One question per chapter, 8 in all, taken from the course's corrected exercises. Answer the ones you can do and skip those from a chapter you have not seen yet: each answer is checked right away, and the solution opens after.
With your free account, each answer is kept and your dashboard files each chapter: passed the check, or to work on. These are the same questions as on the set pages: an answer given here counts there.
These chapters come from an earlier course, and this one relies on them from the first weeks.
360-300 · Index numbers, rates and constant dollars
A rate is a quotient, and what matters is the denominator. The unemployment rate is the most disputed example, because the choice of who counts as unemployed changes the result by several points.
c) The number of jobs stays the same but 12,000 unemployed people leave the labour market. What happens to the unemployment rate?
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Answer
c)404,00020,000≈5.0%: 2.7 points lower without a single job created
c) The labour force falls to 416,000−12,000=404,000 and the unemployed to 20,000. The rate becomes 404,00020,000≈0.0495, that is 5.0% instead of 7.7%. The rate has therefore DROPPED by 2.7 points although no job has been created: people simply stopped looking. This is the exact mechanism of discouragement, and it operates strongly at the end of a recession.
360-300 · Cross-tabulations, chi-square and association
The chi-square adds up the squared differences, each one scaled by what was expected. It then yields a decision and, above all, if you read it cell by cell, it tells you WHERE the relationship is.
χ2=∑E(O−E)2, with df=(r−1)(c−1). Critical values at 5%: 3.841 for 1 degree of freedom, 5.991 for 2, 7.815 for 3, 9.488 for 4.
e) Does rejecting the null hypothesis prove that the relationship is strong? What else must be computed?
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Answer
e)No: rejection says nothing about strength, which is read on Cramér's V
e) No. Rejection says that the observed gap is too large to be attributed to sampling chance alone; it says nothing about the STRENGTH of the relationship. The proof is that, for an identical structure, the chi-square grows in proportion to the sample size: any gap, however tiny, becomes significant with enough respondents. You must therefore also compute a normalized measure of ASSOCIATION, Cramér's V or the contingency coefficient, which does not depend on n.
A poll gives 42% to candidate A and 38% to candidate B, with a margin of 3 points. A four-point lead, a three-point margin: is the lead established? The answer is no, and the reason deserves a computation.
The figure below represents the situation.
a) Write the two confidence intervals, each with a margin of 3.1 points. Do they overlap?
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Answer
a)[38.9,45.1] and [34.9,41.1]: they overlap over 2.2 points
a) Candidate A: [38.9,45.1]. Candidate B: [34.9,41.1]. They overlap over the whole zone from 38.9 to 41.1, a common width of 2.2 points. A hurried reader would conclude that nothing can be said.
Chapter 1 · Comparing two groups: t-test and effect size
When the two measurements are on THE SAME people, before and after, the samples are no longer independent. We no longer compare two means: we test whether the mean of the individual differences is zero.
Ten students, differences between the second and the first exam: 4, −1, 6, 3, 0, 5, 2, 7, 1, 3.
b) Compute the statistic t=sd/nd and its degrees of freedom.
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Answer
b)t=0.81653.0≈3.674, with 9 degrees of freedom
b) Standard error: 102.582=3.16232.582≈0.8165. Statistic: t=0.81653.0≈3.674, with n−1=9 degrees of freedom. Note that the sample here is made of the DIFFERENCES, ten numbers and not twenty: that is what explains the 9 degrees of freedom.
Sums of squares cannot be compared directly: they have to be brought back to their degrees of freedom. The ratio of the two mean squares is the F statistic.
a) Compute the two mean squares and draw up the complete table of the analysis.
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Answer
a)MSbetween≈186.67 and MSwithin≈19.167
a) Mean square BETWEEN: 2373.33≈186.67. Mean square WITHIN: 12230≈19.167. Table: source between, SS=373.33, df=2, MS=186.67; source within, SS=230.00, df=12, MS=19.167; total, SS=603.33, df=14. Careful: the mean squares do NOT add up, unlike the sums of squares and the degrees of freedom.
Software always returns the same table. Five columns, and each answers a precise question.
Output of model 2, on 150 students.
Variable
Coefficient b
Standard error
t
p
Constant
55.40
3.12
17.76
< 0.001
Hours of study
1.05
0.21
5.00
< 0.001
Hours of work
−0.42
0.11
−3.82
< 0.001
e) Can 1.05 and −0.42 be compared directly to say which variable matters more?
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Answer
e)No: different units and different spreads; standardized coefficients are needed
e) No. The two coefficients are expressed in different units, points per hour of study and points per hour of work, and the size of their effect also depends on the SPREAD of each variable: a variable whose values barely vary cannot produce much difference, even with a large coefficient. To compare, use STANDARDIZED coefficients, obtained by multiplying each b by sysxj: they are all expressed in standard deviations and become comparable. The ratio of the t statistics, here 5.00 against 3.82, is an acceptable shortcut for saying which one is more solidly established, but it does not measure the size of the effect.
The nonparametric counterpart of the paired t test. You compute the individual differences, rank their ABSOLUTE VALUES, then give each rank back the sign of its difference.
c) The critical value for n=8 is 3 at the two-sided 5% level. Conclude.
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Answer
c)W=3≤3: only just rejected at the 5% level
c) Keep the smaller one, W=3. The rule is to reject if W≤Wcrit: since 3≤3, you only just REJECT the null hypothesis at the 5% level. The observed improvement is therefore significant, by the narrowest of margins. Note that strict equality is enough to reject, which is not obvious and is read from the definition of the critical value: it is the largest W whose cumulative probability does not exceed 2.5% on each side.
A series measured over time is not read like a sample: successive observations resemble each other, and the overall movement breaks down into four pieces, each of which calls for different treatment.
Quarterly enrolment at a CEGEP over three years. Year 1: 2100, 1500, 2400, 1800. Year 2: 2200, 1560, 2520, 1880. Year 3: 2320, 1640, 2640, 1960.
d) Write the MULTIPLICATIVE model and the ADDITIVE model, and say which one fits here.
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Answer
d)Yt=Tt×St×Ct×It or the sum; here the multiplicative, the gap going from 900 to 1000
d) MULTIPLICATIVE model: Yt=Tt×St×Ct×It, where the seasonal effect is a percentage of the trend. ADDITIVE model: Yt=Tt+St+Ct+It, where the seasonal effect is a gap in units. The multiplicative model fits here: the gap between the high quarter and the low quarter goes from 900 in the first year to 1000 in the third, so it grows with the level of the series. A seasonal gap that is proportional calls for the multiplicative model; a constant gap would call for the additive one.
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