Corrected practice final exam: quantitative methods (360-300)

First dimension, attendance: the attendance rate at classes, as a percentage, a RATIO variable. Second dimension, investment in schoolwork: the number of weekly study hours reported, also a ratio variable. Third dimension, belonging: a score on a four-item Likert scale, treated as an INTERVAL variable by convention although each item taken alone is ordinal. The three do not overlap: a student with perfect attendance may feel attached to nothing.

On a scale from 1 to 5, the reverse-coding formula is $x'=\left(\min+\max\right)-x=6-x$, so a response of 4 becomes 2. Without reversal, an indicator of disengagement would be added to indicators of engagement, which would amount to subtracting instead of adding.

Nothing at all as a mean. The program is a NOMINAL qualitative variable: the codes 1, 2 and 3 are labels, their order is arbitrary, and swapping two programs would change the mean although nothing has changed in the data. The legitimate summaries are the MODE and the breakdown in percentages.

No. The degree Celsius is an INTERVAL scale, whose zero is conventional: ratios have no meaning on it. In kelvins, a ratio scale, $\dfrac{24+273.15}{12+273.15}=\dfrac{297.15}{285.15}\approx 1.042$, that is 4.2% more and not double. A ratio that changes with the unit is not a property of the world. The DIFFERENCE, on the other hand, is 12 degrees in both systems.

The test-retest correlation measures RELIABILITY, specifically temporal stability. It says nothing about VALIDITY: an instrument that was in fact measuring general optimism, and not school engagement, would be just as stable. A reliable but invalid instrument is the most dangerous case, because the consistency of the results gives an illusion of rigour.

$n=\dfrac{1.96^{2}\times 0.25}{0.035^{2}}=\dfrac{0.9604}{0.001225}\approx 784$ respondents. We take $p=0.5$ for lack of prior information, because $p(1-p)$ is maximal there, so the size obtained guarantees the margin whatever the true proportion.

$n_{c}=\dfrac{784}{1+\dfrac{783}{5{,}400}}=\dfrac{784}{1.14500}\approx 685$ respondents. Since the sample represents more than 12% of the population, the correction is legitimate and saves about a hundred questionnaires.

Proportional allocation over 685: $0.55\times 685\approx 377$, $0.32\times 685\approx 219$ and $0.13\times 685\approx 89$. The sum is indeed 685, and this check should always be made, because a misplaced rounding easily slips through.

$\dfrac{685}{0.28}\approx 2{,}447$ invitations. Out of 5,400 students, that means approaching about 45% of the population, which is feasible online but requires planning two reminders, without which the 28% rate will not be reached.

No. A margin of 8 points on a proportion requires $\dfrac{0.9604}{0.0064}\approx 150$ respondents in the third sector alone, yet proportional allocation gives it only 89, which would produce a margin of $1.96\sqrt{\dfrac{0.25}{89}}\approx 0.1039$, that is 10.4 points. Two solutions: OVERSAMPLE that sector by allocating it 150 questionnaires and then reweight at the analysis stage, or raise the total sample to about $\dfrac{150}{0.13}\approx 1{,}154$ respondents. The first is far less costly.

Total: $28+38+26+16+18+12=138$. Densities, frequency divided by width: $\dfrac{28}{4}=7.0$; $\dfrac{38}{4}=9.5$; $\dfrac{26}{4}=6.5$; $\dfrac{16}{4}=4.0$; $\dfrac{18}{8}=2.25$; $\dfrac{12}{16}=0.75$ students per hour.

The DENSEST class is $\left[4, 8\right)$, with 9.5. On a histogram with frequency-height bars, the class $\left[16, 24\right)$ would look comparable to $\left[12, 16\right)$, 18 against 16, although it is nearly half as dense (2.25 against 4.0), and above all the last class, twice as wide again, would occupy a misleading area. That is why the height must equal the density as soon as the widths differ.

Midpoints: 2, 6, 10, 14, 20 and 32. Weighted sum: $2(28)+6(38)+10(26)+14(16)+20(18)+32(12)=56+228+260+224+360+384=1{,}512$. Mean: $\dfrac{1512}{138}\approx 10.96$ hours. The calculation assumes a uniform spread within each class, a fragile assumption for the last one, which is very wide.

Cumulative frequencies: 28, 66, 92, 108, 126, 138. The 69th observation, the middle of 138, falls in the class $\left[8, 12\right)$, where the cumulative count goes from 66 to 92. Interpolation: $8+\dfrac{69-66}{26}\times 4=8+0.46\approx 8.46$ hours.

The mean, 10.96, clearly exceeds the median, 8.46: the distribution is SKEWED TO THE RIGHT. The long tail of heavy volunteers pulls the mean upward without moving the median, the usual shape of any variable bounded at zero with no upper limit. The honest summary of such a profile is the median, possibly accompanied by the quartiles.

$z=\dfrac{79-68}{8}=1.375$. R score: $\left(1.375\times 0.95+0.7\right)\times 5+25=\left(1.30625+0.7\right)\times 5+25=2.00625\times 5+25\approx 35.03$. This is a very good score, at the level required by the most competitive programs.

A Z score of 1.375 leaves about 91.5% of the distribution below it, a value read from the normal table between the known landmarks $z=1$ near rank 84 and $z=1.5$ near rank 93. The percentile rank is therefore about 91 or 92: the student outperforms roughly nine classmates out of ten.

$z=\dfrac{79-74}{8}=0.625$, so $R=\left(0.625+1.2\right)\times 5+25=1.825\times 5+25=34.125$. The same mark in a stronger group gives an almost identical score, 34.1 against 35.0: the Z score falls from 1.375 to 0.625, a loss of 0.75 of a Z point, but the IFG rises by 0.5, which makes up two thirds of the loss. That is exactly the role of the group strength indicator.

Sum of the units: $2+2+2.33+1=7.33$. Weighted sum: $2(32.9)+2(28.0)+2.33(30.5)+1(25.0)=65.8+56.0+71.065+25.0=217.865$. Record: $\dfrac{217.865}{7.33}\approx 29.72$. A simple mean of the four scores would give 29.10: weighting by the units shifts the result by more than half a point, which is far from negligible in a competitive program.

The R score is an INTERVAL scale, built by the affine transformation $5z+25$ whose zero is purely conventional: ratios have no meaning on it. The difference, however, can be interpreted: 32 and 16 are 16 points apart, that is $\dfrac{16}{5}=3.2$ standard deviations, a considerable gap that is enough to describe the situation without using the word “times”.

Fee index: $\dfrac{1{,}894}{1{,}480}\times 100\approx 128.0$. Aid index: $\dfrac{2{,}491}{2{,}350}\times 100\approx 106.0$. Since both series start at 100, the comparison is immediate, whereas raw dollars, 1,894 and 2,491, cannot be compared.

Ratio of the indices: $\dfrac{128.0}{114.0}\approx 1.123$. The fees therefore rose about 12% faster than the general price level: in constant dollars, they actually grew by 12.3%. It is this quotient that makes sense, not the difference 128 minus 114, which would give “index points” depending on the base chosen.

$1{,}894\times \dfrac{100}{114}\approx 1{,}661.40$ dollars of the first year, against the 1,480 actually paid then. The real increase is therefore $\dfrac{1661.40-1480}{1480}\approx 12.3\%$, which confirms the previous calculation by another route.

$\left(1+t\right)^{5}=\dfrac{1894}{1480}=1.27973$, so $1+t=1.27973^{1/5}\approx 1.05055$ and $t\approx 5.06\%$ per year. Watch the exponent: six annual values represent only FIVE intervals of change, and taking 6 would give 4.2%, a classic mistake.

No. To keep up with inflation, the aid would have had to reach $2{,}350\times 1.14=2{,}679$ dollars, whereas it reaches only 2,491: 188 dollars are missing, 7.0% of the required amount. In constant dollars, the aid has therefore FALLEN by $\dfrac{1.060}{1.140}\approx 0.930$, that is 7.0%, while the real fees rose by 12.3%. The cumulative purchasing-power gap on this item thus exceeds 19 points over the period.

Pass rates: $\dfrac{117}{150}=78.0\%$, $\dfrac{114}{160}=71.25\%$ and $\dfrac{47}{90}\approx 52.2\%$. The trend is monotonic and strong: 26 points separate the two extremes, and the main drop occurs at the step to 15 hours. We use row percentages because the hours of work are the independent variable.

Expected frequencies: $\dfrac{150\times 278}{400}=104.25$ and $\dfrac{150\times 122}{400}=45.75$; $\dfrac{160\times 278}{400}=111.2$ and $\dfrac{160\times 122}{400}=48.8$; $\dfrac{90\times 278}{400}=62.55$ and $\dfrac{90\times 122}{400}=27.45$. All of them are well above 5, the observations are independent and the calculation is on counts: the three conditions are met.

Contributions: $\dfrac{\left(117-104.25\right)^{2}}{104.25}=\dfrac{162.5625}{104.25}\approx 1.559$; $\dfrac{162.5625}{45.75}\approx 3.553$; $\dfrac{\left(114-111.2\right)^{2}}{111.2}=\dfrac{7.84}{111.2}\approx 0.0705$; $\dfrac{7.84}{48.8}\approx 0.1607$; $\dfrac{\left(47-62.55\right)^{2}}{62.55}=\dfrac{241.8025}{62.55}\approx 3.866$; $\dfrac{241.8025}{27.45}\approx 8.809$. Total: $\chi^{2}\approx 18.02$, with $df=2$. Since $18.02$ exceeds 13{,}816, the critical value at the 0.1% level with 2 degrees of freedom, we reject independence, with a p-value below 0.001.

The cell “15 h or more, fails at least one course” contributes 8.809, that is $\dfrac{8.809}{18.02}\approx 48.9\%$ of the total. Together with its neighbour in the same row, the profile of heavy workers explains about 70% of the departure from independence: it is there, and nowhere else, that the association is decided.

$V=\sqrt{\dfrac{18.02}{400\times 1}}=\sqrt{0.045}\approx 0.212$, a MODERATE association. Conclusion: “The pass rate in all courses falls from 78.0% among students with no job to 52.2% among those who work 15 hours or more. The association is significant, $\chi^{2}=18.0$ with 2 degrees of freedom, p below 0.001, and of moderate strength, $V=0.21$; since the design is cross-sectional, it establishes no cause-and-effect link.”

$E=1.96\sqrt{\dfrac{0.55\times 0.45}{900}}=1.96\sqrt{0.000275}=1.96\times 0.016583\approx 0.0325$, that is 3.25 points. Interval: $\left[51.75\%, 58.25\%\right]$. The lower bound exceeds 50%, so the majority is secured at the 95% level, with a cushion of about 1.75 points above the bar.

$E=1.96\sqrt{\dfrac{0.62\times 0.38}{150}}=1.96\sqrt{0.0015707}=1.96\times 0.039632\approx 0.0777$, that is 7.8 points, nearly two and a half times the overall margin. The subgroup's interval runs from 54.2% to 69.8%: YES is in the majority there, but its exact lead remains very imprecise.

Variance of the gap: $\dfrac{0.55\times 0.45+0.45\times 0.55+2\times 0.55\times 0.45}{900}=\dfrac{0.2475+0.2475+0.495}{900}=\dfrac{0.99}{900}=0.0011$. Margin: $1.96\sqrt{0.0011}=1.96\times 0.033166\approx 0.065$, that is 6.5 points. The observed gap of 10 points exceeds it by a wide margin: the lead of YES is well established. Note that the margin on the gap is exactly double the simple margin, which is the limiting case of two complementary proportions.

Variance of the difference between two independent samples: $\dfrac{0.49\times 0.51}{900}+\dfrac{0.55\times 0.45}{900}=0.00027767+0.000275=0.00055267$. Margin: $1.96\times 0.023509\approx 0.0461$, that is 4.6 points. The observed rise is 6 points, so the interval runs from $+1.4$ to $+10.6$: it does not contain zero, and the rise is established. Simply looking at the two individual intervals, which overlap, would have led to the opposite conclusion.

$\sqrt{\dfrac{5{,}400-900}{5{,}400-1}}=\sqrt{\dfrac{4500}{5399}}\approx 0.91296$, so $E_{c}=0.0325\times 0.91296\approx 0.02967$, that is 2.97 points. Since the sample represents one sixth of the population, the correction gains nearly three tenths of a point, which strengthens the conclusion of question a) further.

This is a PRE-EXPERIMENTAL before-and-after design with no control group, the weakest of all. Its main threat is REGRESSION TOWARD THE MEAN: the group was selected on an extreme performance, so part of its poor result comes from passing circumstances that will not repeat. Maturation and testing, which act in the same direction, come on top.

$z_{1}=\dfrac{52-70}{12}=-1.5$. Prediction: $z_{2}=0.65\times \left(-1.5\right)=-0.975$, so a mark of $70+\left(-0.975\right)\times 12=70-11.7=58.3$. The group should therefore climb back 6.3 points without anyone doing anything.

The group scores 62, that is 10 points more than on the first exam, but only $62-58.3=3.7$ points more than regression toward the mean alone predicted. Nothing solid can therefore be concluded: 3.7 points over 40 students, with no control group and no measure of uncertainty, do not distinguish a real effect from a fluctuation. The raw figure of 10 points overstates the effect by a factor of nearly three.

It would take a WAITLIST EXPERIMENT: among the 80 weakest students, 40 are drawn at random for the immediate workshop, the other 40 getting it the following term. The two groups are then selected on the same criterion, so subject to the same regression toward the mean, and the difference between them measures the workshop's own effect. This requires the institution to accept delaying the service for half the students, and the two groups not to contaminate each other.

Standard error of the difference: $\sqrt{\dfrac{121}{40}+\dfrac{121}{40}}=\sqrt{3.025+3.025}=\sqrt{6.05}\approx 2.4597$. Margin: $1.96\times 2.4597\approx 4.82$ points. The observed gap is $63.5-60.2=3.3$ points, smaller than the margin: the interval, from $-1.52$ to $+8.12$, contains zero and the effect is not significant. Effect size: $d=\dfrac{3.3}{11}=0.30$, a small effect. The honest conclusion is that the workshop may produce a small to moderate effect, but that 40 students per group are not enough to demonstrate it: about 175 students per group would be needed to detect a $d$ of 0.30 with 80% power, since $n=\dfrac{2\left(1.96+0.84\right)^{2}}{0.30^{2}}\approx 174.2$, rounded up to 175.

FALSE. The chi-square measures SIGNIFICANCE, not strength: with a constant structure, it is proportional to the sample size. Here, $V=\sqrt{\dfrac{18.0}{400}}\approx 0.21$, a moderate association. Correct statement: the association is significant, p below 0.001, and of moderate strength, $V=0.21$.

FALSE. The margin of 3.25 points is that of the full sample of 900 respondents. On 150 respondents, the margin is $1.96\sqrt{\dfrac{0.62\times 0.38}{150}}\approx 7.8$ points. Correct statement: among the 150 students of the technical sector, YES is estimated at 62%, plus or minus 7.8 points.

FALSE. Relative changes are not subtracted, they are divided: $\dfrac{1.28}{1.14}\approx 1.123$, that is a real increase of 12.3% and not 14%. The gap, here 1.7 points, grows with the size of the changes. Correct statement: in constant dollars, fees rose by about 12.3%.

FALSE. Since the group was selected on an extreme performance, regression toward the mean alone predicts a gain of 6.3 points. The net gain is only 3.7 points, measured with no control group and no uncertainty. Correct statement: the group gained 10 points, about 6.3 of them attributable to regression toward the mean alone; concluding would require a control group selected on the same criterion.

FALSE, frequencies cannot be compared directly when the widths differ. The 16-to-24-hour class, 8 wide, has a density of $\dfrac{30}{8}=3.75$ students per hour, against $\dfrac{16}{4}=4.0$ for the 12-to-16-hour class. The second is therefore LESS numerous and DENSER. Correct statement: the first class contains more students because it is twice as wide; at equal width, it contains fewer.

This is a QUASI-EXPERIMENTAL design with voluntary participation, and its main threat is SELECTION BIAS. Students who sign up for support differ from the others in motivation, awareness of their difficulties and availability, all of which are linked to success. The 13-point gap mixes the effect of the program with the effect of these characteristics, and no statistical test can separate them.

$V=\sqrt{\dfrac{8.9}{620\times 1}}=\sqrt{0.014355}\approx 0.120$. The association is therefore WEAK, just above the threshold of negligibility, whereas the report presents it as a demonstration. This is the usual contrast between a spectacular p-value and a modest strength, made possible by a sample of 620 people.

The survey interviews only the PARTICIPANTS, that is, only those who stayed in the program: those who left it, precisely those it did not suit, are not in the database. On top of that comes nonresponse within the participants themselves, which is also selective. We would need to know the response rate, the number of initial enrolments and the number of dropouts: a rate of 91% computed on 40% of the remaining enrolees describes only a satisfied minority, and says nothing about the program.

Because the number of participants measures the program's POPULARITY and not its effectiveness. A program can double its attendance because it is better advertised, because difficulties are increasing, or because it has become mandatory for some files. Worse, the sentence suggests growth to be credited to the program, whereas rising attendance with stable success would be a bad sign. The figure is true, it is simply unrelated to the question asked.

“Among 620 students, the pass rate of program participants reaches 74%, against 61% among non-participants, a gap of 13 points. The association is statistically significant, $\chi^{2}=8.9$, $df=1$, $p=0.003$, but of weak strength, $V=0.12$. Since participation is voluntary, this gap measures both the effect of the program and the pre-existing differences between those who enrol and the others; it cannot therefore be attributed to the program. For the next term we propose random assignment by waitlist among applicants, the only arrangement that will measure the program's own effect.”