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by: MadsSwart

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9

chapters 9 and 10 Psych 2220

Marketplace > Ohio State University > Psych 2220 > chapters 9 and 10
MadsSwart
OSU
GPA 3.54

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About this Document

one-sample t tests paired-sample t tests the steps for each application of equations differences of means
COURSE
Data Analysis in Psychology
PROF.
Joseph Roberts
TYPE
Class Notes
PAGES
9
WORDS
CONCEPTS
Statistics, one-sample t test, paired-sample t tests, steps, ohio state, buckeyes, OSU, Psychology, analysis, data, t-test, distribution of means
KARMA
25 ?

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This 9 page Class Notes was uploaded by MadsSwart on Thursday March 10, 2016. The Class Notes belongs to Psych 2220 at Ohio State University taught by Joseph Roberts in Winter 2016. Since its upload, it has received 143 views.

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Date Created: 03/10/16
One-Sample and Paired-Samples T-tests One-Sample t Test - Compare one sample’s data to population - Know population’s mean but NOT population’s standard deviation X−M ¿ ¿ ¿2 ¿ ∑¿ √¿ ¿ M−μ M M−μ M M−μ M t= s = s = ¿ M √N o Dependent variable is scale  if not, as long as not nominal or ordinal o Data randomly selected  if not, cautious with generalizing o Population normally distributed  if not, either at least 30 participants, or data suggesting no skew in distribution - As sample size increases, t distributions look more like z distribution o z statistic is a single almond single-blade knife o t statistic is a bag of almonds that includes the z statistic almond 8 Steps for the One Sample t Test 2. State Variables a. S= sample standard deviation b. SM= sample standard error c. M = Mean of sample d. µM= Mean of Population e. N= number of scores in population f. n= number of scores in sample g. df = number of scores in population - 1 3. Null and Research hypotheses a. H0: µ1= µ2 b. H1: µ1≠ µ 2 4. Standard Error and Standard Deviation a. Standard Error s sM= √N b. Standard Deviation 2 ∑(X−M) s= N−1 √ 5. Test statistic M−μ M t= s M 6. Critical Values or Cutoffs a. df = number of scores in population - 1 b. use this and p-level (like .05 to look up in t distribution table) 7. Confidence Interval a. For a 95% confidence interval- 2.5% in each tail b. Look up the t statistics that fall at each line blocking out the 2.5% in each tail a. Remember that even though 2.5% is in each tail, the p value is . 05 b. need df  remember this is the number of scores minus 1 (N-1) c. convert t statistics back into raw means a. standard error and t statistic are needed for this 1. both of these were calculated for #4 b. Take the standard error [of the sample], multiply it with the critical value of t, and then respectively add and subtract that number from the sample mean c. M lower samplecrit M d. M upper samplecrit M 8. Effect Size C ohen sd= (M−μ) a. σ 9. Reject if… a. t statistic is not within the critical values found on the t distribution table b. population mean does not fall within the confidence interval Fail to Reject if… a. t statistic is within the critical values found on the t distribution table b. population mean falls within the confidence interval 11 Steps for a Paired-Samples t Test Compare two means in a situation where every participant is in both samples 1. Variables – label what you know just from the problem a. S= sample standard deviation b. SM= sample standard error c. M = Mean of sample d. µM= Mean of Population e. N= number of scores in population f. n= number of scores in sample g. df = number of scores in population - 1 2. Null and Research Hypotheses a. H 0 µ 1 µ 2 b. H 1 µ 1 µ 2 3. Differences a. For each pair, subtract the first from the second b. Once you calculate the difference- this is the set of numbers you will use i. Don’t need the original data anymore 4. Mean Difference a. Calculate the mean for all of the differences i. Subtract this from each difference 1. Square it  this is your squared deviation 5. Standard Deviation: S = standard deviation of the differences D squared deviation SD= √ df 6. Standard Error: sMDstandarderrorof themeandifference S D sM= D √N 7. Test statistic M −0 t = D paired SM D 8. Critical Values and Cutoffs a. use t-distribution table with given i. p level ii. degrees of freedom (N-1) iii. and whether one or two tailed test 1. PAIRED TESTS CAN BE ONE OR TWO TAILED 9. Confidence interval a. For a 95% confidence interval- 2.5% in each tail b. Look up the t statistics that fall at each line blocking out the 2.5% in each tail i. Remember that even though 2.5% is in each tail, the p value is . 05 ii. need df  remember this is the number of scores minus 1 (N-1) c. convert t statistics back into raw means i. standard error and t statistic are needed for this 1. both of these were calculated for #4 ii. Take the standard error [of the sample], multiply it with the critical value of t, and then respectively add and subtract that number from the sample mean M =M −t ∗ s( ) iii. lower samplecrit M iv. M upper samplecrit M 10. Effect Size C ohen sd= (M−μ) a. σ 11. Reject if… a. t statistic is not within the critical values found on the t distribution table b. population mean does not fall within the confidence interval Fail to Reject if… i. t statistic is within the critical values found on the t distribution table ii. population mean falls within the confidence interval - A lot of variability in students SAT scores, is there an improvement from Junior to senior year o Paired samples design o Somewhat consistent SAT scores – valid tests show variability o Like how if you take the SAT over and your score only increases by 10 points – because you have the same level of knowledge - Regression to the mean o Look at an extreme group and realize it is less extreme - T statistic for differences o Null hypothesis is that of no difference o µ1= µ2 o µ1- µ2=0 o µD= 0 - for a sample M=´ x o ´ o M DD - for a population o μ xμ μ o D Time 1 Time 2 Difference time 1 3 9 6 Time 2 3 6 3 Time 3 4 9 5 Time 3 5 6 1 Time 4 4 8 4 Average of times 3.8 7.6 3.8 Standard dev of .83666 1.516575 1.923538 times N=5 M = 3.8 S = 1.92 D D D M ¿ 3.8−0 M Dμ M −μ S 1.92 d= s d= D D MD = sd ¿ √5 tpaired - We use effect size for description not for inference - Sd of heights is easy, but sd of differences is harder Confidence intervals around mean difference - ONE SAMPLE: Mu bounds = M =/- t critical * S m - PAIRED SAMPLE: M sub d =/- t critical * standard error of difference S [¿M D] ¿ Practice: Match the statistic to the condition under which it is used a. One-sample z-test b. One-sample t-test c. Two-sample t-test i. Comparing a sample mean to a population mean with known variance ii. Comparing a sample mean to a population mean when population variance is unknown iii. Comparing two sample means with unknown population variances One sample t test- Represents a single observed mean in terms of its distance from the center of the sampling distribution of the mean as estimated by the observed sample standard deviation, s. One-sample t tests are always two-tailed True - N=15df=14 M=9.164 s=2.377 µM=10.6 - Test for differences between sample and population with known µ and σ  use z statistics, attending to sample size Z (A t statistic is the M same as Z Mor samples but with an estimate M of the actual variability σ M d. Test statistic = sMisstandard error i. = 2.377/√15 =.613 M−μ M−μ M = M= M ii. sM s √N 9.164-10.6=-1.436/.613= -2.34 e. Critical value = 2.145 f. Lower limit of 95% confidence interval= lower samplcritM) =9.164-(2.145*.613)=9.164-1.314=7.85 M =M +t ∗ s g. Upper limit of 95% confidence interval= upper samplcritM) =9.164+1.314=10.478 ' (M−μ) h. Effect size [cohen’s d] ohen sd= σ = (9.164- 10.6)/2.377=-.604 1 2 3 4 5 6 7 8 9 1 1 1 mea 0 1 2 n 10 min before 1 1 1 1 8 1 1 8 1 1 1 1 4 2 0 2 0 0 2 6 4 2 1 min before 1 1 1 1 1 1 1 1 1 1 1 1 8 9 4 6 2 8 6 8 6 9 6 6 Differences 4 7 4 4 4 8 6 1 4 3 2 4 5 0 i. Will there be any difference in anxiety levels at 10 minutes vs 1 minute prior to the exam – using alpha =.05 i. Hypotheses: 1. H0: µ1= 2 2. H1: µ1≠ µ2 ii. t statistic Squared deviation: M =5 1 2 3 4 5 6 7 8 9 10 11 1 df=1 D 2 1 Differen4 7 4 4 4 8 6 10 4 3 2 5 ce D-MD -1 2 -1 1 -1 3 1 5 -1 -2 -3 0 (D-MD) 1 4 1 1 1 9 1 25 1 4 9 0 ∑=57 squared deviation SD= =√(57/11)=√5.182=2.276 √ df s s = D MD √ N = 2.276/√12=2.276/3.464=.657 μD−0 5−0 5 tpaired = = =7.6 SMD .657 .657 iii. Critical t value = 2.201 iv. 95% Confidence Interval 1. M upper samplcrit M =5+(2.201*.657)=5+1.446=6.446 M lower samplcrit(M) 2. =5-1.446=3.554 5. Want to know if chocolate affects mood. Recruit college students to take mood inventory, ingest quarter pound of chocolate, and then complete the mood inventory again. Confidence interval = [2.34, 5.41]. Make a decision about the null hypothesis tested here.  Since 2.34 and 5.41 are rather close and on the positive side of the curve  Zero is not included, which means that there is a statistically significant difference between the populations Reject the null and conclude that chocolate affected mood because zero, or no effect, does not fall within the confidence interval 6. The second step in conducting the single-sample t test is Stating the null hypothesis 7. The correct formula for effect size using Cohen’s d for a single-sample (M−μ) t test: d= s 8. critical cutoff: a. two-tailed test b. p level = .05 c. n=26  df=25 d. N=2500  in t table  2.056  cutoffs: [-2.06,2.06] 9. Know the population mean but not population standard deviation, which statistic do we use to compare a sample to the population?  t 10. Report statistical results in APA format t(5) = 1.2, p > 0.05 11. A researcher is worried that performance on a test will improve because of repeated exposure to the test, which will cloud his ability to assess the impact of two unique educational interventions [program A and program B]. which of the following design options includes counterbalancing? He could vary the order of the educational interventions such that half of the participants experience Program A first and the other half experience Program B first 12. The denominator of the ratio for calculating the dependent samples t statistic is M Do paired SMD estimated standard error of the sampling distribution of the mean difference scores 13. Differen 3 4.5 7 8 9 11.1 µ=7.1 ce Diff -Dµ 3-7.1= 4.5- 7-7.1= 8-7.1= 9-7.1= 4 -4.1 7.1= -0.1 1.1 2.1 -2.6 2ndno.2 16.81 6.76 .01 1.21 4.41 16 ∑=45. 2 Calculate the standard error of the difference scores for a paired- samples t test SD= squared deviat=on5.2=√9.04=3.006 √ df √ 5 s = SD= 3.00= 3.00=1.22 MD √N √6 2.45 14. Calculate the mean of difference scores Befor After Differen e ce 110 103 103- -7 110= 98 104 104- 6 98= 89 97 97-89= 8 114 127 127- 1 114= 3 Mean 102. 107. 5 75 75 15. One of the first steps in calculating the dependent-samples t statistic is Creating a difference score for each participant in the sample

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