Two-Sample, Paired & Chi-Square Inference

1Objectives¶

By the end of this lesson you will be able to:

1. Run and interpret a Welch two-sample t-test with t.test(y ~ g, data=).

2. Run and interpret a paired t-test with t.test(~diff, data=).

3. Run and interpret a two-proportion test with prop.test(c(x1,x2), c(n1,n2)).

4. Run and interpret a chi-square test of independence with xchisq.test(tally(y ~ x, data=)), including its observed/expected/residual display.

5. Read back every test’s decision in one plain-language sentence.

2From one group to two (or more)¶

L10 built the logic of a hypothesis test — state $H_0$ and $H_a$, check conditions, compute a test statistic and a p-value, decide — and already put two comparisons to work: a Welch two-sample t.test(y ~ g, data=) (cat people vs. dog people’s sleep) and a two-proportion prop.test(c(x1,x2), c(n1,n2)) (cat-person rate, Freshmen vs. Sophomores). This lesson keeps that exact logic, gives you one more worked example of each so the pattern sticks, and adds the two shapes of comparison L10 didn’t reach yet — paired measurements and categorical association:

Reload the running dataset if you’re starting fresh:

suppressMessages({library(mosaic); library(BSDA)})
set.seed(2200)
survey <- read.csv("data/survey_sim.csv")
survey$class_year <- factor(survey$class_year,
                             levels = c("Freshman", "Sophomore", "Junior", "Senior"))
survey$pet <- factor(survey$pet, levels = c("Cat person", "Dog person", "Neither"))
survey$major_area <- factor(survey$major_area)

read.csv() reads class_year, pet, and major_area in as plain text (L06); wrapping each in factor() (with levels = spelling out Freshman-before-Sophomore-before-Junior-before-Senior order for class_year) is what lets L07’s tables print in that logical order instead of alphabetical, and is what makes Section 1’s droplevels() meaningful below.

31. Welch two-sample t-test: comparing two independent means¶

Independent samples means the two groups are made of different students — no student appears in both. Are Exam 1 scores different for STEM majors versus Business majors? Start with EDA, restricting survey to just those two majors:

sub2 <- subset(survey, major_area %in% c("STEM", "Business"))
sub2$major_area <- droplevels(sub2$major_area)   # drop the unused major levels
favstats(exam_score ~ major_area, data = sub2)

  major_area  min     Q1 median    Q3  max     mean        sd  n missing
1   Business 42.2 56.150   60.3 66.40 76.7 60.88065  8.811145 31       0
2       STEM 39.7 54.375   58.0 67.85 80.7 60.57917 10.646819 24       0

droplevels() matters here: major_area still remembers its other three levels (Nursing, Kinesiology, Other) even after subset() removes every row that had them. Without droplevels(), favstats(exam_score ~ major_area, data = sub2) would print all five majors — three of them with n = 0 and NaN for every statistic — because it still believes those empty levels exist. Dropping the unused levels first keeps the output to the two majors that actually have data.

The two groups’ means (60.9 vs. 60.6) are close, and the boxes overlap almost completely — EDA already hints this comparison won’t turn up much. Test it the same way L10 tested sleep_hours ~ pet:

t.test(exam_score ~ major_area, data = sub2)

	Welch Two Sample t-test

data:  exam_score by major_area
t = 0.11214, df = 44.307, p-value = 0.9112
alternative hypothesis: true difference in means between group Business and group STEM is not equal to 0
95 percent confidence interval:
 -5.115581  5.718538
sample estimates:
mean in group Business     mean in group STEM 
              60.88065               60.57917 

$H_0$: $\mu_{Business} = \mu_{STEM}$ vs. $H_a$: $\mu_{Business} \ne \mu_{STEM}$. With $t = 0.112$ and $p = 0.911$ — nowhere near a typical $\alpha = 0.05$ — fail to reject $H_0$. The 95% CI for the difference in means, $(-5.12, 5.72)$, contains 0, telling the same story: this sample gives no evidence that STEM and Business majors score differently on Exam

1. Same title, same “Welch” default, same reasoning as L10’s example — different pair of groups.

42. Paired t-test: comparing two measurements on the same subjects¶

Paired data means each subject is measured twice, and you care about the within-subject change — the opposite setup from Section 1’s two separate groups. R ships a classic paired-data example built in, no import needed (same idea as faithful in L06): sleep, ten patients each given two drugs, with extra recording each patient’s extra hours of sleep on that drug.

data(sleep)
str(sleep)

'data.frame':	20 obs. of  3 variables:
 $ extra: num  0.7 -1.6 -0.2 -1.2 -0.1 3.4 3.7 0.8 0 2 ...
 $ group: Factor w/ 2 levels "1","2": 1 1 1 1 1 1 1 1 1 1 ...
 $ ID   : Factor w/ 10 levels "1","2","3","4",..: 1 2 3 4 5 6 7 8 9 10 ...

Twenty rows, but only ten patients (ID 1–10) — group 1 is Drug 1’s measurement and group 2 is Drug 2’s, per patient. That’s the giveaway for paired data: the same ID shows up twice. t.test(~diff, data=) needs the difference already computed as one column, so reshape to one row per patient first:

sleep_wide <- reshape(sleep, timevar = "group", idvar = "ID", direction = "wide")
sleep_wide$diff <- sleep_wide$extra.2 - sleep_wide$extra.1
sleep_wide

   ID extra.1 extra.2 diff
1   1     0.7     1.9  1.2
2   2    -1.6     0.8  2.4
3   3    -0.2     1.1  1.3
4   4    -1.2     0.1  1.3
5   5    -0.1    -0.1  0.0
6   6     3.4     4.4  1.0
7   7     3.7     5.5  1.8
8   8     0.8     1.6  0.8
9   9     0.0     4.6  4.6
10 10     2.0     3.4  1.4

extra.1 and extra.2 are Drug 1’s and Drug 2’s readings for that same patient, side by side; diff is Drug 2 minus Drug 1. Look at diff the way L07 taught — favstats() on it first:

favstats(~diff, data = sleep_wide)

 min   Q1 median  Q3 max mean       sd  n missing
   0 1.05    1.3 1.7 4.6 1.58 1.229995 10       0

Every one of the ten differences is at or above zero — a strong visual hint before any test that Drug 2 tends to add more sleep. A paired t-test is just a one-sample t-test on the differences, testing $H_0: \mu_{diff} = 0$:

t.test(~diff, data = sleep_wide)

	One Sample t-test

data:  diff
t = 4.0621, df = 9, p-value = 0.002833
alternative hypothesis: true mean is not equal to 0
95 percent confidence interval:
 0.7001142 2.4598858
sample estimates:
mean of x 
     1.58 

$p = 0.0028$, well under $\alpha = 0.05$: reject $H_0$. The 95% CI for the mean difference, $(0.70, 2.46)$ hours, does not contain 0 — consistent evidence that Drug 2 is associated with more extra sleep than Drug 1 in this sample, averaging about 1.58 extra hours. Notice ~diff is a one-sided formula (nothing on the left of ~) — the same “just one variable” shape L06 and L07 used for favstats(~x, data=), because after reshaping there is only one variable (the difference) left to test.

53. Two-proportion test: comparing two group proportions¶

L10 already ran prop.test(c(x1,x2), c(n1,n2)) once, comparing the cat-person rate between Freshmen and Sophomores. Here’s a second question, same tool: L07 found that more study time went with higher exam scores. Turn that into a yes/no question — are students who study at or above the class median more likely to pass Exam 1 (score $\ge 60$) than students who study below it?

med <- median(~study_min, data = survey)
survey$pass <- ifelse(survey$exam_score >= 60, "Pass", "Not yet")
survey$study_grp <- factor(ifelse(survey$study_min >= med,
                                   "High study time", "Low study time"),
                            levels = c("Low study time", "High study time"))
tally(pass ~ study_grp, data = survey)

         study_grp
pass      Low study time High study time
  Not yet             35              18
  Pass                25              42

prop.test() wants raw counts, not a table — pull the “successes” ($x$, the Pass count) and totals ($n$) for each group:

hi <- subset(survey, study_grp == "High study time")
lo <- subset(survey, study_grp == "Low study time")
x1 <- sum(hi$pass == "Pass");  n1 <- nrow(hi)    # 42 of 60
x2 <- sum(lo$pass == "Pass");  n2 <- nrow(lo)    # 25 of 60
prop.test(c(x1, x2), c(n1, n2))

	2-sample test for equality of proportions with continuity correction

data:  c out of cx1 out of n1x2 out of n2
X-squared = 8.6511, df = 1, p-value = 0.003269
alternative hypothesis: two.sided
95 percent confidence interval:
 0.09635351 0.47031316
sample estimates:
    prop 1     prop 2 
0.7000000 0.4166667 

$H_0$: the two groups’ true pass rates are equal; $H_a$: they differ. X-squared = 8.65, p = 0.0033 — reject $H_0$. prop 1 (0.70, the high-study-time group, since c(x1, x2) listed it first) and prop 2 (0.417, low-study-time) are the two sample proportions being compared; the 95% CI for their difference, $(0.096, 0.470)$, does not include 0. In this sample, studying at or above the class median goes with a meaningfully higher Exam 1 pass rate.

64. Chi-square test of independence: two categorical variables¶

L07 built a two-way tally() of pet preference by class year and eyeballed that the proportions looked “fairly stable across years.” A chi-square test of independence turns that eyeball check into a formal decision: $H_0$: pet preference and class year are independent (unrelated); $H_a$: they’re associated. Start from the exact same table:

tab <- tally(pet ~ class_year, data = survey)
tab

            class_year
pet          Freshman Sophomore Junior Senior
  Cat person        9        14     11      6
  Dog person       16        11     16     12
  Neither           5        11      6      3

mosaic’s xchisq.test() runs the same test as base R’s chisq.test() but also prints every cell’s expected count and residual — exactly the diagnostic detail you need to say why a result came out the way it did:

xchisq.test(tab)

	Pearson's Chi-squared test

data:  x
X-squared = 5.8412, df = 6, p-value = 0.4412

    9       14       11        6   
(10.00)  (12.00)  (11.00)  ( 7.00) 
[0.100]  [0.333]  [0.000]  [0.143] 
<-0.32>  < 0.58>  < 0.00>  <-0.38> 
       
   16       11       16       12   
(13.75)  (16.50)  (15.12)  ( 9.62) 
[0.368]  [1.833]  [0.051]  [0.586] 
< 0.61>  <-1.35>  < 0.22>  < 0.77> 
       
    5       11        6        3   
( 6.25)  ( 7.50)  ( 6.88)  ( 4.38) 
[0.250]  [1.633]  [0.111]  [0.432] 
<-0.50>  < 1.28>  <-0.33>  <-0.66> 
       
key:
	observed
	(expected)
	[contribution to X-squared]
	<Pearson residual>

Read the grid one cell at a time using the key: at the bottom — for Freshman/Cat person: 9 students observed, (10.00) expected if pet preference and class year were truly unrelated, a [0.100] share of the total chi-square statistic coming from that one cell, and a Pearson residual of -0.32 (observed a bit below expected; residuals beyond about $\pm 2$ are the ones worth a second look — none here come close). X-squared = 5.84, df = 6, p = 0.4412: fail to reject $H_0$. No cell’s residual stands out, and the p-value confirms it — this sample gives no evidence that pet preference and class year are related, matching L07’s informal read exactly.

7Summary¶

• t.test(y ~ g, data=) runs a Welch two-sample t-test on independent groups; droplevels() a grouping factor first if it has unused levels left over from a subset().

• t.test(~diff, data=) runs a paired t-test — a one-sample test on a difference column you compute first (e.g., via reshape() on wide/long data with a shared ID, as with the built-in sleep dataset).

• prop.test(c(x1,x2), c(n1,n2)) compares two group proportions; pull the successes (x1, x2) and totals (n1, n2) from a tally() first.

• xchisq.test(tally(y ~ x, data=)) tests whether two categorical variables are associated, and prints observed counts, expected counts, each cell’s contribution to X-squared, and Pearson residuals — read them with the key: at the bottom of the output.

• Every test here follows the same decision logic as L10: state $H_0$/$H_a$, check the test’s conditions, read the p-value against $\alpha$, decide, and say what that decision means in one plain sentence.

• The full script that generated every figure and every number in this lesson is committed at data/make_L11_figures.R — run it yourself to reproduce all of it exactly.