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Chapter 7 Notes

by: Emma Dahlin

Chapter 7 Notes PSYCH 2220 - 0020

Emma Dahlin
GPA 3.85
Data Analysis in Psychology
Anna Yocom

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About this Document

Detailed notes from in-class and additional notes from textbook concepts.
Data Analysis in Psychology
Anna Yocom
Class Notes
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This 14 page Class Notes was uploaded by Emma Dahlin on Sunday October 11, 2015. The Class Notes belongs to PSYCH 2220 - 0020 at Ohio State University taught by Anna Yocom in Summer 2015. Since its upload, it has received 23 views. For similar materials see Data Analysis in Psychology in Psychlogy at Ohio State University.


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Date Created: 10/11/15
The 2 Table 0 Bene ts of standardization allowing fair comparisons o 2 table 0 provides of scores bt a given 2 score and mean AND in tail of distribution 0 Appendix Table B1 also a pdf on Carmen The Standardized 2 Distribution 0 These are APPROXIMATE percentages use to CHECKyour work not calculate 3 4 9ft 3 4 Wu 2 gs 1 4 are 1 4 2 0 These are the ACTUAL percentages use this table for calculations Appendix Bl meme 11 Eeeenpt from the 3 Table The etehle ht eeidee the hereehteee ett eeeree hetweeh the mean and e gtieeh eeeltue The tutt tehle iheliuttee eeeitiee E etetiettee frer39rt ltil tie eEEJIil The heeetiee eetetietiee ere het ieel suelett heeeuee ett we heee tn 1 ie eheege the eight freh t eeettiee tie heetiee The eereehteee heheeeh the theeh ehe e heeitiee e e tetjietie ie identical te the hereehtege between the men and the hegetitte eeteieh e t thet Eetettetie Remember the hermet ewrtte ie eeh39thttrie ehe ettte elwetre htirrere the etheL 3 ti etweeh Mean emit e eue eeuee 193 3355 99 3339 DD 3113 teen 35135 HIE 3461 Raw Scores 2 Scores and Percentages Step 1 Convert raw score to z score 0 Step 2 Look up area in Table 0 The table presents bt the Mean and z and in tail I I IIIIIEIEEI I IliFquotIU H39Il Hill IL TH E 3 BETH E UTIE H Hq lml mam aiililimlh E yIIrmiliI F t lld l Irenamm litquot i u Jii jiicJl all rumr EiuiL Ili39lgijuinlggij Earjuulgil ihr E Elaiij39 ll Eh Lil Ii39 It 3 il39ill TUEIL 1331 ELIZi Elm intEl Eiiquot 1131 4123 E i 12113 H313 1Er39Ei ail Elsi 1353 3amp3 H hid 3 H343 315913 35 3 5 1 3351 151 e ii 155114 EI 17321 41 15 J 31 Hi51 A WEE 33332 351 dam aEli 1 d 13345 it IRES d a i M IIDEI 33 quotW ELEEE d i 45 H345 W lira s t 11 311 51 4153 31 11Ei 113E 39W 55 dial IELill 111515 715 ERIE u dla i 39 133quot 31131 M EEE ME 354 1915 311539 61 ji i 195 flE 3211 1123 55 WHEE 333415 KW FEES 15H 53 EM 1 53913 ail7nquot 515 Ellii Fi i5 1 4155 55 E u r i 12 an 4112 ampb 21233 193 39M li 55 315 33153 23 Ra Hi d E Ei ELF 531 13 ES 3973 ENE 5 2224 239 EE a i Eb E t REEi W513 E 164 W36 51 225 it 11133 i 212 mm HM 3335 i Hindi 1131quot 311339 EA 1 Iii39i 39 le 11 3133 EE Ell 1 35 E39 33 WEEE 3135 En R n v 5 l i 31 uniSi rjmr 5 may 35 m Sketching the Normal Curve 0 The bene ts of sketching the normal curve Calculations for a Positive 2 Score 500w 3355 SAT N 430100 If Sarah scores a 470 calculate Mean to z score 470430100 4 I 1554 Percentile 5015546554 Percent above 470 3446 percent in tall from table or 100percentile rank Percent at least as extreme as 2 score 2x the amount in the tail bc it is symmetric 3446 x 2 6892 of data is at least as extreme as our 2 score Using a negative 2 Score 0 SAT N 430 100 Miguel earned a score of 380 on the SAT 0 What is his percentile score 2 380430100 O5 0 From table in tail l Percentile 3085 0 What scored higher than Miguel 10030856915 0R 501915 mean to z6915 What is the probability expressed as a of a score on SAT between 230 amp 280 3 34 2 as 14 14 7 2 3 2 l U l 2 2 230430100 20 2 280430100 15 0 go to 2 table and nd 2 and 15 no negative values in 2 table and subtract mean to z 39s o 47724332440 There is a 440 probability of a student scoring bt 230 amp 280 0 Another way you can do it subtract the in tail for each 668 228 From Percentages to 2 Scores 0 Step 1 Use the 2 table in reverse taking a percentage and converting it into a z score 0 Step 2 Convert the z score to a raw score using the formula Calculating a Score from a Percent or Percentile 130090 5000 o SATN430 100 0 Score to be in the top 7 Look in table use value closest to 7 694 and read across and nd 2 of 148 0 Use equation to nd raw score X148100 430 578 0 Your score if you are in the 77th percentile X 74 100430504 More Practice o If the population mean is 10 and the standard deviation is 2 o What is percentile rank of a score of 6 Z6102 2look up percent in tail bc percentile is everything less than that score in tail228 o What percentage of scores would be higher than a score of 6 Take previous score and subtract is from 100 100 2289772 0 Percent between 6 and 11 First nd 2 score of 11 5 and z score of 6 2 and nd the mean to 2 5 of both from table and add them together 447219156687 0 Score to be in top 10 Look at percent in tail closest to 10 and z128then solve for score X1282 10 1256 The Assumptions and the Steps of Hypothesis Testing 0 Requirements to conduct analyses Assumption characteristic about a population that we are sampling necessary for accurate inferences Parametric v Nonparametric Tests Parametric tests inferential statistical test based on assumptions about a population Nonparametric tests inferential statistical test not based on assumptions about the population Three Assumptions for Hypothesis Testing 1 DV assessed with scale measure ratio or interval 2 Participants are randomly selected 3 Population must have approximately normal distribution 0 Usually ok if sample gt 30 o What if these aren t met 0 Hypothesis tests are fairly robuststill very accurate TALE T22 The Three Aaauh ihtiaha fair Hyp theaia Tasting War rhtlat ha awaira at tha aaaurhatiaha tat tha hailathaaia taat that ha ahaaaa aria wa rhuat ha aauttatla ih al39iiaaaihg ta araaaarl witl39i a hypathaaia taat await thautiih aLlr aala titair hat rriaat all at tha EESLtlliitEfItil l ta Nata that in atltlitiari ta lhaaa lhraa aaaiin39iptiai ia far iriarigi iliyaatliaaia taata il ialhdiha lha a laat tha ihrtaaahaiaht aariahla muat ha namihal Tha thraa haaurhahiaha Eraaltirig tha aaahmatiaha l Daaahtlaht aariaala ia an a aaala thaaaura Ltaualla lit it tha aata ara lint alaaaltr iiaiainal at ardihal 2 Participants are raritl ahily aalaata tilt it we aha aautiaua alaat gaharaliaihg 3 F39 lflllll tfl llli diatrihutiaa IS Elia l itti t lltt I39I l iTI l UK If quotHIE SEIIT39IIJIE IFIE IIJE IEE ET Iii331 3t EEEUFEE TABLE 13 39Tha Elia Stacia at l lypathaaia Taatihg War Liaa tha aama aia haaia ataaa with aaah typa at hyaathaaia taat t laai ltit y tha haaulatiaria dialrihutian anti aaaulhatiaha and than altaaaa tha appraariata haaathaaia taat E Stat tha null aril raaaairch hypathaaaa in hath wants and EymEDIil hatatian 21 iatarhiina tha aharaatariatiaa at tha cantaariaan aialrihutian a iatarmina tha critiaal aalLlaa at autatta that inaliaata tha painta hatiahd which wa will raiact tha null ha aathaaia a GalaLilata tha taat atatiatia t1 I EEIIEIE ia hathar ta raiaat at tail ta rajaat tha hull hypathaaia Example of Hypothesis Testing IQ scores are designed to have a mean of 100 and a standard deviation of 15 A school psychologist is convinced that the mean IQ score of the high school seniors in her district is different from 100 She administered an IQ test to random sample of 50 seniors in her district and found their mean IQ was 104 0 Go through six steps of hypothesis testing Six Steps of Hypothesis Testing 1 Identify populations comparison distribution amp assumptionsljwhich test to use 0 Populations represented by 2 groups All HS seniors in her district All HS seniors in US Identify comparison distribution Distribution of means 0 0 Identify hypothesis test amp check assumptions One sample 2 test know population mean amp std deviation Assumptions DV scale Random selection Distributionsample size 2 State null amp research hypotheses O O O O In regard to population use words amp notation Twotailed test quotnondirectionalquot test Words NULL There is no difference bt the average IQ of HS seniors in our district and the average IQ of HS seniors in the US RESEARCH There is a difference bt the average IQ of HS seniors in our district and the average IQ of HS seniors in the US m Ho 11 12 H1 11 12 Onetailed test quotdirectional testquot Words NULL The average IQ of HS seniors in our district is not greater than the average IQ of HS seniors in the US RESEARCH the average IQ of HS seniors in our district is greater than the average IQ of HS seniors in the US m Ho 11 s 12 H1 11 gt 12 3 Determine characteristics of comparison distribution 0 0 Based on null 2 test mean amp std error of distribution I 1m 1 100 O39M 1550 15707 212 39 HM OM 4 Determine critical values cutoffs O O 0 Critical values test statistics needed to reject null Z scores needed to reject null May have one critical value onetail or two critical values both tails depending on our null hypothesis Critical region area beyond critical values tails reject the null if test statistic in this region p level alpha probability used to determine critical values typically 005 may sometimes see 001 reject most extreme 5 of distribution there is a 5 or less chance we would nd results this extreme if the null were true one taied or twotailed matters here One amp TwoTailed Tests Twotailed tests 0 Rejection region divided bt 2 tails pleve 05 put 25 in each tail how to nd in 2 table Tail 25 OR mean to 2 475 critical 2 196 AND 196 Onetailed test 0 Cutoff one tail of distribution plevel05 put all 5 in the one tail so critical 2 165 LESS conservative than twotailed Determining Critical Values for a 2 Distribution One tailed or twotailed test for signi cance V 4750 250 250 5 Calculate test statistic 0 Based on your data 0 Remember we typically test means NOT individual scores 0 Zobs M 1M 0M o 104100212189 0 what is ACTUAL p value P294 6 Make decision amp interpret result o Reject or fail to reject null 0 Compare test statistic to critical values If test statistic is more extremeljreject Statistically signi cant data differ from what we would expect by chance if there were no actual difference if null were true If test statistic is NOT more extremeljfail to reject Making a Decision 3 1 239 1 2 3 1quot lb 026 191 o Fail to reject 0 Zobs 189 lt 196 Zcrit o What is ACTUAL p value 0 p294 LOOKING FOR LESS THAN 25 in twotailed test or 5 SPSS over both tails quotas extreme asquot No evidence from this study to support the research hypothesis that there is a different in average IQ scores bt students at this district and the general population of students 0 Practice 0 A manufacturer of ashlight batteries claims that its batteries will last an average of u 34 hours of continuous use Of course there is some variability in life expectancy with o 3 hours After receiving several complaints about the batteries a consumer protection group predicts that the batteries run less than 34 hours During consumer testing a sample of n30 batteries lasted an average of only 325 hours Conduct a hypothesis test with d 05 to determine whether the batteries have an average life span that is lower than that speci ed by the manufacturer What can you conclude Onetailed test bc it is saying it is lowerthan average life span ID populations comparison distribution and assumptionswhich test to use Populations represented by 2 groups 0 1 Available consumer batteries from this manufacturersampled o 2 All ashlight batteries from this manufacturer ID comparison distribution 0 Distribution of means lD hypothesis test amp check assumptions 0 One sample 2 test know population mean amp std deviation 0 Assumptions DV scale Random selection Distributionsample size State null amp research hypotheses In regard to population use words amp notation o Onetailed test quotdirectional testquot Words 0 NULL The average consumer battery life is at least as long as the average battery life of all batteries from this manufacturer 0 RESEARCH the average consumer battery life is less than the average battery life of all batteries from the manufacturer Notation O H0IJ1 Z 12 0 H1 pl lt H2 0 OR 0 H0IJ1 Z 0 H1J1 lt 3 Determine characteristics of comparison distribution Based on null 2 test mean amp std error of distribution 0 W u 34 0 0M 330 055 0 HM 34 CM 4 Determine critical values plevel 005 onetailed 0 look up 5 in tail critical values 2 165 5 Calculate test statistic Zobs o zobs 32534055 273 6 Make decision amp interpret result Reject null zobs 273 lt 165 zcrit Average life span of consumer batteries is less than 34 hours According to data consumer batteries run on average signi cantly less than the manufacturer s claim REVIEW OF CONCEPTS from textbook 2 Table 0 Raw scores 2 scores and percentile rankings are three ways to describe the same score within a normal distribution 0 2 table is how we transition from one way of naming a score to another 0 negative 2 statistics not included in table bc all we have to do is change sign from negative to positive normal curve is symmetric one side mirrors other 0 if we know the mean and the standard deviation of a population we can convert a raw score to a z score and then use the 2 table to determine percentages below above or at least as extreme as this 2 score 0 for a positive 2 score we double the percentage above that z score to get the percentage of scores that are at least as extremethat is at least as far from the mean as the z score is 2 Table and Distribution of Means only addition is that we now need to calculate the mean and the standard error for the distribution of means before taking that information to the 2 table 0 same conversions can be conducted on sample mean instead of on score we just use mean and standard error of distribution of means instead of distribution of scores The Assumptions and Steps of Hypothesis Testing The formal process of hypothesis testing is based on particular assumptions about the data Assumptionscharacteristics that we ideally require the population from which we are sampling to have so that we can make accurate inferences Three Assumptions for Conducting Analyses o Dependent variable is on a scale measure 0 Participants are randomly selected 0 Population distribution is approximately normal Six Steps of Hypothesis Testing 0 Identify the population distribution and assumptions and then choose the appropriate hypothesis test 0 State the null amp research hypotheses in both words and symbolic notation 0 Determine the characteristics of the comparison distribution 0 Determine the critical values or cutoffs that indicate the points beyond which we will reject the null hypothesis 0 Calculate the test statistic 0 Decide whether to reject or fail to reject the null hypothesis Critical value test statistic value beyond which we reject the null hypothesis often called the cutoff Critical region area in the tails of the comparison distribution in which the null hypothesis can be rejected p level probability used to determine the critical values or cutoffs in hypothesis testing often called alpha statistically signi cant nding is statistically signi cant if the data differ from what we would expect by chance if there were in fact no actual difference parametric statistics are those that are based on assumptions about the population distribution nonparametric statistics have no such assumptions 0 Parametric statistics are often robust to violations of the assumptions Directional hypothesis predicts either an increase or a decrease but not both Nondirectional hypothesis predicting a difference in EITHER direction When we study a sample rather than an individual the comparison distribution is the distribution of means Onetailed testljhypothesis test in which the research hypothesis is directional positing either mean decrease or mean increase in the dependent variable but not both as a result of the independent variable o Rarely seen in research literature only used when researcher is absolutely certain that effect cannot go in the other direction or they would not be interested in result if it did 0 Twotailed testljhypothesis test in which the research hypothesis does not indicate a direction of the mean different or change in the dependent variable but merely indicates that there will be a mean difference 0 Research convention is to set cutoffs to a p level of 005 o 25 on both tails 0 Critical values are 196 and 196 Variability decreases when you have distribution of means that s why you need to use standard error if something is statistically signi cant it means it is unlikely to occur by chance critical values associated with 2tailed test with alpha of 05196 and 196


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