒 𝑠𝑡𝑎𝑛𝑑𝑎𝑟𝑑 𝑑𝑒𝑣𝑖𝑎𝑡𝑖𝑜𝑛 = 𝑀 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 − 𝜇 𝑛𝑜 𝑡𝑟𝑒𝑎𝑡𝑚𝑒𝑛𝑡 𝜎 d = 47−40 8 = 7 8 =0.875 This is a large effect. Remember: These are thresholds. Any effect less than d = 0.2 is a trivial effect and should be treated as having no effect. Any effect between d = 0.2 and d = 0.5 is a small effect. And between d = 0.5 and d = 0.8 is a medium effect.

Computing Power Question 9: A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of µ = 100 and a standard deviation of σ = 20. The researcher expects a 10-point treatment effect and plans to use a two-tailed hypothesis test with α = 0.05. Compute the power of the test if the researcher uses a sample of n = 25 individuals.

Computing Power Question 9 Answer: Step #3: Calculate the z-score Step #1: Calculate standard error for sample 𝜎 𝑀 = 𝜎 𝑛 = 20 25 = 20 5 =4 Step #2: Locate Boundary of Critical Region 1.96 * 4 = 7.84 points Thus, the critical boundary corresponds to M = 100 + 7.84 = 107.84. Step #3: Calculate the z-score 𝑧= 𝑀−𝜇 𝜎 𝑀 = 107.84 −110 4 = −2.16 4 =−0.54 z = 1.96, for α = 0.05 Any sample mean greater than 107.84 falls in the critical region.

Computing Power Step #4: Interpret Power of the Hypothesis Test Find probability associated with a z-score > - 0.54 Look this probability up as the proportion in the body of the normal distribution (column B in your textbook) p(z > -0.54) = 0.7054 Thus, with a sample of 25 people and a 10-point treatment effect, 70.54% of the time the hypothesis test will conclude that there is a significant effect.

Computing Power Question 10: A researcher is evaluating the influence of a treatment using a sample selected from a normally distributed population with a mean of µ = 80 and a standard deviation of σ = 20. The researcher expects a 12-point treatment effect and plans to use a two-tailed hypothesis test with α = 0.05. Compute the power of the test if the researcher uses a sample of n = 25 individuals.

Computing Power Question 10 Answer: Step #3: Calculate the z-score Step #1: Calculate standard error for sample 𝜎 𝑀 = 𝜎 𝑛 = 20 25 = 20 5 =4 Step #2: Locate Boundary of Critical Region 1.96 * 4 = 7.84 points Thus, the critical boundary corresponds to M = 80 + 7.84 = 87.84. Step #3: Calculate the z-score 𝑧= 𝑀−𝜇 𝜎 𝑀 = 87.84 −92 4 = −4.16 4 =−1.04 z = 1.96, for α = 0.05 Any sample mean greater than 87.84 falls in the critical region.

Computing Power Question 10 Answer: Step #4: Interpret Power of the Hypothesis Test Find probability associated with a z-score > - 1.04 Look this probability up as the proportion in the body of the normal distribution (column B in your textbook) p(z > -1.04) = 0.8508 Thus, with a sample of 25 people and a 12-point treatment effect, 85.08% of the time the hypothesis test will conclude that there is a significant effect.

Frequently Asked Questions FAQs What is power? Power is the probability that a hypothesis test will reject the null hypothesis, if there is a treatment effect. There are 4 steps involved in finding power. Step #1: Calculate the standard error. Step #2: Locate the boundary of the critical region. Step #3: Calculate the z-score. Step #4: Find the probability. Using the example from the lecture notes, let’s go through each step. β is the probability of a type II error (false negative). Therefore, power is 1 – β.

Frequently Asked Questions FAQs The previous slide was based upon a study from your book with μ = 80, σ = 10, and a sample (n=25) that is drawn with an 8-point treatment effect (M=88). What is the power of the related statistical test for detecting the difference between the population and sample mean?

Frequently Asked Questions FAQs Step #1: Calculate standard error for sample In this step, we work from the population’s standard deviation (σ) and the sample size (n)

Frequently Asked Questions FAQs Step #2: Locate Boundary of Critical Region In this step, we find the exact boundary of the critical region Pick a critical z-score based upon alpha (α =.05)

Frequently Asked Questions FAQs Step #3: Calculate the z-score for the difference between the treated sample mean (M=83.92) for the critical region boundary and the population mean with an 8-point treatment effect (μ = 88).

Frequently Asked Questions FAQs Interpret Power of the Hypothesis Test Find probability associated with a z-score > - 2.04 Look this probability up as the proportion in the body of the normal distribution (column B in your textbook) p = .9793 Thus, with a sample of 25 people and an 8-point treatment effect, 97.93% of the time the hypothesis test will conclude that there is a significant effect.