COMPUTE traditional = sexrol1 + sexrol2 + sexrol3 + sexrol4 + (6-sexrol5).

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1 Handige Syntax: Week 1: There are several ways in SPSS to obtain these estimates, such as via FREQUENCIES, DESCRIPTIVES or MEANS (see help function SPSS or De Vocht's SPSS book). ML estimates have attractive properties (a) they are EFFICIENT, that is have smallest variance (b) they are UNBIASED for large values of n (c) they have approximately NORMAL sampling distributions. Testen tegen bepaalde waarden T-TEST /TESTVAL=21 /MISSING=ANALYSIS /VARIABLES=ageleft /CRITERIA=CIN (.90). Remember that spss always reports a two-sided p-value. That is p = Pr(z < & z > 5.616) <.001. Because we have a one-sided test we should report the p-value as p <.001/2. Het kiezen van specifieke groepen zoals in dit geval alleen mannen USE ALL. COMPUTE filter_$=(sex = 0). VARIABLE LABEL filter_$ 'sex = 0 (FILTER)'. VALUE LABELS filter_$ 0 'Not Selected' 1 'Selected'. FORMAT filter_$ (f1.0). FILTER BY filter_$. EXECUTE. Uitzetten en testen van filter FILTER OFF. FREQ VAR=sex. Tijdelijk uitzetten TEMPORARY. SELECT IF (sex =0). COMPUTE traditional = sexrol1 + sexrol2 + sexrol3 + sexrol4 + (6-sexrol5). Hercoderen educatie RECODE educ (1=6) (2=10) (3=10) (4=11) (5=12) (6=14) (7=16) (8=18) (9=22) INTO educyr. EXECUTE. independent samples t-test T-TEST GROUPS = sex(0 1) /MISSING = ANALYSIS /VARIABLES = educyr /CRITERIA = CI(.95).

Testen tegen bepaalde waarden T-TEST /TESTVAL=21 /MISSING=ANALYSIS /VARIABLES=ageleft /CRITERIA=CIN (.90). Remember that spss always reports a two-sided p-value. That is p = Pr(z < -5.616 & z > 5.

2 Week 2 Crosstab met chi square en gamma en tau-b CROSSTABS /TABLES=supvis BY class /FORMAT= AVALUE TABLES /STATISTIC=CHISQ GAMMA BTAU /CELLS= COUNT EXPECTED ASRESID /COUNT ROUND CELL. Week 3 In spss select GRAPHS/SCATTER/SIMPLE/DEFINE and let hrinc correspond with the Y-axis and educyr with the X-axis. Doing this, and pasting the syntax, leads to GRAPH /SCATTERPLOT(BIVAR)=educyr WITH hrinc /MISSING=LISTWISE. ANALYZE/ESSION/LINEAR/ define hrinc as the dependent (= response) variable and educyr as the independent (= explanatory) variable. ESSION /MISSING LISTWISE /STATISTICS COEFF R ANOVA /DEPENDENT hrinc /METHOD=ENTER educyr. Week 4 Scatterplot met multiple regressie GRAPH /SCATTERPLOT(BIVAR)=educyr WITH hrinc BY sex /MISSING=LISTWISE. Het computen van Dummy variabelen COMPUTE white = (class = 1). COMPUTE blue = (class = 2). COMPUTE farm = (class = 3). EXECUTE. I Haz No Cluer ESSION /STATISTICS COEFF OUTS R ANOVA CHANGE /DEPENDENT hrinc /METHOD=ENTER educyr sex /METHOD=ENTER blue farm. Wie weet komt het nog van pas First, declare the category "miscellaneous" of the variable religion as missing; too heterogeneous to use in the analyses. MISSING VALUES religion (6). COMPUTE none = religion = 1. COMPUTE catholic = religion = 2. COMPUTE reformed = religion = 3. COMPUTE rereformed = religion = 4. COMPUTE islam = religion = 5.

Doing this, and pasting the syntax, leads to GRAPH /SCATTERPLOT(BIVAR)=educyr WITH hrinc /MISSING=LISTWISE.

3 EXECUTE. Oefentoets week 5 Exercise 1 Use the file donations2.sav. Construct a new variable relicat with categories: (1) no religion (2) catholic and (3) protestant. All other values have to be left out of the analysis. Construct a categorical variable doncat based on the continuous variable donations with three categories such that each category contains about the same amount of respondents. Construct a contingency table of relicat against doncat. FREQ religion. RECODE religion (1=1) (2=2) (3, 4=3) (else=sysmis) INTO relicat. VAL LAB relicat 1 "no religion" 2 "catholic" 3 "protestant". RANK donation (A) /NTILES (3) INTO doncat. (a) Resulting table: CROSS doncat BY relicat /STA = CHI /CELL = COUNT COLUMN. (a) Is there an association in this table? YES, because chi^2 significantly different from 0 (b) How would you describe this association? Most of the non-religious are in the lower tricile of donation, most of the catholics are in the middle and most of the protestants are in the upper tricile. Consequently, protestants donate the most, then catholics and then the non-religious who donate not so much Exercise 2 Use the file donations2.sav. If we regress percdon on relicat, we have to construct dummy variables "norel", "catholic", and "protestant" from the relicat variable, and choose a reference category, e.g. "norel". The resulting regression coefficients of catholic and protestant can be interpreted as the difference in predicted percdon between catholic and norel and between protestant and norel. For the difference between catholics and protestants, we have to choose a different reference category. It is perfectly possible to do this. COMPUTE norel = relicat = 1. COMPUTE catholic = relicat = 2. COMPUTE protestant = relicat = 3. /DEP percdon /METHOD=ENTER catholic protestant.

Construct a categorical variable doncat based on the continuous variable donations with three categories such that each category contains about the same amount of respondents.

4 /DEP percdon /METHOD=ENTER norel protestant. to be discussed after Christmas holidays:. A more straightforward solution is to use GLM univariate and ask for a post-hoc test. UNIANOVA percdon BY relicat /POSTHOC = relicat ( BONFERRONI ) /DESIGN = relicat. In the multiple comparisons table, you will find the differences, and the p-values. You may check these results with: Exercise 3 Use the file sbs399.sav. This file contains 17 items with respect to the presence or absence of luxury goods in the household (0 = not reported; 1 = item not present; 2 = item present; 3 = not appropriate). Construct a new variable luxe that counts how many items are present in the household. (Use the COUNT option in the TRANSFORM menu). COUNT luxe = good01 good02 good03 good04 good05 good06 good07 good08 good09 good10 good11 good12 good13 good14 good15 good16 good17 (2). (a) What is the average, the standarddeviation and the standard error of this new variable luxe?. DESC luxe /STA = MIN MAX MEAN STDDEV SEMEAN. (b) Give the 95% confidence interval for the average. by hand from the 'desc' command results: = = ( 4.37; 4.54 ) or in SPSS with: EXAM VAR=luxe /COMPARE group /STAT = DESC /PLOT = NONE /CINTERVAL 95 /MISSING LISTWISE /NOTOTAL. or as in week 1:. T-TEST /TESTVAL=0 /MISSING=ANALYSIS /VARIABLES=luxe /CRITERIA=CI(.9500). (c) Construct a new variable ( left ) from the variable polit such that

You may check these results with: Exercise 3 Use the file sbs399.sav.

5 pvda, d66, groenlinks and sp get score 1 on this new variable, all other parties score 0, and don t know is defined as missing value. Chosse adequate variable and value labels. RECODE polit (1, 3, 5, 11 = 1) (13=sysmis) (else=0) INTO left. VAR LAB left "Left-wing parties". VAL LAB left 1 "Left wing" 0 "Other". (d) Test whether the average amount of luxury goods in the household differs between left wing and right wing voters. T-TEST GROUPS=left(0 1) /MISSING=ANALYSIS /VARIABLES=luxe /CRITERIA=CI(.9500). Other option (not yet discussed in class). mean luxe by left /sta=anova. (e) Test the hypothesis that a higher net income leads to a larger number of luxury goods in the household. Control for age, the number of children, the number of years of education, and occupational prestige. H_0 : b_hrinc = 0 H_a : b_hrinc > 0 We need a one-sided test. /STAT COEFF OUTS R ANOVA CHANGE /DEP luxe /METHOD = ENTER hrinc /METHOD = ENTER age kids educ pres. one sided H_a so halve the p-value (f) If a higher net income leads to a larger number of luxury goods in the household, is that effect different for left wing and right wing voters? Sketch a graph for the results. COMPUTE inc_left = hrinc left. /STAT COEFF OUTS R ANOVA CHANGE /DEP luxe /METHOD = ENTER hrinc /METHOD = ENTER age kids educ pres /METHOD = ENTER left /METHOD = ENTER inc_left. Interaction not really significant. To draw a graph run. GRAPH /SCATTERPLOT(BIVAR) = hrinc WITH luxe BY left /MISSING = LISTWISE. and then open the graph by double-clicking and from menu "elements" pick option called "Add fit line at subgroups"

(d) Test whether the average amount of luxury goods in the household differs between left wing and right wing voters. T-TEST GROUPS=left(0 1) /MISSING=ANALYSIS /VARIABLES=luxe /CRITERIA=CI(.9500).

6 Exercise 4 (a) Is it possible that in a contingency table of satisfaction (1 = low; 2 = average; 3 = high) against income (1 = low; 2 = average; 3 = high) Pearson s Chi2 will be not significant and Kendall s tau-b significant? Yes/No; why? It is not possible to get simultaneously Chi^2 = 0 and tau-b ~= 0 because ordinal association, measured by tau-b, is a "specific type" of statistical association which is captured by Chi^2. Therefore, if Chi^2 = 0 then all measures of association will be 0. The converse is not true. However it is still possbile to get "not significant" Chi^2 test (stating there is not enough evidence to reject hyp. stating statistical independence) and, simultaneously, a value of tau-b which "is significantly" different from 0. This slight contradiction is a result from the fact that, according to A&F, "ordinal measures are more powerful in detecting the association than the Chi^2" in the situations with two ordinal variables. (b) Give a 3 x 3 table with frequencies such that gamma will (almost) be zero and Pearson s Chi2 much larger than zero Here we have = 4 concordant pairs and 22 = 4 disconcordant pairs, ergo gamma is equal to 0. Simultaneously, the variables are strongly statisticly dependent. Once you know the value of one of the variables you can uniquely identify the value of the other. Consequently the value of Chi^2 will be high. (c) If we multiply all cellfrequencies in a contingency table by two, what happens with the p-values of gamma and Pearson s Chi2? The value of chi^2 will INCREASE with a factor 2, and consequently its p-value will decrease The value of gamma will NOT change, but its standard error will decrease, consequently the p-value will decrease (d) Give a 3 x 3 table with frequencies such that the number of discordant pairs (D) equals zero, and the number of concordant pairs (C) is greater than zero All three pairs are concordant

It is not possible to get simultaneously Chi^2 = 0 and tau-b ~= 0 because ordinal association, measured by tau-b, is a "specific type" of statistical association which is captured by Chi^2.

Simple Linear Regression, Scatterplots, and Bivariate Correlation

1 Simple Linear Regression, Scatterplots, and Bivariate Correlation This section covers procedures for testing the association between two continuous variables using the SPSS Regression and Correlate analyses.