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PSY 2110 Exam 2 Study Guide

by: Madie Ritter

PSY 2110 Exam 2 Study Guide 2110

Marketplace > Ohio University > Psychology > 2110 > PSY 2110 Exam 2 Study Guide
Madie Ritter

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About this Document

This study guide covers all of the essential information covered on exam 2.
PSY 2110 (Behavioral Statistics)
S. Tice-Alicke
Study Guide
Psychology, Statistics
50 ?




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This 5 page Study Guide was uploaded by Madie Ritter on Sunday October 2, 2016. The Study Guide belongs to 2110 at Ohio University taught by S. Tice-Alicke in Fall 2016. Since its upload, it has received 23 views. For similar materials see PSY 2110 (Behavioral Statistics) in Psychology at Ohio University.


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Date Created: 10/02/16
Weeks4, 5,6,and7 EXAM 2 STUDY GUIDE 1. PRIOR KNOWLEGDE Useful things to remember whilestudying forthis exam A. Important symbols (Greek symbolsare alwaysassociated with thepopulation and Englishletters are associated withsamples) i. N: population size ii. n: sample size iii. ????: sample mean iv. µ: population mean v. ???? : sample variance 2 vi. ???? : population variance vii. ????: sample standard deviation viii. ????: population standard deviation ix. ????: z-score x. ???? = ???? : true standard errorof themean ???? √???? xi. ???? = ???? : estimated standard errorof themean (used whensigma (????)is ???? √???? unknown) xii. ????????: degrees offreedom xiii. s : estimated standard errorof thedifferences x1−x2 2. PROBABILITY A. Independent events i. Whenone eventhasno effectontheoccurrenceof theother B. Dependent events i. Whenone event’soutcomeis relatedto the outcomeoftheother C. Multiple eventswithoutreplacement (forindependent events only) i. Whencalculatingthe probability formultiple eventswithoutreplacement, be sure to changethe numbers accordingly 1. i.e. What is the probability of drawing 3 cards out of a standard deck of 52 cards, without replacement, and have all 3 cards turn up red? 26 25 24 a. 52× 51 × 50 = .1176 i. decreasing the numbers by one each time ensures that you’re account for the drawing of a red card each time D. Mutually exclusiveevents i. Whenthe occurrenceofone eventmakes theoccurrenceof theotherimpossible 1. i.e. if it is Monday it cannot be Tuesday E. Conditional probability i. Theprobability that oneeventwill occurgiventhe occurrenceofsome other event 3. HYPOTHESIS TESTING A. Z-scorehypothesistesting withsample of x-values i. Fivestep process: 1. State the null and alternative hypotheses a. In words i. When writing the null hypothesis (H ) be sore to: 1. States there isno differencebetweenthe twogroups ii. When writing the alternative (research) hypothesis (H ) be 1 sure to: 1. Foraone-tailed test: statethat thetwo groups arenot equal 2. Foratwo-tailedtest: statethe directionofthe inequality of the groups (moreorless) b. In symbols i. The null hypothesis 1. ???? ???? ???? ???????????????????? 1(????????????????????????????) = ???? ????????????????????2(????????????????????) ii. The alternative (research) hypothesis 1. Foraone-tailed test: a. ???? 1 ???? ???????????????????? 1 ???????????????????????????? )< ???????? > ???? ???????????????????? 2(????????????????????) 2. Foratwo-tailedtest: a. ???? 1 ???? ???????????????????? 1 ???????????????????????????? )≠ ???? ???????????????????? 2 ???????????????????? ) 2. Set up criteria for making a decision about the null hypothesis (H ) o 2 a. i. Critical value (cv): marks off the region of rejection 1. Foralpha setat .01: a. One-tailed i. Therewill be1% ofthe datain the tail forthe region of rejection.Meaning youwillneed to findthe corresponding ???? to .0100 on the z-score table. ???????? b. Two-tailed i. There willbe .5%of thedata ineachtail ofthe regions of rejection.Meaning youwillneed to findthe corresponding ???? ????????to .0050 onthe z-scoretable. 2. Foralpha setat .05: a. One-tailed i. Therewill be5% ofthe datain the tail forthe region of rejection.Meaning youwillneed to findthe corresponding ???? ???????? to .0500 on the z-score table. b. Two-tailed i. Therewill be2.5% ofthe data ineachtail ofthe regions ofrejection.Meaning youwill need to findthe corresponding ???? ???????? to .0250 on the z-score table. 3. Compute the appropriate statistical test ????−µ ???? a. ???? = ???? ????ℎ???????????? ???? =???? √???? ???? 4. Make a decision about the null hypothesis a. To do this, compare the computed value in step 3 to the value from the z-score table found in step 2. If the computed value falls into the region of rejection, then you would reject the null hypothesis. If it does not fall in the region of rejection, then you would obtain the null hypothesis. 5. Conclusion 3 a. If analyzing a one tailed test, be sure to state whether there is a significant difference ornot. If analyzing a two tailed test and there is a difference, be sure to take is a step further and state the direction (more or less). B. Single sample t-test i. this methodis used whensigma (????) isunknown ????−µ ???? ii. ???? = ????ℎ???????????? ???? = ???? ???????? √???? 1. ???????? = ???? −1 iii. degrees of freedom(df) 1. the number of components in a calculation that are free to vary iv. confidenceintervals 1. Builds off of point estimation to arrive at a range of values that we are confident contains the population parameter 2. ????????.95???????? .99= ???? ± (???? ???????????? )???? i. When using .95, alpha will be equal to .05 and when using .99, alpha will be equal to .01 ii. ALWAYS use the ???? ???????? from a two-tailed test C. Errors i. Type1 1. The error of rejecting H when it is true o ii. Type2 1. The error of not rejecting H whon it is false iii. Power 1. The probability of correctly rejecting a false null hypothesis (H ) o D. Sampling distributions i. Sampling distribution ofthemean 1. Refers to the distribution of all possible random sample means when an infinite number of random samples of the same size “n” are randomly selected from one raw score population ii. Standard error 1. Just by luck of the draw, the sample value may not be equal to the population value iii. Central limit theorem 1. Defines three properties of sampling distribution of the mean a. Regardless of the shape of the raw score distribution, the sampling distribution of means is nearly a normal distribution b. The mean of the sampling distribution of means is always equal to the mean of the raw score population 4 c. There is a mathematical relationship between variability of raw scores and variability of means E. Independent measures t-test i. Atest used to determine if there isa significantdifferencebetweenthe population means oftwo randomindependent samples ????1−????2 ii. ???? = ????????1−????2 1. ???????? = ???? +1 −2 2 ′ ????1 22 2. ????ℎ???????????? ???? ???? 1???? 2????ℎ???????? ???????????? ℎ???????????? ???????????????????? ???? ???? = √ ???? + ???? 1 2 2 2 ′ ???? ???????? 3. ????ℎ???????????? ???? ????1−????2 ????ℎ???????? ???????????? ℎ???????????? ???????????????????????????? ???? ???? = √ ????1 + ????2 2 (????1−1)????1+(???? 21)???? 22 a. where ???? ???? = √( ) ???? 1???? 22 (????1−1)????1+(???? 21)???? 22 1 1 4. OR ????ℎ???????? ???????????? ℎ???????????? ???????????????????????????????? ???? = √( ′ )( + ) ????1+???? 22 ????1 ????2 2 iii. Pooledvariance(???? ) ???? 1. A weighted average of the sample variances 2 (????1−1)????1+(???? 21)???? 22 2. ???? ???? = √( ) ???? 1???? 22 iv. Confidenceinterval 1. ???????? = ???? −???? 1 2) ±(???? ???????????? ????1−????2) v. Homogeneity of variance 1. The assumption that two population variances are equal ???? 12 = ???? 22 a. If one sample variance (???? ) is no more than 4 times the other sample variance (???? ) and the sample sizes are roughly equal, then you did not violate this assumption 5


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