Practical — Statistical Analysis using SPSS (STS-207)

Topics Covered

The Data Set

Practical 1: Data Entry, Import, Export and File Handling

1. Question

Read the twenty cases of the data set into SPSS from customers.csv; define every variable in Variable View — its label, value labels, missing-value code and Measure; derive the ordinal grade from score; check the coding with frequency tables; and save the file as .sav and as an Excel workbook.

2. Aim

Set up a data file that every later procedure can trust: the right Measure for each variable, labelled codes and declared missing values.

3. Steps

1. File → Import Data → CSV Data (GET DATA) reads the file; name the data set.
2. In Variable View give each variable its label, its value labels, its missing-value code and its Measure: Nominal, Ordinal or Scale.
3. Transform → Recode into Different Variables makes grade from score, each class closed on the left; label it, set it Ordinal, and run EXECUTE.
4. Analyze → Descriptive Statistics → Frequencies on grade, gender and buy checks the coding against the data.
5. File → Save As writes the .sav; SAVE TRANSLATE (File → Export) writes the Excel copy.
6. Press Paste, not OK, in every dialog, so that each step is written into the Syntax Editor.

4. Programme

* Read a CSV file.  SPSS syntax comments begin with * and end with a full stop.
GET DATA /TYPE=TXT /FILE='C:\data\customers.csv'
  /DELIMITERS=',' /FIRSTCASE=2 /VARIABLES=
  id F3.0 age F3.0 income F4.0 visits F3.0 score F4.0 gender A1 buy F1.0.
DATASET NAME customers WINDOW=FRONT.

* Define the variables properly -- this is the step most often skipped.
VARIABLE LABELS income 'Monthly income, thousands' score 'Assessment score'.
VALUE LABELS buy 0 'No' 1 'Yes'.
VARIABLE LEVEL id gender buy (NOMINAL) / age income visits score (SCALE).
MISSING VALUES income (-99).

* Derive the ordinal grade, closing each class on the LEFT.
RECODE score (LO THRU 56.999=1) (57 THRU 64.999=2) (65 THRU HI=3) INTO grade.
VALUE LABELS grade 1 'Low' 2 'Med' 3 'High'.
VARIABLE LEVEL grade (ORDINAL).
EXECUTE.

FREQUENCIES VARIABLES=grade gender buy.
* grade: Low 6, Med 9, High 5.   gender: F 10, M 10.   buy: No 12, Yes 8.

* Export.
SAVE OUTFILE='C:\data\customers.sav' /COMPRESSED.
SAVE TRANSLATE OUTFILE='C:\data\customers.xlsx' /TYPE=XLS /VERSION=12
  /FIELDNAMES /CELLS=VALUES.

5. Execution and Results

Three file-handling traps. EXECUTE is needed after transformations such as RECODE and COMPUTE before the new values exist — SPSS defers them until a procedure reads the data. /CELLS=VALUES exports the codes and /CELLS=LABELS exports the labels; choosing the second and then re-importing turns a numeric variable into a string. And a declared missing value (MISSING VALUES income (-99)) is excluded from analysis, while an undeclared \(-99\) is treated as income of minus ninety-nine thousand and will quietly ruin every mean.

Practical 2: Data Visualization

1. Question

For the data set, draw a pie chart of grade, a bar chart of gender, a histogram of score with a normal curve, a scatter plot of score against income, boxplots of score by gender, and an ogive of score.

2. Aim

Produce each prescribed chart from syntax, so that every one can be drawn again exactly.

3. Steps

1. Graphs → Legacy Dialogs → Pie and Bar for the categorical variables, counting cases.
2. Graphs → Legacy Dialogs → Histogram for score, with Display normal curve ticked.
3. Graphs → Legacy Dialogs → Scatter/Dot, score on the vertical axis and income on the horizontal.
4. Analyze → Descriptive Statistics → Explore for the boxplots of score by gender.
5. The ogive, which has no button: the cumulative count of score drawn as a line.
6. Paste each dialog into the Syntax Editor instead of pressing OK.

4. Programme

GRAPH /PIE=COUNT BY grade.
GRAPH /BAR(SIMPLE)=COUNT BY gender.
GRAPH /HISTOGRAM(NORMAL)=score.
GRAPH /SCATTERPLOT(BIVAR)=income WITH score /MISSING=LISTWISE.
EXAMINE VARIABLES=score BY gender /PLOT=BOXPLOT /STATISTICS=NONE /NOTOTAL.

* the ogive, which SPSS has no button for: cumulative percentage against the class
FREQUENCIES VARIABLES=score /FORMAT=NOTABLE /NTILES=10.
GRAPH /LINE(SIMPLE)=CUM(COUNT) BY score.

5. Execution and Results

SPSS draws each chart in the Output Viewer. The charts are not reproduced here, because SPSS is not run on this site, but the figures they are drawn from can be checked: the normal curve on the histogram has the mean 60.75 and standard deviation 9.4361 of score (Practical 3), the pie has the slices Low 6, Med 9 and High 5 (Practical 1), and the two bars are 10 and 10.

GRAPH /HISTOGRAM(NORMAL) superimposes a normal curve with the sample's own mean and standard deviation. It is a picture, not a test: the Shapiro–Wilk statistic in EXAMINE ... /PLOT NPPLOT is the test, and on a sample of twenty it has very little power, so a histogram that “looks normal” and a Shapiro–Wilk that does not reject are both weak evidence.

Practical 3: Descriptive Statistics

1. Question

Find the mean, median, standard deviation, minimum, maximum and skewness of age, income and score; the median and mode of grade; and the summaries of score for each gender.

2. Aim

Choose the right one of SPSS's three descriptive procedures for each summary, and know which conventions its output uses.

3. Steps

1. Analyze → Descriptive Statistics → Descriptives for the scale variables, with skewness and kurtosis ticked under Options.
2. Frequencies for grade, with the median and the mode under Statistics: it is the only one of the three that gives the mode.
3. Explore for score by gender, with boxplots and normality plots.
4. Before quoting the skewness, convert between SPSS's coefficient and the moment coefficient, by the factor under the table.

4. Programme

DESCRIPTIVES VARIABLES=age income score
  /STATISTICS=MEAN STDDEV MIN MAX SKEWNESS KURTOSIS.
FREQUENCIES VARIABLES=grade /STATISTICS=MEDIAN MODE.
EXAMINE VARIABLES=score BY gender /PLOT BOXPLOT NPPLOT /STATISTICS DESCRIPTIVES.

5. Execution and Results

Two conventions to state in the report. SPSS's Std. Deviation divides by \(n-1\); and the skewness SPSS prints is not the moment coefficient \(g_1 = m_3/m_2^{3/2}\) but the sample-adjusted version

so at \(n = 20\) it is larger by a factor \(\sqrt{380}/18 = 1.082977\), or \(8.3\%\). The table above gives the moment coefficients, so that they match the R and Python pages; SPSS will print \(0.379812\), \(0.419544\) and \(0.607924\) for the three variables, and the difference is the convention and not an error.

Practical 4: Parametric Tests

1. Question

Test whether the mean score differs between men and women, and run a one-way analysis of variance of score by grade. Say what each result means.

2. Aim

Run the two-sample \(t\) test in the right order, the variances first, and recognise an analysis that SPSS will run but that answers nothing.

3. Steps

1. Analyze → Compare Means → Independent-Samples T Test: score as the test variable, gender as the grouping variable with the groups M and F.
2. Read Levene's test first, then the \(t\) row it points to.
3. Analyze → Compare Means → One-Way ANOVA: score by grade, with the descriptives, the homogeneity test and Tukey's and the LSD post hoc tests.
4. Before reading the ANOVA, ask whether the groups could differ by construction.

4. Programme

T-TEST GROUPS=gender('M' 'F') /VARIABLES=score /CRITERIA=CI(.95).
ONEWAY score BY grade /STATISTICS DESCRIPTIVES HOMOGENEITY /POSTHOC=TUKEY LSD ALPHA(.05).

5. Execution and Results

The two means differ by \(0.1\) and the variances by a factor of \(3.32\). A \(t\) test of means answers a question this data does not raise; the difference here is entirely one of spread, which is why the variance line is read first and reported whatever it says.

The ANOVA of score by grade is on the page as a trap. It gives \(F = 19.756\) on \((2, 17)\) with Sig. \(= 0.000\), and it is worthless: grade was recoded out of score, so the groups differ by construction. SPSS will print the table without complaint. A procedure that runs is not a result.

Practical 5: Non-Parametric Tests

1. Question

Run the sign and Wilcoxon signed-rank tests on a pair of before-and-after variables, the Mann–Whitney test of score by gender, the runs test and the one-sample Kolmogorov–Smirnov test on score, and the \(\chi^{2}\) test of independence of gender and grade.

2. Aim

Find each prescribed non-parametric test in SPSS, and read the warnings its output prints.

3. Steps

1. Analyze → Nonparametric Tests → Legacy Dialogs → 2 Related Samples, with Sign and Wilcoxon ticked, for the paired variables.
2. 2 Independent Samples → Mann-Whitney U, score by gender.
3. Runs, cut at the median; and 1-Sample K-S against the normal, for score.
4. Analyze → Descriptive Statistics → Crosstabs, gender by grade, with Chi-square under Statistics and the expected counts under Cells.
5. Read the footnote under the Chi-Square Tests table before the Sig. column.

4. Programme

NPAR TESTS /SIGN=before WITH after (PAIRED) /WILCOXON=before WITH after (PAIRED).
NPAR TESTS /M-W=score BY gender('M' 'F').
NPAR TESTS /RUNS(MEDIAN)=score.
NPAR TESTS /K-S(NORMAL)=score.
CROSSTABS TABLES=gender BY grade /STATISTICS=CHISQ /CELLS=COUNT EXPECTED ROW.

5. Execution and Results

Read the footnote under the Chi-Square Tests table. On this data it says that 6 cells (100.0%) have expected count less than 5, and it is right: the six expected counts are \(3.0\), \(4.5\), \(2.5\), \(3.0\), \(4.5\), \(2.5\). The Pearson \(\chi^{2}\) is \(1.867\) on 2 df with Sig. \(= 0.393\), and that significance cannot be quoted as it stands. Pool Med and High, or read Fisher's Exact Test, which SPSS prints in the same table for a \(2\times2\) and offers under /METHOD=EXACT otherwise.

SPSS's /RUNS defaults to cutting at the median, not at zero. To run the test on the signs of regression residuals — the use it is really for — save the residuals from REGRESSION, then use /RUNS(0).

Practical 6: Design and Analysis of Experiments

1. Question

Write the UNIANOVA syntax for a response \(y\) in a completely randomised design, a randomised block design, a Latin square, and the \(2^{2}\) and \(2^{3}\) factorials.

2. Aim

Express each design through the /DESIGN subcommand of one procedure, and know what leaving a term out does.

3. Steps

1. Analyze → General Linear Model → Univariate: \(y\) as the dependent variable, the treatment and blocking factors as fixed factors.
2. Under Model choose Build terms and enter exactly the terms of the design, from the table below.
3. Ask for the treatment means compared with a Bonferroni adjustment, the effect sizes and the homogeneity test.
4. Paste, and check the /DESIGN line before running.

4. Programme

UNIANOVA y BY block treat
  /METHOD=SSTYPE(3) /INTERCEPT=INCLUDE
  /EMMEANS=TABLES(treat) COMPARE ADJ(BONFERRONI)
  /PRINT=ETASQ HOMOGENEITY DESCRIPTIVE
  /DESIGN=block treat.

UNIANOVA y BY A B C
  /METHOD=SSTYPE(3)
  /DESIGN=A B C A*B A*C B*C A*B*C.

5. Execution and Results

The Tests of Between-Subjects Effects table has one row for each term named in /DESIGN, and an Error row that holds everything left out. The sums of squares for each design are worked by hand on the pages linked above, and the SPSS table can be checked against them.

Leaving an interaction out of /DESIGN does not delete it — it pools it into the error. That is exactly right for a randomised block design, where the block-by-treatment interaction is the error term, and exactly wrong for a factorial, where it is the effect of interest. The default /DESIGN that the menus write includes every interaction; the randomised block analysis requires you to take them out deliberately.

Type III sums of squares are SPSS's default and are the right choice for a balanced design, where all four types coincide. For an unbalanced design they do not, and the difference is the subject of STS-203, Unit 1, where the naive sums of squares are shown producing an impossible negative interaction. Say which type you used.

Practical 7: Regression Analysis

1. Question

(a) Regress score on age, income and visits, and choose the model by all possible subsets and by forward, backward and stepwise selection. (b) Model the probability that buy is Yes from income by logistic and by probit regression, and compare the two fits.

2. Aim

Select a regression model by five methods and show that they agree; then fit the two binary-response models and say why one is usually preferred.

3. Steps

1. (a) Analyze → Regression → Linear: score as dependent, the three predictors entered together, with the descriptives, the collinearity diagnostics and the \(R^{2}\) change.
2. (a) Fit every subset of the three predictors, and tabulate \(R^{2}\), adjusted \(R^{2}\), Mallows's \(C_p\), AIC and BIC for each.
3. (a) Run the regression again with Method = Forward, Backward and Stepwise, stating the entry and removal thresholds.
4. (b) Analyze → Regression → Binary Logistic: buy as dependent, income as covariate.
5. (b) The probit model on the same 0/1 data, through Generalized Linear Models with a binomial distribution and a probit link.
6. (b) Compare the coefficients, the likelihoods, the fitted probabilities and the classification.

4. Programme

REGRESSION /DESCRIPTIVES MEAN STDDEV CORR
  /STATISTICS COEFF OUTS R ANOVA COLLIN TOL CHANGE
  /DEPENDENT score
  /METHOD=ENTER age income visits
  /RESIDUALS DURBIN
  /SAVE RESID.

REGRESSION /STATISTICS COEFF OUTS R ANOVA CHANGE
  /DEPENDENT score
  /METHOD=FORWARD age income visits.       /* PIN = .05 by default   */

REGRESSION /DEPENDENT score
  /METHOD=BACKWARD age income visits.      /* POUT = .10 by default  */

REGRESSION /DEPENDENT score
  /METHOD=STEPWISE age income visits.      /* both, in alternation   */

* the two thresholds are settable, and the values used must be reported
REGRESSION /CRITERIA=PIN(.05) POUT(.10) /DEPENDENT score
  /METHOD=STEPWISE age income visits.

LOGISTIC REGRESSION VARIABLES buy
  /METHOD=ENTER income
  /PRINT=GOODFIT CI(95)
  /CRITERIA=PIN(0.05) POUT(0.10) ITERATE(20) CUT(0.5).

PROBIT r OF n WITH income
  /MODEL=PROBIT /PRINT=ALL.
* PROBIT wants grouped data -- r successes out of n trials at each level of the
* covariate.  For ungrouped 0/1 data use GENLIN with a probit link instead:
GENLIN buy (REFERENCE=FIRST) WITH income
  /MODEL income DISTRIBUTION=BINOMIAL LINK=PROBIT
  /CRITERIA METHOD=FISHER(1) SCALE=1
  /PRINT SOLUTION FIT.

NOMREG outcome (BASE=FIRST ORDER=ASCENDING) WITH income age
  /MODEL /PRINT=CLASSTABLE FIT PARAMETER LRT.
PLUM grade WITH income /LINK=LOGIT /PRINT=FIT PARAMETER TPARALLEL.
* TPARALLEL tests the proportional-odds assumption PLUM rests on.
* If it is rejected, PLUM is the wrong model and NOMREG is the fallback.

5. Execution and Results

(a) Model selection

This is the part of the course that is not taught elsewhere on the site, so it is worked in full. Regress score on age, income and visits. The three predictors are correlated with the outcome at \(0.5082\), \(0.6536\) and \(0.0535\), and age and income are correlated with each other at \(0.7795\).

All possible subsets — with three predictors there are eight models, and every criterion can be computed for each:

Read the two \(R^{2}\) columns against each other. Plain \(R^{2}\) rises with every variable added — from \(0.4272\) to \(0.4307\) to \(0.4308\) — because it must; adding a column can never increase the residual sum of squares. Adjusted \(R^{2}\) falls, from \(0.3954\) to \(0.3637\) to \(0.3241\), because it charges for each one. The third decimal of \(R^{2}\) is not evidence of anything.

Mallows's \(C_p\) is read against \(p\), not against zero: a model with negligible bias has \(C_p \approx p\). Here income alone has \(C_p = 0.100\) against \(p = 2\), the full model has \(C_p = 4.000\) against \(p = 4\) — which it must, identically — and the intercept-only model has \(10.109\) against \(1\), which is the signal that it is badly biased.

All four criteria choose the same model. Highest adjusted \(R^{2}\), lowest \(C_p\) relative to \(p\), lowest AIC and lowest BIC all point at income alone.

Stepwise alternates the two and reaches the same place. The final model is

with standard errors \(5.857109\) and \(0.131217\), so \(t = 6.854517\) and \(3.664175\).

Why age is dropped although it correlates with score at \(0.5082\). In the model age + income its coefficient is \(-0.002852\) with a standard error of \(0.264487\), so \(t = -0.0108\) and the sign has even flipped. The reason is the correlation of \(0.7795\) between age and income: once income is in the model, age has nothing left to explain. A predictor's simple correlation with the outcome tells you nothing about whether it belongs in a model that already contains its neighbours, and this is the whole reason selection procedures exist.

Two warnings that belong in the report. Stepwise selection with three candidates on twenty cases is already at the edge: the rule of thumb is ten or more cases per candidate predictor. And the \(p\) values printed for the final model are not honest, because the model was chosen by looking at them; they are conditional on a selection that the same data performed.

(b) Logistic and probit

The two fits are all but identical, and that is the normal situation: the logistic and normal distribution functions differ only in the tails. The coefficients differ by a scale factor and nothing else — \(0.250124/0.142527 = 1.755\), and the textbook rule of thumb is \(\beta_{\text{logit}} \approx 1.6\) to \(1.8\) times \(\beta_{\text{probit}}\). The fitted probabilities are so close that both classify all twenty cases identically.

So why choose one? Because \(\exp(\beta_1)\) is an odds ratio under the logit and means nothing under the probit. Here \(e^{0.250124} = 1.284184\): each extra thousand of income multiplies the odds of buying by \(1.28\). Nothing in the probit output has an interpretation that simple, which is why logistic regression is the default everywhere and probit survives mainly in bioassay, where the normal tolerance distribution has a physical meaning.

The classification table, and the five ratios read off it, are worked in STS-108, Practical 6 — accuracy \(0.900\) against a no-information rate of \(0.600\), precision and recall both \(0.875\), specificity \(0.9167\), AUC \(0.921875\). SPSS prints the classification table automatically and the ROC curve under Analyze → ROC Curve.

Practical 8: Multivariate Data Analysis

1. Question

Write the SPSS syntax for a linear discriminant analysis of buy on age, income and visits; a principal components analysis, with varimax rotation, of age, income, visits and score; and a hierarchical and a \(k\)-means cluster analysis of age, income and visits.

2. Aim

Know where SPSS keeps each multivariate method, and which option changes its answer.

3. Steps

1. Analyze → Classify → Discriminant: buy as the grouping variable (0, 1), the three predictors entered together, prior probabilities from the group sizes, and Box's \(M\).
2. Analyze → Dimension Reduction → Factor: principal components from the correlation matrix, eigenvalues over 1 kept, varimax rotation.
3. Analyze → Classify → Hierarchical Cluster (average linkage between groups, squared Euclidean distance, the dendrogram), then K-Means Cluster with three clusters.
4. For each method, decide and record the option in the last column of the table below.

4. Programme

DISCRIMINANT GROUPS=buy(0 1) /VARIABLES=age income visits
  /METHOD=DIRECT /PRIORS=SIZE /STATISTICS=MEAN STDDEV BOXM TABLE.

FACTOR /VARIABLES age income visits score
  /MISSING LISTWISE /ANALYSIS age income visits score
  /PRINT INITIAL KMO EXTRACTION ROTATION
  /CRITERIA MINEIGEN(1) ITERATE(25)
  /EXTRACTION PC /ROTATION VARIMAX /METHOD=CORRELATION.

CLUSTER age income visits /METHOD BAVERAGE /MEASURE=SEUCLID
  /PLOT DENDROGRAM /PRINT SCHEDULE.
QUICK CLUSTER age income visits /CRITERIA=CLUSTER(3) MXITER(20) /PRINT ANOVA.

5. Execution and Results

The procedures run on this data and print their tables: the Classification Results for DISCRIMINANT, Total Variance Explained and the Rotated Component Matrix for FACTOR, the Agglomeration Schedule and the dendrogram for CLUSTER, and the Final Cluster Centers for QUICK CLUSTER. They are not reproduced here, since SPSS is not run on this site, and on twenty cases they should not be interpreted.

Three warnings, all of which SPSS will let you walk past.

• FACTOR /EXTRACTION=PC is not factor analysis. SPSS's default under the Factor menu is principal components, which is a different model with a different meaning — components absorb the specific variance that a common factor sets aside. The comparison on one correlation matrix is worked in STS-202, Unit 4.
• Standardize before clustering, or do not, but say which. Euclidean distance on age (range 35) and visits (range 8) is dominated by age; the clusters are then a partition of age alone.
• Twenty cases will not support any of these. The usual minimum for a factor analysis is five to ten cases per variable, and for a discriminant analysis the smallest group must exceed the number of predictors comfortably. The procedures above run on this data set because the syntax has to be shown running; their output on twenty cases should not be interpreted, and saying so is part of the report.

Practical 9: Statistical Quality Control

1. Question

Write the SPSS syntax for an \(\bar X\) and R chart, with \(C_p\) and \(C_{pk}\), for measurements taken in rational subgroups; and for a \(p\) chart of the defectives found in samples of varying size, with the run rules applied to both.

2. Aim

Draw the control charts from SPCHART, and read them with the run rules.

3. Steps

1. Choose the chart from the kind of data, by the table below.
2. Analyze → Quality Control → Control Charts, the chart type, and the variable that identifies the subgroups.
3. Under Control Rules tick all the rules; under Statistics give the specification limits, which \(C_p\) and \(C_{pk}\) need.
4. Read the chart for the flagged rule violations first, and only then for capability.

4. Programme

SPCHART /XR=measure BY subgroup
  /STATISTICS=CP CPK /RULES=ALL.

SPCHART /P=defectives BY sample (inspected)
  /SIGMAS=3 /RULES=ALL.

5. Execution and Results

SPSS draws each chart with its centre line and its \(3\sigma\) limits, and lists under it any points that break a rule.

/RULES=ALL applies the Western Electric run rules — a point beyond \(3\sigma\), two of three beyond \(2\sigma\), four of five beyond \(1\sigma\), eight in a row on one side — and flags them on the chart. A chart read only for points outside the limits misses every slow drift, which is the failure mode these rules exist to catch.

\(C_p\) and \(C_{pk}\) require the specification limits, which are an engineering input and not in the data. /STATISTICS=CP CPK without them produces nothing. And process capability means nothing until the process is in control: computing \(C_{pk}\) from a chart with points outside the limits describes a process that has no single standard deviation to speak of.

How Marks Are Lost

What the Practical Record Should Contain