Q1. What is your household's annual after-tax income (from any source)? a) $0 - $50,000; b) $50,001 - $100,000; c) $100,001 - $150,000; d) $150,001 - $200,000; e) $200,001 or more;

It should come as not surprise to anyone studying sociology that how people operationalize gender has changed over time. Until recently, the conventional operationalization went something like this:

Imagine you are filling out a questionnaire and one of the questions is about age, like this:

	Age	Year at university	Employment	GPA	Major (by Faculty)	Language spoken at home
Arjun	20	1	yes	3.6	Business	Punjabi
Benjamin	25	3	yes	3.2	Science	German
Cecilia	22	4	no	4.0	Health	Spanish
Xingxing	19	1	yes	3.3	Arts	Mandarin

	Variable 1	Variable 2	Variable 3	Variable 4	Variable 5	Variable 6
Respondent #1	Response1.1	Response2.1	Response3.1	Response4.1	Response5.1	Response6.1
Respondent #2	Response1.2	Response2.2	Response3.2	Response4.2	Response5.2	Response6.2
Respondent #3	Response1.3	Response2.3	Response3.3	Response4.3	Response5.3	Response6.3
Respondent #4	Response1.4	Response2.4	Response3.4	Response4.4	Response5.4	Response6.4

	X1	X2	X3	X4	X5	X6
I1	x11	x21	x31	x41	x51	x61
I2	x12	x22	x32	x42	x52	x62
I3	x13	x23	x33	x43	x53	x63
I4	x14	x24	x34	x44	x54	x64

	X1	X2	X3	X4	X5	X6	X7	...	XK
I1	x11	x21	x31	x41	x51	x61	x71	...	xk1
I2	x12	x22	x32	x42	x52	x62	x72	...	xk2
I3	x13	x23	x33	x43	x53	x63	x73	...	xk3
I4	x14	x24	x34	x44	x54	x64	x74	...	xk4
I5	x15	x25	x35	x45	x55	x65	x75	...	xk5
I6	x16	x26	x36	x46	x56	x66	x76	...	xk6
I7	x17	x27	x37	x47	x57	x67	x77	...	xk7
...	...	...	...	...	...	...	...	...	...
IN	x1n	x2n	x3n	x4n	x5n	x6n	x7n	...	xkn

N= number of elements in the dataset K= number of variables in the dataset

  Snapshot of APS 2012's Data View in SPSS:     Snapshot of APS 2012's Variable View in SPSS:  

Table 2.1 Educational Attainment by Frequency 

Degree	Count (a.k.a. frequency)
No degree	1
Secondary/High School	6
Associate's	3
Bachelor's	5
Master's	2
PhD	1
Didn't answer	3
TOTAL	21

Table 3.1 Religious Affiliation of Canadian Prime Ministers (Wikipedia 2017)

Religious affiliation	Frequency
Anglican	4
Baptist	3
Evangelical	1
Presbyterian	3
Roman Catholic	10
United Church of Canada (prev. Methodist)	2
TOTAL	23

  What is the most popular religious affiliation of Canadian Prime Ministers as of 2019? Or, what religious affiliation is most frequently reported by Canadian Prime Ministers so far? In other words, what religious affiliation do Canadian Prime Ministers most have tended to have?   Surprising no one with any knowledge about Canada, the largest category among the religious denominations, or the one that Canadian Prime Ministers most frequently subscribe to -- i.e., the category with the highest frequency -- is "Roman Catholic", with 10 of the Canadian Prime Ministers identified as such. (And are you surprised that Canada has only had Christian Prime Ministers?)

N=3   a) Let's say we have three students at different levels of their studies: one is a first-year, the second one a fourth-year, and the third a third-year. Before we do anything else, we need to establish the correct order. We rearrange the students properly:   (1) a first-year student (2) a third-year student           ← median (3) a fourth-year student   The case in the middle is Case #2, the second one on the list (as there is one student below and one student above), i.e., the third-year student. Thus we have established that the median category is "third year of study". That is, half of the students are below the third year of study and half are above (as odd as it sounds when we only have three cases).

N=8   c) What if I complicate matters further? What if I add three more students to the group, say, two second-years and a fourth-year? Their order will be:   (1) a first-year student (2) a first-year student (3) a second-year student                 The median is between (4) a second-year student (new)      ← this case (5) a second-year student (new)      ← and this case (6) a third-year student (7) a third-year student (new) (8) a fourth-year student   If you go by the same logic as above, you'll quickly find that there is no "middle" student: unlike before, the students now are an even number. The middle of the group actually falls between Cases #4 and Case #5, the fourth and the fifth cases on the list (so that four are below and four above it). Since both the fourth and the fifth students are second-year, we can conclude that, again, the median is "second-year of study".  Had the fourth and the fifth student been different years of study, we could say that the median was between their respective categories.

a) N=3   (1) a first-year student (2) a third-year student (3) a fourth-year student   According to the formula,   $$\frac{N+1}{2}=\frac{3+1}{2}=\frac{4}{2}=2 $$   That is, the "numbered position of the median case in the ordered list of cases" is equal to 2; the middle case is Case #2, the second one on the list, or like we established before, the third-year student.   b) N=5   (1) a first-year student (2) a first-year student (new) (3) a second-year student  (new) (4) a third-year student (5) a fourth-year student   According to the formula,   $$\frac{N+1}{2}=\frac{5+1}{2}=\frac{6}{2}=3 $$   That is, the "numbered position of the median case in the ordered list of cases" is equal to 3; the middle case is Case #3,  the third one on the list, or again, the second-year student.   c) N=8   (1) a first-year student (2) a first-year student (3) a second-year student (4) a second-year student (new) (5) a second-year student (new) (6) a third-year student (7) a third-year student (new) (8) a fourth-year student   According to the formula,   $$\frac{N+1}{2}=\frac{8+1}{2}=\frac{9}{2}=4.5 $$   That is, the "numbered position of the median case in the ordered list of cases" is equal to 4.5. Considering we have discrete numbers (after all, the cases are individuals), there is no case number 4.5. Instead, we say that the median falls between Case #4 and Case #5, the fourth and fifth cases on the list, or between two second-year students, so it is "second year of study".

Imagine you talk to seven of your friends and ask them about the number of siblings they have. Let's say these are the responses you receive: 2, 1, 4, 2, 1, 0, 3. That is, two friends report having two siblings each, two friends report having one sibling each, and three of your friends report having four, zero, and three siblings each.   To find the median, the first thing we need to do is put the responses in order:   (1) 0 (2) 1 (3) 1 (4) 2 (5) 2 (6) 3 (7) 4   Whether you visually identify Case #4 as the middle case (three cases below and three cases above it) or use the formula ($\frac{N+1}{2}=\frac{7+1}{2}=\frac{8}{2}=4$) to obtain the same result, it is clear that the median is "two siblings": half of your friends in this example have fewer than two siblings, and half have two or more siblings.  

  If you recall our Example 3.3 (A) from the previous Section 3.2 ( https://pressbooks.bccampus.ca/simplestats/chapter/3-2-median/ ‎), you imagined yourself asking seven of your friends about the number of siblings they had. We imagined the responses as follows: 2, 1, 4, 2, 1, 0, 3. We had to put these values in order to be able to find the median, but the mean works either way, whether the values are in order or not.   To calculate the average number of siblings your imagined friends have, we simply add all responses together and divide them by the total number of friends, i.e., by 7:   $$\frac{(2+1+4+2+1+0+3)}{7}=\frac{13}{7}=1.86$$   That is, your imagined friends have 1.86 siblings on average (or not quite but closer to two, rather than one siblings on average). We could also say that the mean of number of siblings is 1.86.

  Depending on the courses you take in a semester, what you pay for books will vary but let's say we're interested in how much you pay for books in a typical semester. Perhaps you are very-well organized and want to finish your degree as quickly as possible so you have decided to take five courses per semester. For simplicity's sake, let's assume your were assigned one book per course. These are the books' prices: \$120, \$230, \$300, \$65, \$30. How much did you pay for a book on average?   $$\frac{(120+230+300+65+30)}{5}=\frac{745}{5}=149$$   That is, despite the fact that some of your books were expensive (like the \$300 one), and some relatively cheap (like the \$30 one), the average price you paid for a book in that semester was \$149.

  Arranging the raw data from Example 3.4 (A) above, we again get the following table.

Value	Frequency
0	1
1	2
2	2
3	1
4	1
Total	7

According to the formula for the mean, we need to add all values together and then divide their sum by their total number. When the values are disaggregated (i.e., raw), we can proceed to adding them up right away. However, when they are grouped by frequency, we first need to multiply each value by its respective frequency, and then add the value-times-frequency products together, before dividing them by the total number, like this:   $$\frac{\sum\limits_{i=1}^{N}{x_i}}{N}=\frac{(0+1+1+2+2+3+4)}{7}=\frac{0(1)+1(2)+2(2)+3(1)+4(1)}{7}=\frac{13}{7}=1.86=\overline{x}$$   Again, the average number of siblings of these seven friends is 1.86, as previously calculated.

  Imagine you are doing a survey for one of your class assignments and one of the questions is about age. You aggregate the data by frequency and you get the following table.   Table 3.5 Mean for Age of Classmates

Value	Frequency
19	1
20	10
21	12
22	8
25	2
27	1
35	1
TOTAL	35

By the formula, we have: $\frac{\sum\limits_{i=1}^{N}{x_i}}{N}=\frac{19(1)+20(10)+21(12)+22(8)+25(2)+27(1)+35(1)}{35}=\frac{19+200+252+176+50+27+35}{35}=\frac{759}{35}=21.69=\overline{x}$   Or, now you know that the average age of your classmates in that class is 21.69 years, or a bit less than 22 years.

Recall Example 3.5 from Section 3.4 (https://pressbooks.bccampus.ca/simplestats/chapter/3-4-mean/)  where we calculated the mean price of textbooks we imagined you paid in a particular semester. The books' prices were \$120, \$230, \$300, \$65, \$30. The cheapest book (i.e., the lowest value, $x_{min}$) was \$30 and the most expensive book (i.e., the highest value, $x_{max}$) was \$300. Thus   $$x_{max}-x_{min} = 300 - 30 = 270 = range$$   That is, now we have found that the range of textbook prices for that semester was \$270, with prices you paid ranging between \$30 and \$300.

  If you recall, we had imagined you as a research assistant (RA) on a research project and you had worked 20 weeks in total in the last two semesters, ten weeks in each semester. The maximum hours per week you could work was 15, limited by the nature of your contract.   As there are a lot of calculations to be done, to simplify our job, let's imagine further that we're interested in only one of the two semesters you had worked, and these are only the hours in the ten weeks of that one semester:   3, 3, 5, 7, 8, 10, 12, 12, 13, 14   Considering that for Step 1A we need the distances of each of these ten values to the mean, we'll calculate the mean as a preliminary requirement.[footnote]Since N=10 or more makes for quite the long equations if the values are listed (summed) one by one separately, from now on I will group values by frequencies in the calculations I do as a matter of principle. (I.e., instead of 3+3, here I have (3)2, instead of 7+7+7, I would have (7)3, etc.) Coincidentally, this is exactly what we do when working with data organized in a frequency table.[/footnote]   \begin{equation*} \begin{aligned} & \frac{\sum\limits_{i=1}^{N}{x_i}}{N} = \\ &= \frac{(3)2+5+7+8+10+(12)2+13+14)}{10} = \\ &=\frac{(6+5+7+8+10+24+13+14)}{10}=\frac{87}{10}=8.7=\overline{x} \end{aligned} \end{equation*}   Armed with the mean of 8.7 hours, we can now proceed to calculate the distance of every value to the mean (i.e., subtract the mean from each value to obtain the difference). I list the values and their respective distances from the mean in the table below.   Table 4.3 Step 1A Calculating Distances To the Mean

$x_i$	$(x_i - \overline{x})$
3	(3 - 8.7) = -5.7
3	(3 - 8.7) = -5.7
5	(5 - 8.7) = -3.7
7	(7 - 8.7) = -1.7
8	(8 - 8.7) = -0.7
10	(10 - 8.7) = 1.3
12	(12 - 8.7) = 3.3
12	(12 - 8.7) = 3.3
13	(13 - 8.7) = 4.3
14	(14 - 8.7) = 5.3

Again, as usual, $x_i$ is the value of each and any Case #$i$ (from 1 to 10), and $(x_i - \overline{x})$ is the distance (i.e., difference) between each and any Case #$i$ (from 1 to 10) to the mean.

  A reminder: what we are trying to get is a dispersion measure giving us an average distance of the cases to the mean; something to account for the variability of all cases, not just a few (unlike the range and IQR). To make the calculations look more orderly, I add a third column to Table 4.3 above, one with the squared distances. Thus, our mysterious Step 1B is squaring each individual distance.   Table 4.4 Step 1B Squaring Individual Distances

$x_i$	$(x_i - \overline{x})$	$(x_i - \overline{x})^2$
3	(3 - 8.7) = -5.7	(-5.7)2 = 32.5
3	(3 - 8.7) = -5.7	(-5.7)2 = 32.5
5	(5 - 8.7) = -3.7	(-3.7)2 = 13.7
7	(7 - 8.7) = -1.7	(-1.7)2 = 2.9
8	(8 - 8.7) = -0.7	(-0.7)2 = 0.5
10	(10 - 8.7) = 1.3	(1.3)2 = 1.7
12	(12 - 8.7) = 3.3	(3.3)2 = 10.9
12	(12 - 8.7) = 3.3	(3.3)2 = 10.9
13	(13 - 8.7) = 4.3	(4.3)2 = 18.5
14	(14 - 8.7) = 5.3	(5.3)2 = 28.1

  We are thus ready for Step 2: summing up the (now-squared) distances from the mean:   $\sum\limits_{i=1}^{N}{(x_i-\overline{x})^2} =(32.5)2+13.7+2.9+0.5+1.7+(10.9)2+18.5+28.1=\\ = 152.1 =$      ← Sum of Squares    As you can see above, the sum of the squared distances from the mean is called the sum of squares (sometimes indicated by SS).   Finally, to get the average distance from the mean we need Step 3: to divide the sum of squares by the total number, N:   $\frac{\sum\limits_{i=1}^{N}{(x_i-\overline{x})^2}}{N}= \frac{152.1}{10} = 15.21 =\sigma^2 =$      ← variance   That is, the variance of your hours worked per week is 15.21, or the average of the squared distances from the mean is 15.21. (Note that we cannot say 15.21 hours as now we are working in squared units.)  

Imagine you are enrolled in a class along with 49 other students, so the total class size is 50. Let's say as a class assignment (perhaps in a research methods class) you are tasked with taking a sample of your class and administering a survey to your sample. In this sense, your class is your population of interest. For simplicity's sake, we focus on one possible question, say, age of respondent. You want to know the average age of the study population but, instead of asking all 50 of your classmates, you draw a random sample of them for the purposes of estimating the class's average age.[footnote]Of course, with a population of only 50 in real life you can just collect the information from everyone. I'm using a small-size example for teaching purposes only and to make calculations manageable. The principle of sampling applies equally not only to a population of 50 but of any size -- and when your study population's size is in the millions, you wouldn't attempt to survey all of them (barring the already discussed case for censuses).[/footnote] Now despite that I still haven't said anything about sample size (but we're getting there), I'll assume that a sample size of 10 (i.e., 20 percent of the population) would sound reasonable enough to you. The random draw yields the following ten classmate's ages: 19, 19, 20, 20, 20, 21, 21, 22, 23, 28 Based on these values, the average age of the sample, $\overline{x}$, is $\overline{x}=\frac{\sum\limits_{i=1}^{N}{x_i}}{N}=\frac{(19)2+(20)3+(21)2+22+23+28}{10}= \frac{213}{10}=21.3$ I.e., your sample's average age is 21.3 years. Considering that these ten people were randomly drawn, and that they are, well, only ten, can we assume that the average age of your entire class of 50 is 21.3 years? While this is a good -- educated even -- guess and a good starting point, it is unlikely that, had you polled everyone in the class, your calculation would have produced exactly 21.3. After all, polling 10 people is not the same as polling 50; in the latter case your calculation would include a lot more information than in the former. Thus, it's also reasonable to expect that there will be some difference between the mean based on the sample, $overline{x}$, and the true population mean, μ. The how about if you decided to draw another random sample of ten people out of your class? Would you expect to have the exact same mean of 21.3 years? Unless you somehow end up with the exact same ten people who were in the first sample (and after Chapter X on probability you should know how minuscule that probability is), it is again unlikely you'd get the same mean. We could easily imagine that the new, second sample's ages might look like this: 18, 19, 19, 19, 20, 20, 22, 22, 24, 25 Based on these ten new values, the average age of the second sample (let's call it $\overline{x_2}) is: $\overline{x_2}=\frac{\sum\limits_{i=1}^{N}{x_i}}{N}=\frac{18+(19)3+(20)2+(22)2+24+25}{10}= \frac{208}{10}=20.8$ I.e., your second sample's average age is 20.8 years, despite it being drawn from the same population. Your two samples (of the same size) yielded two close -- but still different -- numbers. As well, following the same logic as before, it's just as unlikely that the population mean μ (your class's average age) is 20.8 years as it was unlikely that it's 21.3 years (the sample is still only 10 people).

Let's say that university students on average spend \$250 for a statistics textbook, with a standard deviation of \$100 -- i.e., we assume to know the population parameters: μ = 250 and σ = 100 We draw a random sample of N=1,600 students. We want to know the probability for that sample to have a specific mean price paid for statistics textbooks. To get that probability, we first need the standard error, $\sigma_\overline{x}$: $\sigma_\overline{x}$ $=\frac{\sigma}{\sqrt{N}}=\frac{100}{\sqrt{1600}}=\frac{100}{40}=2.5$ Next, we can draw the sampling distribution: bell-shaped, centered on μ, and with a (standard deviation called) standard error of \$2.5. Applying what we know about the normal distribution in terms of the probability under the curve, we get the following Fig. XX. Fig. XX The Sampling Distribution of the Mean Price of Statistics Textbooks   That is, we see that 68% of the sample mean prices of statistics textbooks in hypothetical repeated sampling would fall between \$247.5 and \$252.5[footnote]That is, 250-2.5=247.5 and 250+2.5=252.5.[/footnote] (i.e., within 1 standard error away from the mean, denoted with green in Fig. XX) and 95% of the sample means will fall between \$245 and \$255[footnote]That is, 250-2(2.5)=250-5=245 and 250+2(2.5)=250+5=255.[/footnote] (i.e., within 2 standard errors away from the mean, denoted with blue in Fig. XX). Since this is just a heuristic way to imagine the sampling distribution, we can restate our finding more correctly: a single, one-off sample mean will fall between \$247.5 and \$252.5 68% of the time, and between \$245 and \$255 95% of the time. Or, even more precisely, we have a 68% probability that the average paid price for statistics books obtained from a random sample of 1,600 students will be between \$247.5 and \$252.5, and a 95% probability that it will be between \$245 and \$255. This means that we have a 95% chance that the sample mean, $\overline{x}$, will fall within \$10 (i.e., ±\$5) of the population mean, μ. Quite good as far as predictions go, eh?

Imagine you are interested in the average annual income in a medium-size city. You randomly select N=1,600 people, and ask them about their annual income. You then calculate the mean of the resulting variable as \$50,000, and the standard deviation as \$12,000. I.e., $\overline{x}=50,000$ and s = 12,000 As a first guess, you could say that the average annual income in the city is \$50,000. However, since we know this is an estimate, and random error exists, you can do better: you can also provide information about how certain you are about your estimate along with some margins for error. To do that, you need to draw the sampling distribution of the mean. Following the CLT, you draw the sampling distribution as a normal curve centered on \$50,000. At this point, you also need information about the sampling distribution's dispersion, i.e., its standard error. You substitute the s you do know for the σ you don't[footnote]Recall that a "hat" over a symbol indicates it being estimated.[/footnote]: $\hat\sigma_\overline{x}$ $=s_\overline{x}$ $=\frac{s}{\sqrt{N}}= \frac{12000}{\sqrt{1600}}=\frac{12000}{40}=300$ Fig. XX shows the resulting sampling distribution. Fig. XX. Average Annual Income   Based on the figure above (and following the same logic as in the previous Example XX), you find that the average annual income of the city's population will be between \$49,400 and \$50,600 with 95% probability.[footnote]We get these bounds (i.e., within two standard errors away from the mean) through 50,000-2(300)=50,000-600=47,400 and 50,000+2(300)=50,000+600=50,600.[/footnote] That is, you can be 95% confident that the city's average annual income will be within \$1,200 of the sample average of \$50,000, or, that the city's average annual income is \$50,000 ±\$600, with 95% certainty.

Going back to our Average Annual Income example, we had that $N=1600$ $\overline{x}=50000$ $s=12000$ We had also already calculated its 95% CI:

• 95% CI: $\overline{x}\pm1.96\times\hat\sigma_\overline{x}$ $=50000\pm1.96\times300=50000\pm588= (49412; 50588)$.

What would happen, if we increased the sample size to, say, N=10,000? As usual, we start with calculating the standard error: $\hat\sigma_\overline{x}$ $=s_\overline{x}$ $=\frac{s}{\sqrt{N}}= \frac{12000}{\sqrt{10000}}=\frac{12000}{100}=120$ Then, the new 95% CI would be

• 95% CI: $\overline{x}\pm1.96\times\hat\sigma_\overline{x}$ $=50000\pm1.96\times120=50000\pm235= (49765; 50235)$.

To be sure, the larger-N confidence interval *is* smaller; we did gain precision. But consider these numbers for what they actually are, in actual dollar terms, had this been a real-life research instead of a hypothetical example. With a sample of N=1,600 we found that, with 95% certainty, the average annual income for the population is between \$49,412 and \$50,588. We now find that had we a sample of N=10,000, the average annual income of the population would be between \$49,765 and \$50,235. The precision "gain"between the two sample sizes is \$353 on both error bounds; i.e., our estimate of average annual income of the population becomes ±\$353 more precise (a total "gain" of \$706). At the same time, consider that surveying a sample size of N=10,000 would cost more than six times more than surveying one of N=1,600 (as 10,000 is 6.25 times more than 1,600) -- would this be worth it, to only be able to improve your estimate by \$350, give or take, on both sides, when the actual sums we are dealing with are in the tens of thousands dollars magnitude?

Imagine you have a friend who is partial to the colour black so much so that they always wear a monochromatic, all-black outfit. Then one day you notice your friend is wearing a single article of a different colour, say, dark purple. Arguably, that's more variability than wearing all-black, but the outfit will still be predominantly black. Then on the next day, there are two pieces of purple amid all the black, then three, then four, and so on. At what point would your friend's outfit stop being "predominantly black" and would become "predominantly purple"? And what would happen eventually, if the exchanging-black-for-purple trend continues? The answer to the latter question is obvious: the end point of such a trend would be for the outfit to become monochromatic again, this time all-purple. Now think about variability. At what point was there the greatest and at what point was there the least amount of variability in your imaginary friend's outfit? To make it easier, let's add a numerical aspect to what we have imagines, and say that your friend's outfit consisted of 10 articles of clothing (and accessories) to start with, and then your friend swapped a black article for a purple article on each successive day, for ten days straight after that. Table XX illustrates. Table XX Black and Purple Articles of Clothing

	Black Articles	Purple Articles
Initial state	10	0
Day 1	9	1
Day 2	8	2
Day 3	7	3
Day 4	6	4
Day 5	5	5
Day 6	4	6
Day 7	3	7
Day 8	2	8
Day 9	1	9
Day 10	0	10

Again, on what day(s) would your friend's outfit be the least and the most variable in terms of colour? Looking at Table XX, it's not difficult to spot that the least variable were your friend's initial (all-black) outfit and what they wore on Day 10 (all-purple), both consisting of a single colour. There is a slight variability on Days 1 and 9 (when there was a single article of different colour); then more variability on Days 2 and 8 (when there were two articles of different colour); then even more variability on Days 3 and 7 (when your friend had three different-coloured articles); and yet even more variability on Days 4 and 6 (when there were four articles of different colour). The most outfit was most variable on Day 5, when it was half-black and half-purple, neither colour predominating. Going by "half-black and half-purple", let's restate the information in Table XX in terms of proportions, as this will help us generalize the logic without the constraint of an actual count (of 10 articles of clothing, or anything else). Table XX Black and Purple Articles of Clothing, Proportions

	Black Articles	Purple Articles
Initial state	1	0
Day 1	0.9	0.1
Day 2	0.8	0.2
Day 3	0.7	0.3
Day 4	0.6	0.4
Day 5	0.5	0.5
Day 6	0.4	0.6
Day 7	0.3	0.7
Day 8	0.2	0.8
Day 9	0.1	0.9
Day 10	0	1

One convenient way to quantify what we found in terms of the least and the largest variability is through multiplying the proportions in the two columns, like so: Table XX Black and Purple Articles of Clothing, Variability

	Black Articles	Purple Articles	Variability
Initial state	1	0	1(0)=0
Day 1	0.9	0.1	0.9(0.1)=0.09
Day 2	0.8	0.2	0.8(0.2)=0.16
Day 3	0.7	0.3	0.7(0.3)=0.21
Day 4	0.6	0.4	0.6(0.4)=0.24
Day 5	0.5	0.5	0.5(0.5)=0.25
Day 6	0.4	0.6	0.4(0.6)=0.24
Day 7	0.3	0.7	0.3(0.7)=0.21
Day 8	0.2	0.8	0.2(0.8)=0.16
Day 9	0.1	0.9	0.1(0.9)=0.09
Day 10	0	1	0(1)=0

That is, starting from zero, variability is the highest at precisely the half-and-half point, when neither outcome/category (in our example, neither colour) predominates. Now we are ready for the formula to measure the dispersion of a proportion. I demonstrate it by restating Table XX, by designating black as 1 and purple as 0, and taking black as the colour of interest (i.e., all proportion will be expressed in terms of black). Table XX Black and Purple Articles of Clothing, Generalized

	Black Articles	Non-black Articles	Variability
Initial state	1	0	1(0)=0
Day 1	0.9	(1-09)	0.9(1-0.9)=0.09
Day 2	0.8	(1-0.8)	0.8(1-0.8)=0.16
Day 3	0.7	(1-0.7)	0.7(1-0.7)=0.21
Day 4	0.6	(1-0.6)	0.6(1-0.6)=0.24
Day 5	0.5	(1-0.5)	0.5(1-0.5)=0.25
Day 6	0.4	(1-0.4)	0.4(1-0.4)=0.24
Day 7	0.3	(1-0.3)	0.3(1-0.3)=0.21
Day 8	0.2	(1-0.2)	0.2(1-0.2)=0.16
Day 9	0.1	(1-0.1)	0.1(1-0.1)=0.09
Day 10	0	(1-0)	0(1-0)=0

 

Let's say we want to know what proportion of Canadian workers work part time, and that we are especially interested in what Statistics Canada calls "the core ages" 25 to 54 (Statistics Canada, 2017 [https://www150.statcan.gc.ca/n1/pub/71-222-x/71-222-x2018002-eng.htm]). We conduct a survey of N=1,600 Canadian individuals aged 25-54 and find that 12% of our respondents work part time. As usual, we want to estimate the proportion of all Canadians aged 25-54 who work part time. We start with calculating the standard error: $\hat\sigma_p=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.12(0.88)}{1600}}=\sqrt{\frac{0.106}{1600}}=\frac{0.325}{40}=0.008$ Then, a 95% confidence interval for the proportion would be:

• 95% CI: $p \pm1.96\times\hat\sigma_p=p \pm1.96\times0.008=0.12 \pm0.016=(0.104; 0.136)$

That is, we can estimate with 95% certainty that (that is, 95% of the time such a study is undertaken) between 10.4% and 13.6% of the Canadian workers aged 25-54 work part time. Alternatively, we can say with 95% certainty that 12% ±1.6% points of Canadian workers aged 25-54 work part time.

Let's say a large, nationally-representative study of N=10,000 finds that women in Canada occupy 36% of managerial positions. [https://www.expertmarket.com/female-managers] What would be the estimate for Canada as a whole? The estimated standard error of the proportion would be: $\hat\sigma_p=\sqrt{\frac{p(1-p)}{N}}=\sqrt{\frac{0.36(0.64)}{10000}}=\sqrt{\frac{0.230}{10000}}=\frac{0.48}{100}=0.005$ As in the previous examples, the 95% confidence interval for the proportion would be:

• 95% CI: $p \pm1.96\times\hat\sigma_p=p \pm1.96\times0.005=0.36 \pm0.01=(0.35; 0.37)$

That is, we can estimate with 95% certainty (i.e, 95% of the time such a study is undertaken) that between 35% and 37% of managerial positions in Canada are occupied by women. Alternatively, we can say with 95% certainty that women occupy 36% ±0.01% points of managerial positions in Canada.[footnote]In this chapter I have presented the most commonly used interpretation of confidence intervals, and the one most frequently taught to introductory statistics students. I should point out, however, that this is one of those instances (of which I spoke in the introduction to this book) where the reality is a bit different than what is being taught. The interpretation presented here is easier to understand and follows a logic that is more intuitive to students than what confidence intervals really tell us. Briefly, the range of plausible values we find are just that -- values that the population could have, as we have't ruled them out yet, and 95% (or 99%) of the time such studies will not be able to rule these plausible values out (van der Zee, 2017 [How (Not) To Interpret Confidence Intervals, in the hyperlink]). This, technically speaking, is somewhat different than the "95% (or 99%) certainty that the population mean/proportion will be between the calculated error bounds" version we usually work with. If you'd like to go down that particular rabbit hole, go here. For everyone else, the interpretation of confidence intervals presented so far in this chapter should be enough.[/footnote] If you find this a bit too precise to believe, note the quite large sample size of N=10,000. As established above, confidence intervals based on large N and around proportions indicating not very strong variability (after all, the sample statistics indicated that managerial positions are predominantly occupied by men) tend to have small standard errors (due to the relatively small numerator (the variability) and the large denominator (the sample size)).

Hair colour and eye colour:

• H₀: Hair colour and eye colour are not associated; e.g., dark-haired individuals are equally likely to have blue eyes as blond people are.
• H_a: Hair colour and eye colour are associated; e.g., dark-haired individuals' and blond individuals' likelihood of having blue eyes is different.

In Chapter 7 we imagined that you asked 35 students in your class[footnote]Note that this of course is not a random sample; we're using it here only for illustrating how hypothesis testing works so we are effectively pretending it's random. In a real-life study, you shouldn't use non-probability samples for statistical inference.[/footnote] whether they liked the campus cafeteria: 12 of your classmates said yes (i.e., 34.3%), 7 (out of 15) first-years and 5 (out of 20) second-years (46.7% of all first-years and 25% of all second-years, respectively). We want to know whether the observed in the sample difference in proportions (0.467-0.25=0.217) is statistically significant: can it be generlized to a larger student population, or is it due to a regular sampling variability?

• H₀: The proportion of first year students who like the cafeteria is the same as the proportion of second year students who do; $\pi_1=\pi_2$.
• H_a: The proportion of first year students who like the cafeteria is different than the proportion of second year students who do; $\pi_1\neq\pi_2$.

Substituting these numbers in the formula we have: $t=\frac{(p_1 -p_2)}{\sqrt{p(1-p)(\frac{1}{N_1}+\frac{2}{N_2})}}=\frac{0.467-0.25}{\sqrt{0.343(1-343)(\frac{1}{15}+\frac{1}{20})}}=\frac{0.217}{0.162}=1.34$ With a t=1.34, df=34, and p=0.189 (i.e., p>0.05) we fail to reject the null hypothesis: at this point we don't have enough evidence to conclude there is a difference between the proportions of first and second year students who like the campus cafeteria. The 21.7 percentage points difference is not statistically significant, and has a high enough probability of being due to random chance. We can check this with a confidence interval too:

• 95% CI: $(p_1 -p_2)\pm1.96\times\sqrt{\frac{p_1(1-p_1)}{N_1}+\frac{p_2(1-p_2)}{N_2}}=0.217\pm1.96\times\sqrt{\frac{0.467(0.533)}{15}+\frac{0.25(0.75)}{20}}=0.217\pm0.316=(-0.099; 0.533)$

In other words, the difference between the proportion of first years and the proportion of second years who like the cafeteria could be anywhere between -9.9 percentage points and 53.3 percentage points with 95% confidence (or 19 out of 20 such samples will have a difference within this pretty large interval). The difference can be in favour of second years or in favour of the first years (notice the negative lower bound); it can even be 0. Thus, since a difference of 0 (i.e., no difference) is a plausible value, we cannot reject the null hypothesis. We conclude that we don't have enough evidence of an association between year of study and opinion on the campus cafeteria.

To use the imaginary data from before, we had 12 people who admitted liking the campus cafeteria food out of the 35 polled. (Since we're interested only in one of the variables, here we ignore whether the students who like the cafeteria are first- or second-years.) As such, we have the following table: Table XX Approval of the Campus Cafeteria, Observed Count (Univariate)  

Yes	12
No	23
Total	35

If you didn't know anything about the campus cafeteria and had no observations about it whatsoever -- i.e., had you been an impartial observer, as it were -- wouldn't you expect to see an approximately 50/50 split of the 35 students into the two categories? After all, there are only two groups, and an unbiased (random) distribution would be exactly like everyone flipping a coin as a manner of deciding in which group they end up. Thus, the expected count here is simply N divided by the number of groups/categories (denoted by k): $f_e=\frac{N}{k}=\frac{35}{2}=17.5$ Table XX adds the expected count in brackets next to the observed count. Table XX Approval of the Campus Cafeteria, Observed and Expected Count (Univariate)

Yes	12    (17.5)
No	23    (17.5)
Total	35

Then, according to the formula, this is what we have for each of the two groups:

• Yes-group: $\frac{(f_o-f_e)^2}{f_e}=\frac{(12-17.5)^2}{17.5}=\frac{30.25}{17.5}=1.73$
• No-group: $\frac{(f_o-f_e)^2}{f_e}=\frac{(23-17.5)^2}{17.5}=\frac{30.25}{17.5}=1.73$

Finally, to get the χ² we only need to add these two numbers together: $\chi^2=\Sigma\frac{(f_o -f_e)^2}{f_e}= \frac{(12-17.5)^2}{17.5}+\frac{(23-17.5)^2}{17.5}=1.73+1.73=3.46$ The degrees of freedom in a one-way test is k-1, where k is the number of categories/groups. In this case we have k=2, so df=1. With a χ² =3.45, df=1, and a p=0.06[footnote]You can check the significance of any χ² with a convenient online calculator, like this one here.[/footnote] (i.e., p>0.05), we fail to reject the null hypothesis. At this time, we do not have enough evidence to conclude that the observed distribution of the students is unusual enough to suggest a pattern which is different than a random variation of a 50/50 split. As such, this distribution is not statistically significant -- we cannot conclude that the students lean one way or the other in their opinion about the campus cafeteria.

While we already know that year of study and opinion on he campus cafeteria are not statistically associated from the t-test in Example XX, I will further use the imaginary data in original contingency table from Example XX to demonstrate a two-way χ²-test. Then, this was the table we had is Section 7.7.2. Table XX Do You Like The Campus Cafeteria? (Revisited)

	First Year Students	Second Year Students	Total
YES	7	5	12
NO	8	15	23
Total	15	20	35

Our hypotheses are:

• H₀: Liking the cafeteria or not is not associated with one's year of study; first- and second-year students are equally likely to lie the cafeteria, or π₁=π₂.
• H_a: Liking the cafeteria is associated with one's year of study; first-year students and second-year students differ in their liking of the cafeteria, or π₁≠π_2.

To compute the χ², we need the expected count for each cell. Unlike the one-way χ² case, however, determining the expected count in a contingency table is a bit more complicated than dividing the N on the number of groups and expecting the same (expected) number in each cell. Instead, we multiply the respective group/category sizes (i.e., the raw total and the column total at the margins) and divide the product by N (the full total)[footnote]We do that to account for the different group/category sizes.[/footnote]: $f_e=\frac{N_j\times N_k}{N}$ where j is the size of the respective group and k is the size of the respective category[footnote]Recall that to differentiate between the groups/categories of the two variables, we refer to one variable having groups and the other having categories: so that we can say we compare the groups of one variable on the categories of the other.[/footnote]. Thus we have the following:

• First-years who said "Yes": $f_e=\frac{N_j\times N_k}{N}=\frac{15\times 12}{35}=5.14$
• Second-years who said "Yes": $f_e=\frac{N_j\times N_k}{N}=\frac{20\times 12}{35}=6.86$
• First-years who said "No": $f_e=\frac{N_j\times N_k}{N}=\frac{15\times 23}{35}=9.86$
• Second-years who said "No": $f_e=\frac{N_j\times N_k}{N}=\frac{20\times 23}{35}=13.14$

Table XX adds the expected count in brackets next to the observed count. Table XX Do You Like The Campus Cafeteria? (Observed and Expected Frequencies)

	First Year Students	Second Year Students	Total
YES	7   (5.14)	5   (6.86)	12
NO	8   (9.86)	15   (13.14)	23
Total	15	20	35

Now we only need calculate the four elements of the χ2 and add them altogether at the end.

• First-years who said "Yes": $\frac{(f_o-f_e)^2}{f_e}=\frac{(7-5.14)^2}{5.14}=0.67$
• Second-years who said "Yes": $\frac{(f_o-f_e)^2}{f_e}=\frac{(5-6.86)^2}{6.86}=0.5$
• First-years who said "No": $\frac{(f_o-f_e)^2}{f_e}=\frac{(8-9.86)^2}{9.86}=0.35$
• Second-years who said "No": $\frac{(f_o-f_e)^2}{f_e}=\frac{(15-13.14)^2}{13.14}=0.26$

Finally, $\chi^2=\Sigma\frac{(f_o -f_e)^2}{f_e}=0.67+0.5+0.35+0.26=1.78$ The degrees of freedom are, again, df=(rows-1)(columns-1), so here df=(2-1)(2-1)=1(1)=1. That is, with χ²=1.78, df=1, and p=1.18 (i.e., p>0.05), we do not have enough evidence to reject the null hypothesis. At this time, we cannot claim there is an association between year of study and opinion on the cafeteria, i.e., the 0.217 difference in proportions we observe in the sample (7/15 versus 5/20, or 0.467 versus 0.25) is not statistically significant. Of course, we already knew this from the t-test in Example XX[footnote]You may find it curious to know that the correspondence of results between the t and the χ² goes even further: in the binary variables' case, squaring the t-value will give you exactly χ²: t²=χ². In our examples, t=1.34, and 1.34²=1.79 which, if it wasn't for rounding, would be the same as χ². Even their respective degrees of freedom are the same, 1.8. This of course isn't the case when at least one of the discrete variables has more than two categories.[/footnote], so no surprises here.

Our exploration in Section 7.7.2 left us with the following table. Table XXB  Speaking Aboriginal Language Ability by Gender, APS 2012 Source: Statistics Canada (2019). Our hypotheses are:

• H₀: Gender and the ability to speak an Aboriginal language are not associated; women and men are equally likely to speak an Aboriginal language, or π_f=π_m.
• H_a: Gender and the ability to speak an Aboriginal language are associated; women and men are not equally likely to speak an Aboriginal language, or π_f≠π_m.

SPSS calculates χ² as 31.78. With χ² =31.78, df=1, and p<0.001, we have enough evidence to reject the null hypothesis and conclude that Indigenous women and men tend to differ in their ability to speak an Aboriginal language. The 3.6 percentage points difference (i.e., 45%-41.4%) in favour of women being more likely to speak an Aboriginal language is statistically significant and therefore generalizable to the larger Indigenous population. 

A lot has been written about Canada's selective immigration practices: the Canadian government is committed to getting "the best and the brightest" immigrants through a point system which awards more points the more education the prospective immigrant has. [CITATIONS] Be that as it may, how does the rest of the Canadian population (the one born in Canada) compare to the supposedly highly-educated foreign-born? With the help of NHS 2011 (Statistics Canada, 2019), we can find out. (Note that once again, I'm using about 3% random sub-sample of the data, for an N=21,577.) For this example I use the variable citizenship which has three categories: born in Canada, naturalized Canadian, and not a Canadian citizen. For education, I use the same recoded variable I used in Example XX in the previous section, namely degree. Degree has six categories, ranging from (1) "no high school degree" to (6) "PhD" (for full category listing, see Example XX). Table XX cross-tabulates citizenship and degree in a busy-looking 3x6 table (that's 18 cells!). Table XX Degree by Canadian Citizenship Status Source: Statistics Canada (2019). What do we see? Let's carefully examine the evidence[footnote]Don't forget to focus on the percentages, not the number count in each cell! Recall that you can only compare relative frequencies (relative to group size, that is).[/footnote]. While all citizenship groups follow a similar vertical "spread" (i.e., relatively few people without degrees, most people with high/secondary school and some post-secondary school certificates and short of Bachelor's degrees, then decreasing proportions in the higher education categories), this isn't what we are interested in. Recall that we are looking for a pattern between the two variables's categories/groups -- we are comparing groups on their levels of education. As such, we see that fewer naturalized Canadians (18.6%) and fewer still non-Canadian citizens (15.8%) have no degrees compared to Canadian citizens (20.4%). Furthermore, in the three highest education categories (Bachelors, Master's, and PhD), both naturalized Canadians and non-Canadian citizens outperform those born in Canada (while the non-Canadian citizens even outperform naturalized Canadians in turn): 16.7% of naturalized Canadians and 20.5% of non-Canadian citizens have Bachelor's degrees compared to only 12.3% of Canadians born in the country; 11.2% of naturalized Canadians and 13.5% of non-Canadian citizens have Master's degrees compared to only 5.5% of the ones born in Canada; and, finally, 1% of naturalized Canadians and 1.4% of non-Canadian citizens have PhD's compared to 0.4% of those born in Canada. Thus, the table suggests a pattern -- Canadians born elsewhere and non-Canadian citizens seem to have more education than the Canadian-born. Whether this pattern showing difference in proportions in the education degrees among the different citizenship status groups is statistically significant (i.e., generalizable to the Canadian population) remains to be checked -- through a χ²-test. These are our hypotheses:

• H₀: Citizenship status and educational degree are not associated; Canadian-born, naturalized citizens, and non-Canadian citizens are on average similarly educated, and are equally likely to to be highly educated.
• H_a: Citizenship status and educational degree are associated; Canadian-born, naturalized citizens, and non-Canadian citizens have different levels of education on average, and are not equally likely to be highly educated.

I would guess you'd rather not calculate the expected frequencies and their differences from the observed frequencies for all 18 cells (but if you want to do it, who am I to stop you), so I'll report the SPSS output instead. With χ²=449.543, df=10[footnote]Df=(rows-1)(columns-1)=(6-1)(3-1)=5(2)=10.[/footnote], and p<0.001, we have enough evidence to reject the null hypothesis and conclude that citizenship status and educational degree are statistically significantly associated: people born in Canada, naturalized Canadians, and non-Canadian citizens differ in their levels of education. It seems indeed that Canadians born in the country are on average less educated than both naturalized Canadians and non-Canadian citizens, perhaps as a result of the selective criteria for Canadian immigration.

Table XX lists the years of schooling (our variable y) of seven respondents in the GSS 2018 (NORC, 2019) and the years of schooling of their respective fathers (our variable x)[footnote]Here parental education is the independent variable and respondent's education is the dependent variable, so they are denoted as x and y, respectively, according to convention. [/footnote]. While inference with N=7 is not a serious proposition, the small observation count allows for a quick calculation for demonstration purposes only. (After all, we already know the correlation coefficient of these exact same two variables from Section 7.2.3; there the SPSS-calculated r=0.413.) The rest of the columns in Table XX list the necessary computations (obtaining distances from the mean, squaring distances, summing distances, etc.) to produce SS_x, SS_y, and SP_xy. Table XX Calculating Pearson's r

$x$	$y$	$(x-\overline{x})$	$(x-\overline{x})^2$	$(y-\overline{y})$	$(y-\overline{y})^2$	$(x-\overline{x})(y-\overline{y})$
12	8	(12-12.4) = -0.4	-0.42 = 0.2	(8-13.6) = 5.6	5.62 = 31.4	(-0.4)(5.6)=2.2
6	12	(6-12.4) = -6.4	-6.42 = 41	(12-13.6) = -1.6	-1.62 = 2.6	(-6.4)(1.6)=10.2
12	19	(12-12.4) = -0.4	-0.42 = 0.2	(19-13.6) = 5.4	5.42 = 29.2	(-0.4)(5.4)=-2.2
16	16	(16-12.4) = 3.6	3.62 = 13	(16-13.6) = 2.4	2.42 = 5.8	(3.6)(2.4)=8.6
15	12	(15-12.4) = 2.6	2.62 = 6.8	(12-13.6) = -1.6	-1.62 = 2.6	(2.6)(-1.6)=-4.2
12	12	(12-12.4) = -0.4	-0.42 = 0.2	(12-13.6) = -1.6	-1.62 = 2.6	(-0.4)(-1.6)=0.6
14	16	(14-12.4) = 1.6	1.62 = 2.6	(16-13.6) = 2.4	2.42 = 5.8	(1.6)(2.4)=3.8
12.4	13.6		SSx=63.7		SSy=79.7	SPxy=19.3

Then, according to the formula for r we have: $$r=\frac{SP_{xy}}{\sqrt{SS_x SS_y}}=\frac{19.3}{\sqrt{63.7\times79.7}}=\frac{19.3}{71.3}=0.271$$ Obviously, this r=0.271 is not the same as the SPSS-produced r=0.413 we had from Section 7.2.3; in fact, it would be very surprising if they were the same, considering the former is based on N=7 while the latter is based on N=1,687. The exact value of r in the above calculation (r=0.271) doesn't matter, and doesn't serve any purpose and shouldn't be interpreted as it exists only as the end result of our demonstration.

As usual, it helps to know what we are testing exactly:

• H₀: There is no correlation between parental and offspring education; ρ=0.
• H_a: There is a correlation between parental ad offspring education; ρ≠0.

Then, for N=1,687 and r=0.413, we have: $$t=\frac{r\sqrt{N-2}}{\sqrt{1-r^2}}=\frac{0.413\sqrt{1687-2}}{\sqrt{1-0.413^2}}=\frac{0.413(41.1)}{0.911}=18.633$$ With t=18.633, df=1,685, and p=0.00001 (i.e., p=0.00001<0.5), we can reject the null hypothesis that parental and offspring education are not correlated. At this time, we have enough evidence to conclude that there is a moderately weak (r=0.413), statistically significant correlation between parental education and offspring education in the US population[footnote]Purely for demonstration purposes, we could also calculate the t for the 7 respondents whose responses we used to calculate r=0.271: $$t=\frac{r\sqrt{N-2}}{\sqrt{1-r^2}}=\frac{0.271\sqrt{7-2}}{\sqrt{1-0.271^2}}=\frac{0.271(5)}{0.963}=1.407$$ In this case, we could interpret the results like this: "With t=1.407, df=5, and p=0.218 (i.e., p=0.218>0.5), we cannot reject the null hypothesis that parental and offspring education are not correlated. At this time, we do not have enough evidence to conclude that there is a statistically significant correlation between parental education and offspring education in the US population." However, we cannot trust this "inference" as it's only based on N=7.[/footnote].

Imagine you are given written, take-home assignment in some class. Your professor has stipulated that there are three part of the assignment, each worth 30 points, and that you'll receive 10 points just for turning in your work. In this case, your assignment mark is entirely a function of your submitted work. You'll be getting 10 points to start with, then 30 points for fulfilling each of the three requirements. The class grades on the submitted assignments could thus be 10 points  (0 completed requirements), 40 points (1 completed requirement),  70 points (2 completed requirements), and 100 points (3 completed requirements). Figure XX plots this. Figure XX Assignment Mark as a Function of Completed Requirements

As you can see, the relationship between the two variables, assignment requirements completed and assignment mark, is simply $$y=10+30x$$ as helpfully shown in the graph itself. This is a summary form of having to write out all the observations:

• when x=0, $y=10+30x=10+30\times 0=10+0=10$
• when x=1, $y=10+30x= 10+30\times 1=10+30=40$
• when x=2, $y=10+30x= 10+30\times 2=10+60=70$
• when x=3, $y=10+30x= 10+30\times 3=10+90=100$[footnote]Of course, to draw a line you only really need two points. Thus if you only take x=0/y=10 and x=3/y=100 and connect these points with a line, the line will also pass through x=1/y=40 and x=2/y=70. This is a useful property if you need to draw a line by hand.[/footnote].

  The Vancouver Sun recently reported that the average monthly rent of a two-bedroom apartment in Vancouver, BC was \$2,915, at the time of writing the highest in all Canada. (REFERENCE https://vancouversun.com/news/local-news/vancouver-two-bedroom-apartments-now-cost-close-to-3000-report) While the standard deviation was not reported, for the purposes of this exercise we can imagine it as \$150.

Research has shown that, despite similar lower body strength, women have less upper body strength than men, on average.  [LIST CITATIONS FROM HERE https://health.howstuffworks.com/wellness/diet-fitness/personal-training/men-vs-women-upper-body-strength.htm AND THE FOLLOWING] One such study examined differences in upper body strength in a sample of Caucasian and East-Asian college students engaged in weight-lifting classes in American colleges (Chen, Liu and Yu, 2012)  [https://content.sciendo.com/view/journals/ssr/21/3-4/article-p153.xml, pdf here https://www.degruyter.com/downloadpdf/j/ssr.2012.xxi.issue-3-4/v10237-012-0015-5/v10237-012-0015-5.pdf]. While the study examined numerous aspects of the difference in strength, I'll take only one of the researchers' findings to illustrate my point: triceps strength in arm extension. The reported means were 46.2 pounds for women versus 87.4 pounds for men in the Caucasian sample, and 39.6 pounds for women versus 82.1 pounds for men in the East-Asian sample Chen, Liu and Yu, 2012, p.156). Consider what we are discussing here: We have two variables of interest[footnote]You could argue that race/ethnicity is also there. As reported in the study, however, race/ethnicity was a secondary variable bringing more detail to the study, though which the authors were able to demonstrate that upper-body strength differences based on sex exist in both race/ethnic groups considered.[/footnote], gender and upper-body strength. Gender is a nominal discrete (and, in this study, binary) variable while upper-body strength (through various measurements in pounds) is a ratio continuous variable. The hypothesized association between the two posits that some categories of the discrete variable (e.g., men) tend to go with specific values of the continuous variable (e.g., higher values on upper body-strength). That is, if both men and women had the same means for, in this case, triceps strength in arm extension, gender and upper-body strength would be unrelated, as one's sex wouldn't be predictive of one's upper-body strength at all. In effect, we are comparing the mean values (of a continuous variable) across groups (i.e., the categories of a discrete variable). Now, as far as a numerical description of that comparison goes, we have the two means (of men and of women) and we can thus calculate the difference of means: $\overline{x}_{men}$ $-\overline{x}_{women}$ $=87.4-46.2=41.2$ (Caucasian sub-sample) $\overline{x}_{men}$ $-\overline{x}_{women}$ $=82.1-39.6=42.5$ (East-Asian sub-sample) Thus, what we observe here in this sample is a 41.2 pounds difference in the upper-body strength (as measured by triceps strength in arm extension) between Caucasian men and women and a difference in upper-body strength of 42.5 pounds between East-Asian men and women. Again, note that the fact that we see these differences in the sample does not mean they exist in the population -- they may, or they might not. We wouldn't know this unless we test if the differences are generalizable to the population[footnote]If you are interested, the authors of the study did test these differences (with a t-test, discussed later) and found them generalizable to the population indeed (Chen, Liu and Yo, 2012).[/footnote]. We'll get to testing later, for now we are only interested in the differences descriptively, i.e., that they exist in the sample.

Statistics Canada's National Household Survey 2011 (NHS 2011) was designed to replace the until-that-time mandatory long form of the Census[footnote]For the problematic nature of the (Harper) Government's decision in 2010 to make the survey voluntary and its related implications, see for example here. The mandatory long-form census was restored in 2016 by the Liberal Government. My usage of the data here is strictly for demonstration purposes and as such shouldn't be taken as an endorsement of the NHS 2011.[/footnote]. For this example, I'm using a random sample of about 3% of the NHS 2011 iindividual data (aka a Public Use Microdata File, or PUMF), resulting in N=22,123. I'm interested in whether men and women's income for the year preceding the survey differed, i.e., whether the variables gender (called sex in the dataset) and total income (i.e., income from all possible sources) appear associated. With the help of SPSS, I plot the data. The resulting boxplots graph is given in Figure XX below. Figure XX Gender Differences in Total Income, NHS 2011 Source: Statistics Canada, 2019. Boxplots are charts visually incorporating a lot of statistical information in one neat little package; I encourage you to make use of them when exploring your data as they can be quite useful. What do we see in Figure XX our case? Obviously, we have two groups to compare (as per the two categories in the nominal variable gender), women and men, and therefore the graph presents two boxplots. (Had we multiple categories in our discrete variable, we'd have had multiple boxplots.) How to read a boxplot. Each boxplot consists of the eponymous "box" and two so-called "whiskers" protruding from it. The "box" (in green above) represents the middle 50 percent of the data (i.e., the two middle quartiles, or the IQR); the lower whisker represents the first/bottom quartile of the data, and the upper whisker represents the last/top quartile of the data. The dark line bisecting the box indicates the median. The two ends of the whiskers are the lowest and the highest values. Note, however, that the quartiles (as represented by the whiskers) exclude outliers as to not visually distort  the "regular" spread of the data. As such, the chart plots run-of-the-mill outlier cases as small circles (above they are in red)) outside of the whiskers; extreme outliers are indicated by stars (in black above)[footnote]Also note that to make the boxplot readable in an appropriate size, in Figure XX I cut some extremely extreme outliers off at the top of the male boxplot.[/footnote]. Now that you know how to read them, compare the two boxplots above. First, we see that the median  for men is higher than the median for women (again, these are the dark lines within the boxes); as well, total income appears to be more spread out for men than for women (the whiskers in the men's boxplot reach further, indicating larger range and IQR. Further, while both men and women appear to have outliers, the men's group seems to include more extreme outliers and at higher values than those observed in the women's group[footnote]You might have noticed that the first quartile (i.e., the bottom "whisker") includes negative values. Statstics Canada uses several income variables for which this is the case. Negative income exists as an accounting possibility: when one's annual expenses end up larger than one's annual income (e.g., like for a self-employed individual whose business hasn't been as successful, etc.). In many real-life research, negative income values are frequently dropped/removed if that course of action is justified by the study's design, research question, and purposes. In the case of this example I have no reason to do that, hence I left the negative income values in the data.[/footnote]. All this points to the conclusion that men in the sample had higher (median, and quite likely average) total income for 2010 than women did, despite that the individuals with the lowest incomes also appear to be men. As useful the general information we gleaned from the boxplots, we should look at the precise numbers too. SPSS calculates the mean total income as \$32,465 for women and \$48,866 for men -- that is, there is \$16,401 difference in mean total income in favour of men. In this sample of 22,123 people, men's average total income is \$16,401 more than women's. We could also compare the medians (especially useful when dealing with income variables): SPSS gives the median total income of women in the sample as \$23,000, while the median total income for men is \$35,000 -- a difference of medians of \$12,000, again in favour of men.

Imagine you are frustrated within the food options available in your campus cafeteria and you wonder if others share your thoughts on the matter (perhaps in order to gauge support for changes you'd like to see enacted or similar type of activism). Before you devote time to do an actual random sample study (now that you know), you do a quick exploratory poll of your classmates in one of your classes . You ask 35 people whether they like the campus cafeteria, and in the process, you get the inkling that second-year students seem to have different opinion about the food options than the first-year students in the class. You plot your results: Table XXA Do You Like The Campus Cafeteria?

	First Year Students	Second Year Students	Total
YES	7	5	12
NO	8	15	23
Total	15	20	35

I'm certain you know how to read this: 7 first-year and 5 second-year students like the cafeteria, while 8 first-year and 15 second-year students do not. There is a total of 12[footnote]As 7+5=12.[/footnote] students who like the cafeteria and 23[footnote]As 8+15=23.[/footnote] students who do not.  You talked to 15[footnote]As 7+8=15.[/footnote] first-year and 20[footnote]As 5+15=20.[/footnote] second-year students, a total of 35 students. Can you compare the relevant numbers as they are presented in the table? And, for that matter, what are the relevant numbers? Let's answer both questions in turn. Recall from Chapter XX on frequencies: No, you cannot compare the numbers as stated since the two groups you have to compare are different size. The relevant comparison is between the different year students who like the cafeteria -- first-years vs. second-years -- as this is what you want to know. It's true that 2 more first-years like the food in the cafeteria than the second year students (7>5) but at the same time you had 5 more second-year students in your sample (20>15). To take into account the differing group size, you need to compare proportions (or percentages): the proportion of first-year students who like the cafeteria against the proportion of second-year students who like the cafeteria. You therefore calculate the respective proportions, turning them into percentages at the end:

• $\frac{7}{15}=0.467$, or 46.7% of first-years like the cafeteria
• $\frac{5}{20}=0.250$, or 25% of second-years like the cafeteria
• $\frac{8}{15}=0.533$, or 53.3% of first years do NOT like the cafeteria
• $\frac{15}{20}=0.750$, or 75% of the second-years do NOT like the cafeteria
• $\frac{12}{35}=0.342$, or 34.2% of ALL students like the cafeteria
• $\frac{23}{35}=0.718$, or 71.8% of ALL students do NOT like the cafeteria

To summarize the information neatly, we modify our table to this: Table XXB Do You Like The Campus Cafeteria? (Column Percentages)

	First Year Students	Second Year Students	Total
YES	46.7%	25%	34.3%
NO	53.3%	75%	71.8%
Total	100%	100%	100%

So far so good? From Table XXB now we clearly see that your initial hunch was right: there does seem to be a difference in the opinions of your classmates based on which year they are in their studies. That is, while you do have support for anti-cafeteria activism (only 34.3% of your classmates like the campus cafeteria, while 71.8% dislike it) first-year students seem to like the cafeteria a lot (almost twice) more than second-year students do: 46.7% of first-years like the food options in the cafeteria compared to only 25% of the second-years, a difference of 21.7 percentage points.

Statistics Canada's Aboriginal People Survey (APS) 2012 is a nationally representative survey of First Nations peoples (living off reserve), Métis and Inuit, 6 years of age and older (Statistics Canada, 2019)[footnote]One could perhaps see the APS 2012 as an effort by Statsistics Canada to address some of the volunary NHS 2011 issues with coverage/non-response of the listed population groups.[/footnote]. Language is a key element in retaining, preserving, and transmitting culture; as such, the ability of Indigenous peoples to speak their ancestral languages is of special interest given the recommendations of the Truth and Reconciliation Commission's (TRC) final report (2015). For the purposes of this example, I am interested if there are gender differences in the ability to speak an Aboriginal Language among the collected sample. Table XX shows the cross-tabulation of gender (called sex in the APS) and speaking Aboriginal language variables. (Both variables are binary in the survey.) Table XXA Speaking Aboriginal Language Ability by Gender, APS 2012 Source: Statistics Canada (2019) As you can see, working with real, large N data makes proportions even more indispensable for making sense of the table.   We need to compare the fraction of women who speak an Aboriginal language (or languages) to the fraction of men who are able to do that. To make things easier, I followed the rules of thumb I listed in the Watch Out!! XX Box above: the groups-to-be-compared are in the columns, and we need to compare them horizontally[footnote]A point to be made here is that when working with binary data, it's enough to focus on one of the categories on which you compare the groups, as the other category would be a complement of the first as we are working with proportions. That is, here we need consider only the YES category (as the NO category is it's exact opposite, i.e., "1- YES") due to the fact that we're interested in those who can speak the language, not those who don't.[/footnote]. Therefore, I need column percentages. Table XXB does just that. Table XXB  Speaking Aboriginal Language Ability by Gender, APS 2012 (Column Percentages) Source: Statistics Canada (2019) SPSS provides both the original cell count (i.e., frequency) and the respective percentage below it[footnote]Yet another useful rule of thumb: make sure that SPSS lists "% within [groups-to-be-compared]" as this indicates that the correct percentages appear in the table. In this case, SPSS tells us that it has listed "% within Sex of respondent", i.e., the ones we need in order to compare the two gender groups.[/footnote]. We can thus easily see that while only 41.4% of men in the sample can speak an Aboriginal language, 45% of women in the sample can do that; i.e., there is a gender difference of 3.6 percentage points in favour of women[footnote]You might think it a small difference, but the magnitude of the difference is not the most important thing when establishing statistical associations. More on the topic in Section XX below.[/footnote].