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Intermediate Educational Statistics

by: Valerie Yundt

Intermediate Educational Statistics EPSY 5381

Marketplace > Texas Tech University > Educational Psychology > EPSY 5381 > Intermediate Educational Statistics
Valerie Yundt
GPA 3.99


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Class Notes
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This 50 page Class Notes was uploaded by Valerie Yundt on Thursday October 22, 2015. The Class Notes belongs to EPSY 5381 at Texas Tech University taught by Staff in Fall. Since its upload, it has received 26 views. For similar materials see /class/226398/epsy-5381-texas-tech-university in Educational Psychology at Texas Tech University.

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Date Created: 10/22/15
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t blow 2 j K 1 Iii 3 V w Mm as i fart a 3 I 4 2 LE1 fax a x k T 92 55V T r Tu K if N 4 a E 1 r r K w 5335 A 5 3 Av A W 3 5 w a z a sh C s f figfsi m a Effigy xix Hui 3 Mix a ffgfk 5 mus 5 g g Mi fn ijik xx f2 r m 5 r ML a E w E a m c 1 33 x 3 4 x U 5 0i 1 33 f v m 556 X E 5uQ Cf lt1 f E f kg VinCOV 05 5 W43 lt9 is 35 m gislii 5 jva R 3 aw 73 k a f v M as bk W Xi WEW K a f 3 i E 2 if amp a i fix 1 iVAm 353 AM a 5541 a i ii iag as g Eia igxiiazx 5 3 3 i r in 615562 if f g3 33 g j a v lt1 r 3139 quot Q m 8 WW its 17 WWW d quot Wmmw NM Mmmh wmw aw t quot3 a gaff53W ampw L f 7 i 5f 5 Sim n 4 m if 3 3 3 3 4 sfquot 3 3 I iii if I 5 Q J VI 1 W E gtw f Cg 256W W i M3MWW quot57 45 FM 93 A gt Midi n Lx39 mfg a 2 E galK quot quot if z a I 5 a f p I I z r yawm 0 lt3 1 5 15633 iii 9 is 2w fig 39 g if WW a f as xix 4 k c 1 is a 2 C 1 C Eliquotx 139 w V y W C 7quot x 9 Q i mquot i m a K x EMS V 4 k i 4 y K 2 A Ki 3 i M 3 w affkf 3 Page a magi xmv r RE 32 quotm WWMMMWW W f M MNXK f x i t r V i s g r 5quot z g z 3 57 f Chapter 14 r r WWWMWWMS Hypothesis Testing KSample Case Analysis of Variance OneWay Classi cation This chapter presents the procedure for testing the hypothesis that K population means are equal where Kgt2 In other words this chapter addresses what to do when the independent variable has more than two levels You would use a t test in the event there were only two levels Problems with Multiple t Tests If there are more than two levels of the independent variable the research could conduct multiple t tests however this is not appropriate The problem with computing multiple independent t tests for comparing K sample means is that as the number of t tests increases the Type I error rate increases The Type I error rate is determined by 1 l itc where on level of signi cance for each separate t test and c number of independent t tests The Variables in ANOVA Independent variable the variable that forms the groupings This variable is categorical Dependent variable the measure on which the groups are being compared This variable is continuous In AN OVA one way classi cation only one independent variable is considered but there can be two three six or theoretically any nite number of levels of the independent variable For example The researcher may be interested in differences in students math scores dependent vari ble based on the amount of tutoring the student received independent variable or treatment There is only one independent variable tutoring but four levels may exist such as 30 minutes of tutoring one hour one and a half hours and two hours Changes in the dependent variable in ANOVA are or are presumed to be the result of changes in the independent variable Concepts Underlying ANOVA The null hypothesis is that the population means from which the K samples are selected are equal Ho unu2u5 ui Ha u at pk for some i k In order to test the null hypothesis this procedure analyzes the variance of the scores on the dependent variable The total variation of the scores is partitioned into two components The variation of the scores Within the K groups and the variation between the group means and the mean of the total group grand mean Small observed differences between the sample means and the grand mean are attributable to sampling uctuation However sizable differences between the group means and the grand mean resulting in a substantial 883 are an indication that the null hypothesis may not be true Two Estimates of the Population Variance In ANOVA the betweengroups and within groups variance estimates are found by dividing 88 and SSw by the degrees of freedom associated with each of these estimates The resulting between groups and within groups variance estimates are commonly called mean squares Testing the Null Hypothesis The computed mean squares are actually statistics and have counterparts in the population called expected mean squares Discussion of expected mean squares relates the analysis to the linear model in ANOVA See page 358 The F Ratio The difference between the two variances can be tested by forming a ratio of variances and using the F distribution as the sampling distribution Assumptions Underlying AN OVA There are three primary assumptions relevant to AN OVA 1 The observations are random and independent samples from the populations 2 The distributions of the populations from which the samples are selected are normal 3 The variances of the distributions in the populations are equal Homogeneity of variance Generally failure to meet these assumptions changes the Type I error rates To assure that the assumptions are met 1 Use random sampling and appropriate sampling techniques 2 Use a goodnessof t test to check the normality assumption 3 Utilize Bartlett s test for homogeneity of variance ANOVA is robust with respect to violations of the assumptions except in the case unequal variances with unequal sample sizes l of A Measure of Association quotg Rejecting the null hypothesis in ANOVA indicates that there are signi cant differences among the sample means a greater difference than would be expected on the basis of chance With larger sample sizes statistically significant ndings may hold little practical signi cance A 3 Omega squared is a measure of the strength of the association between the independent 2 39 and dependent variables in AN OVA Omega squared indicates the proportion of p I 5 x g k S V F A E a 1 x if H y 5 3 NE aquot KKK iv 1 in preparing a planning re port the director of a community continuing education program classi es course requests a s 1 job related 2 consumer related 3 health related 4 recre ational or 5 other Over ve registration periods the following numbers of special requests were received Use one way ANOVA procedures to determine whether or not the registration data suggest differences in overall community interest Use the 05 level of signi cance 39 Course request classi cations 3 2 l 1 2 3 4 5 r 26 1 18 28 23 g 30 2 25 40 30 35 33 20 26 24 29 ll 16 31 25 24 5 11 20 18 22 5 5 5 5 25 Z 90 145 120 140 110 605 180 290 240 280 220 g 146413 1726 4421 2954 4026 2530 39 ik 15657 16200 42050 28800 39200 24200 2T2nk 15045 a Hui l 1 3 14 5 r 39 ii 8 H pi at pk for some 139 k quot WigVa f freedom F 287 A a 2226 b For K l 4 and N K 20 degrees 0 tv 2 L0 c39 ssnz glUfno TzN 15045 l464l 404 mm Mommy we ssW Tr xikz mlnk 1565715045 612 I1l i1 kl I C A ssT 35 quot xi TZN l5657 i4641016 kl il Summary ANOVA Source SS df MS F F 330 287 quot Between 404 4 Wm g 101 5quot Within 612 20 dmom 3060 3 Total 1 01 6 24 ror me results 01 the oneway ANOVA presented in Exercise 1 use the Tukcy method to test tor pairwwe dili erences among the population means at the 05 level of signi cance Al though not rcqu1red for the Tukey or TukeyKramer procedures the means are ranked from low to h1gh 2 2 Xi Xk Q 2 1x 18 2 4 162 it 6 2 243 0815 I 4 28 10 6 4 405 31243 162 9 29 7 5 1 445993quot 283 202 040 Mswn 306075 247 plt05 Forr 5 and dfw 2ch 423 Therefore pl at 1 all other pairwise compariso s are not statistically signi cant 39 Ex 3 a f 8 fl X3 4 A i 41 2 5 RX a Z 281 a 3 Again referring to the data used 1n Exercxses l and 2 periorm the NewmanKeuls procedure to test for pairwise dif gzrences between means fisc 05 as your level of signi cance a 3 2 59 Q Qcv XI 1 xi 2 22 162 295 X 24 2 243 081 358 295 4 28453 10 6 4 405 243 162 396 358 295 2396 11 7 5 1 445 283 202 040223 423 396 358 295 iMSWn 430605 247 plt01 p 75 114 and p at 114 all other pairwise comparisons are not statistically signi cant 5 For 110 data in Exercise 7 oi39Chaptcr l4 perform 1hc TukcyKramcr procedure 10 lost f or pair wise mean di 39crcnccs Set the level of signi cance at 05 Mel10d I II 111 ilk 5 4 6 N 15 x 2500 2900 3600 Sunuuary ANOVA Smut SS 1139 MS F FW Bctwccn 34173 2 17087 707 389 Within 29000 12 2417 Total 63 173 14 XI nk Xi Xk denominator Q J 250 5 39 290 4 4 233 172 X 360 6 1 1 7 21 1 224 521 313 1 1 r denominator MSw i l plt05 Forr 3111111 dfw 12 QCV 377 39 Therefore 11 ti 3 all other pairwise comparisons are not statistically 51 gm cant 7 Assume that in Exercise 6 the drug dosages were measured on a ratio scale and that the drug dosages increase from Group 1 and Group 4 Use orthogonal polynomial contrasts to test for linear quadratic and cubic trends at the 05 level of signi cance Also graph your results Group 1 2 3 4 X 15 12 75 9 n 10 IO 10 10 7 3 Linear 3 i l 3 Quadratic l l l 1 Cubic 1 3 3 1 g b Fl 315112175392 inear quot39 3682a1 L9 10 10 10 10 2252 6878 736 2 F l15 112 17o519 quadratic 39 368 1 LL 1 10 10 10 10 2 4395 1376 1472 F H 1lt1slt312 3gt7s09 cu 1C 368l 9 2 L 10 10 10 10 2 75 7 64 736 chws 4l l for df l 36 c What are your conclusions All three contrasts are signi cant the linear quadratic and cubic trends are all signi cant The researcher would want to describequot all three but concentrate on the cubic trend E15 u 39D 012 E E9 2 6 E 23 6 l L l i 2 3 4 5P5 5363 g D QCWL WW guy6 Maw4 AIS 3 PairJ 7 A f S geeCS GD 6 Vk 5 kw WP 23 0lt 4 a Macwfm tg K 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