Skip to content

11-01: Exercises — One-Sample t-Test

Notes reference: 11-01: One-Sample t-Test

Q1: Critical t values

Find the critical value(s) for each.

Test α n Critical t
Two-tailed 0.05 20 ?
Right-tailed 0.05 20 ?
Left-tailed 0.01 12 ?
Two-tailed 0.01 31 ?
Two-tailed 0.10 8 ?

Solution

Test α df = n−1 Critical t
Two-tailed 0.05 19 ±2.0930
Right-tailed 0.05 19 +1.7291
Left-tailed 0.01 11 −2.7181
Two-tailed 0.01 30 ±2.7500
Two-tailed 0.10 7 ±1.8946 
=T.INV.2T(0.05, 19)      ' 2.09302   two-tailed
=T.INV(0.95, 19)         ' 1.72913   right-tailed
=T.INV(0.01, 11)         ' -2.71808  left-tailed
=T.INV.2T(0.01, 30)      ' 2.75000
=T.INV.2T(0.10, 7)       ' 1.89458

qt(0.975, 19); qt(0.95, 19); qt(0.01, 11); qt(0.995, 30); qt(0.95, 7)

T.INV takes a left-tail probability; T.INV.2T takes the total two-tailed area. T.INV(0.975, 19) and T.INV.2T(0.05, 19) both give 2.093.

Q2: Two-tailed t-test from summary statistics

A cereal box should contain 500 g. A sample of 25 boxes gives x̄ = 495.8 g with s = 9.4 g. Test at α = 0.05.

Solution

STEP 1  H₀: μ = 500        H₁: μ ≠ 500        TWO-tailed,  α = 0.05

STEP 2  σ unknown (s computed from the data) → t-test
        n = 25, assume fill weights are approximately normal
        df = 24

STEP 3  SE = 9.4/√25 = 9.4/5 = 1.88
        t  = (495.8 − 500)/1.88 = −4.2/1.88 = −2.23404

STEP 4  Critical values: ±t(0.025, 24) = ±2.06390
        |−2.23404| > 2.06390                   →  REJECT H₀

        p-value = 2 × P(T₂₄ < −2.23404) = 2(0.017562) = 0.035124
        0.0351 ≤ 0.05                          →  REJECT H₀

STEP 5  There is sufficient evidence at the 5% level to conclude that the
        mean fill weight differs from 500 g. The boxes are UNDERFILLED
        by about 4.2 g on average.

REPORT:  t(24) = −2.23, p = .035, d = −0.45, 95% CI [491.92, 499.68]

    CI:  495.8 ± 2.06390(1.88) = 495.8 ± 3.880 = (491.92, 499.68)
         500 lies OUTSIDE — the same conclusion.
    d :  (495.8 − 500)/9.4 = −0.447   (small-to-medium)

=9.4/SQRT(25)                       ' 1.88
=(495.8-500)/1.88                   ' -2.234043
=T.INV.2T(0.05, 24)                 ' 2.063899
=T.DIST.2T(ABS(-2.234043), 24)      ' 0.035124
=495.8-T.INV.2T(0.05,24)*1.88       ' 491.920
=495.8+T.INV.2T(0.05,24)*1.88       ' 499.680

xbar <- 495.8; s <- 9.4; n <- 25
se <- s/sqrt(n); t <- (xbar - 500)/se
c(se = se, t = t, p = 2*pt(-abs(t), n-1),
  lo = xbar - qt(0.975, n-1)*se, hi = xbar + qt(0.975, n-1)*se)

Q3: One-tailed t-test from raw data

A supplier claims mean delivery time is at most 3 days. A customer records 12 deliveries:

3.2  4.1  2.8  3.9  4.5  3.1  3.7  4.2  2.9  3.8  4.0  3.5

Test at α = 0.05 whether deliveries take longer than claimed.

Solution

STEP 1  H₀: μ ≤ 3         H₁: μ > 3          RIGHT-tailed,  α = 0.05

STEP 2  n = 12, σ unknown → t-test, df = 11
        Check a boxplot first: no outliers, roughly symmetric ✓

STEP 3  Σx = 43.7        x̄ = 43.7/12 = 3.64167
        Σ(x − x̄)² = 3.24917  →  s² = 3.24917/11 = 0.29538,  s = 0.54349

        SE = 0.54349/√12 = 0.54349/3.46410 = 0.15689

        t  = (3.64167 − 3)/0.15689 = 0.64167/0.15689 = 4.09000

STEP 4  Critical value: t(0.05, 11) = 1.79588
        4.09000 > 1.79588                      →  REJECT H₀

        p-value = P(T₁₁ > 4.09000) = 0.000895
        0.00090 ≤ 0.05                         →  REJECT H₀

STEP 5  There is very strong evidence at the 5% level that mean delivery
        time exceeds the claimed 3 days — by about 0.64 days on average.

EFFECT SIZE  d = 0.64167/0.54349 = 1.181   — very large.

=AVERAGE(A2:A13)                                     ' 3.641667
=STDEV.S(A2:A13)                                     ' 0.543488
=(AVERAGE(A2:A13)-3)/(STDEV.S(A2:A13)/SQRT(12))      ' 4.090003
=T.DIST.RT(4.090003, 11)                             ' 0.000895
=T.INV(0.95, 11)                                     ' 1.795885

x <- c(3.2,4.1,2.8,3.9,4.5,3.1,3.7,4.2,2.9,3.8,4.0,3.5)
t.test(x, mu = 3, alternative = "greater")
#  t = 4.09, df = 11, p-value = 0.0008953
#  95 percent confidence interval: 3.3599  Inf
#  mean of x: 3.641667

boxplot(x, horizontal = TRUE)        # assumption check FIRST
shapiro.test(x)

x = np.array([3.2,4.1,2.8,3.9,4.5,3.1,3.7,4.2,2.9,3.8,4.0,3.5])
stats.ttest_1samp(x, 3, alternative="greater")

Q4: t or z?

Decide which test applies and give the critical value at α = 0.05, two-tailed.

  1. n = 45, σ = 6.2 known
  2. n = 45, s = 6.2 computed
  3. n = 9, s = 2.1, population known to be normal
  4. n = 9, s = 2.1, population strongly skewed
  5. n = 250, s = 30

Solution

1. Z-TEST,  critical ±1.9600
2. t-TEST,  df = 44,  critical ±2.0154
3. t-TEST,  df =  8,  critical ±2.3060
4. NEITHER is safe. n = 9 from a strongly skewed population fails both
   the normality assumption and the CLT.
   → transform (log), use the Wilcoxon signed-rank test, or bootstrap.
5. t-TEST,  df = 249, critical ±1.9695  ≈ z

NOTE ON ROW 5: at df = 249 the t and z critical values agree to two
decimals. Large-sample t and z tests give practically identical answers —
which is why "use t whenever you estimated σ" costs you nothing.

Q5: Interpret a software output

    One Sample t-test

data:  scores
t = -1.6248, df = 29, p-value = 0.115
alternative hypothesis: true mean is not equal to 75
95 percent confidence interval:
 70.7836 75.4831
sample estimates:
mean of x
 73.13333
  1. What were the hypotheses?
  2. What is n?
  3. Decide at α = 0.05.
  4. Is the conclusion consistent with the CI?
  5. Estimate s.

Solution

1.  H₀: μ = 75      H₁: μ ≠ 75     (two-tailed, from "not equal to 75")

2.  df = n − 1 = 29   →   n = 30

3.  p = 0.115 > 0.05  →  FAIL TO REJECT H₀.
    There is not sufficient evidence that the mean differs from 75.

4.  YES. The 95% CI (70.78, 75.48) CONTAINS 75, which is exactly the
    condition for failing to reject at α = 0.05.

5.  Work backwards from the CI half-width:
        E = (75.4831 − 70.7836)/2 = 2.34975
        t(0.025, 29) = 2.04523
        SE = E / t = 2.34975/2.04523 = 1.14889
        s  = SE × √n = 1.14889 × √30 = 1.14889 × 5.47723 = 6.2926

    Cross-check with the printed t:
        t = (73.13333 − 75)/1.14889 = −1.6248   ✓ matches exactly.

Q6: Assumption checks

Before running a one-sample t-test on 22 observations, what do you check and how?

Solution

1. INDEPENDENCE — from the DESIGN, not the data. Was it a random sample?
   Are the observations from distinct units? No software test for this.

2. NORMALITY of the population (matters at n = 22).
   Boxplot     → symmetric? outliers?
   Q-Q plot    → do the points hug the line?  ← the most informative check
   Shapiro-Wilk → H₀: normal. A large p is reassuring.

3. OUTLIERS — the t-test is robust to mild non-normality but NOT to
   outliers, because a single extreme value moves both x̄ and s.

x <- rnorm(22, 100, 15)

par(mfrow = c(1, 3))
hist(x, breaks = 8, col = "#8A5FBF", border = "white", main = "Histogram")
boxplot(x, col = "#0FA3A3", main = "Boxplot")
qqnorm(x, pch = 19); qqline(x, col = "#0FA3A3", lwd = 2)
par(mfrow = c(1, 1))

shapiro.test(x)
boxplot.stats(x)$out              # any flagged outliers?

fig, axes = plt.subplots(1, 3, figsize=(12, 3))
axes[0].hist(x, bins=8, color="#8A5FBF", edgecolor="white")
axes[1].boxplot(x)
stats.probplot(x, dist="norm", plot=axes[2])
stats.shapiro(x)

=SKEW(A2:A23)         ' near 0 if symmetric
=KURT(A2:A23)         ' near 0 if normal (Excel reports EXCESS kurtosis)
' Q-Q plot: sort into column B, then C2: =NORM.S.INV((ROW()-1.5)/22)
'           scatter B against C — a straight line means normal

Q7: When the assumptions fail

You have 14 observations with one clear outlier and obvious right skew. Give three defensible options.

Solution

OPTION 1 — INVESTIGATE THE OUTLIER
   Data-entry error, wrong units, or a different population? If it is a
   documented error, fix or remove it and say so. If it is genuine, keep it.

OPTION 2 — TRANSFORM
   Right-skewed positive data often becomes symmetric under a log
   transform. Run the t-test on log(x) — but note that you are then
   testing the GEOMETRIC mean, and the CI back-transforms to a ratio,
   not a difference.
        t.test(log(x), mu = log(mu0))

OPTION 3 — NONPARAMETRIC TEST
   The Wilcoxon signed-rank test tests the MEDIAN and needs only
   symmetry of the differences, not normality.
        wilcox.test(x, mu = mu0)
        stats.wilcoxon(x - mu0)

OPTION 4 — BOOTSTRAP
   Resample with replacement and build a percentile CI. Assumes nothing
   about the distribution's shape.
        boot_means <- replicate(10000, mean(sample(x, replace = TRUE)))
        quantile(boot_means, c(0.025, 0.975))

WHAT NOT TO DO
   Do not silently delete the outlier because the p-value improves.
   Do not run the t-test and omit the diagnostics from the write-up.
   Report the sensitivity: "with and without the outlier, the conclusion
   is / is not the same."

Q8: Report it properly

Given x̄ = 68.4, s = 11.2, n = 36, μ₀ = 65, two-tailed, α = 0.05, produce a complete write-up.

Solution

COMPUTATION
    df = 35
    SE = 11.2/√36 = 11.2/6 = 1.86667
    t  = (68.4 − 65)/1.86667 = 3.4/1.86667 = 1.82143
    critical t(0.025, 35) = 2.03011
    p = 2 × P(T₃₅ > 1.82143) = 2(0.038538) = 0.077076
    d = (68.4 − 65)/11.2 = 0.3036
    95% CI = 68.4 ± 2.03011(1.86667) = 68.4 ± 3.789 = (64.61, 72.19)

DECISION
    |1.82143| < 2.03011  and  p = 0.077 > 0.05   →  FAIL TO REJECT H₀

WRITE-UP
    "A one-sample t-test found no significant difference between the
     sample mean (M = 68.4, SD = 11.2, n = 36) and the reference value
     of 65, t(35) = 1.82, p = .077, d = 0.30, 95% CI [64.61, 72.19].

     The confidence interval includes 65, so a mean of 65 remains
     plausible. However, the interval also extends to 72.2, so a
     practically meaningful increase has NOT been ruled out — the study
     is simply not precise enough to distinguish the two. A larger
     sample would be needed to resolve the question."

WHY THAT LAST PARAGRAPH MATTERS
    "p > 0.05" is NOT evidence that μ = 65. Failing to reject is not
    accepting. The confidence interval is what tells the reader how much
    uncertainty remains — which is why it belongs in every report.

⬅️ Previous: 10-02: Exercises — One-Sample Z-Tests ➡️ Next: 11-02: Exercises — Two-Sample t-Test