Homework — Are Commutes Faster Than 10 Minutes? (One-Tailed t-test, n = 23)

Context & Research Question

We study students’ one-way commute times to Classroom #288. We want to know if the true mean is less than 10 minutes.

H₀: μ = 10
H₁: μ < 10 (left-tailed)
α: 0.05
n = 23; σ unknown → use one-sample t-test with df = n − 1 = 22.

Data (23 one-way times in minutes)

Copy into Excel (one per line → B2:B24)

Suggested cell setup (single sheet)

• Put μ0 = 10 in E2 • Put α = 0.05 in E3 • (Optional for power) Put μ1 in E4 (e.g., 9.5 or 9.0)

Interpretation target: “Do dorm students get to #288 in under 10 minutes on average?”

What to Deliver (Student Tasks)

Compute sample mean and st.dev. (x̄, s) from the 23 values.
Run a left-tailed one-sample t-test at α=0.05. Report t, df, p-value, decision, and a short interpretation.
Compute a 95% CI for μ (two-sided t-interval) and comment briefly.
Error rates / Power. For μ₁=9.5 and μ₁=9.0, estimate β and Power (normal approximation; optional exact via noncentral t).
Sample size planning. For 80% power (one-sided, α=0.05) to detect δ=0.5 minute faster (μ₁=9.5), find n (round up).
One-paragraph report (2–3 sentences) summarizing your decision, CI, and power for μ₁=9.5.

Excel Cheat-Sheet — One-Tailed t (single sheet)

/* Data in B2:B24; μ0 in E2; α in E3; optional μ1 in E4. All on one sheet. */

n           = COUNT(B2:B24)
xbar        = AVERAGE(B2:B24)
s           = STDEV.S(B2:B24)
df          = n - 1
SE_t        = s / SQRT(n)

t_stat      = ( xbar - E2 ) / SE_t

/* Left-tailed p-value for H1: μ < μ0 */
p_value     = T.DIST(t_stat, df, TRUE)

/* Critical t for left tail; will be negative */
t_crit_left = T.INV(E3, df)

/* Equivalent X̄ threshold for rejection */
xbar_crit   = E2 + t_crit_left * SE_t

Decision    = "Reject H0 if t_stat < t_crit_left" 
            = "Reject H0 if xbar < xbar_crit"

/* 95% two-sided CI */
t_star_2s   = T.INV.2T(0.05, df)
CI_lower    = xbar - t_star_2s * SE_t
CI_upper    = xbar + t_star_2s * SE_t

/* β & Power (normal approx; plug-in s) for μ1 in E4: left-tailed */
beta_approx = 1 - NORM.DIST( xbar_crit, E4, s/SQRT(n), TRUE )
Power       = 1 - beta_approx

/* Optional exact one-tailed power via noncentral t (Excel 365+):
   λ = (μ1 - μ0) / (s/SQRT(n))
   Power_exact = T.DIST.NC( t_crit_left, df, λ, TRUE )
   Beta_exact  = 1 - Power_exact
*/

/* Sample size for one-sided 80% power to detect δ = |μ1 - μ0| */
n_needed = ROUNDUP( ((NORM.S.INV(1 - E3) + NORM.S.INV(0.80)) * s / δ)^2 , 0 )

Show instructor solution (computed for the dataset)

Using the provided data (n=23), unknown σ → one-sample t, α=0.05 (left-tailed):

x̄ = 9.243, s = 0.873, df = 22
SE = s/√n = 0.873/√23 = 0.1821
t = (9.243 − 10) / 0.1821 = −4.155
p-value (left-tailed) = 0.000207
Decision: p < 0.05 → Reject H₀. Evidence the true mean is under 10 minutes.
95% two-sided CI for μ: 9.243 ± t*·SE, with t* = t_0.975,22 ≈ 2.074 → [8.866, 9.621].
Critical x̄ at α=0.05 (left): t_crit = t_0.05,22 ≈ −1.717; x̄_crit = 10 + (−1.717)·0.1821 ≈ 9.687. Our x̄ = 9.243 < 9.687 ⇒ reject.
β/Power (normal approx; plug-in s):
- μ₁=9.5 → β ≈ 0.152, Power ≈ 0.848
- μ₁=9.0 → β ≈ 0.00008, Power ≈ 0.9999
n for 80% power to detect δ=0.5 (one-sided, α=0.05; z_α=1.645, z_0.8=0.842; using s≈0.873): n = ((1.645+0.842)·0.873/0.5)² ≈ 18.9 → 19.

One-paragraph template:

We tested H₀: μ = 10 vs H₁: μ < 10 at α = 0.05 using a one-sample t-test (n = 23). The sample mean was x̄ = 9.243 (s = 0.873, SE = 0.182), yielding t = −4.155 and a left-tailed p-value = 0.000207, so we reject H₀. A 95% two-sided CI is [8.866, 9.621], which lies below 10. For a practically relevant shift of μ₁ = 9.5, Power ≈ 85%; to reliably detect a 0.5-minute improvement at 80% power we’d need about n ≈ 19 students.

How this differs from the earlier two-sided z-test

Tail: Now left-tailed (μ < μ₀). Before was two-sided (μ ≠ μ₀). One tail uses all α in one direction (stronger for that direction).
Distribution: Now t because σ is unknown and n=23. Before you used z with a historical σ.
Degrees of freedom: t uses df = 22, which makes the critical value slightly more extreme than z for small n.
p-value formula: one-sided uses T.DIST(t,df,TRUE). Two-sided z used 2*(1-NORM.S.DIST(|z|,TRUE)).
Decision rule (x̄): Reject if x̄ < μ₀ + t_crit·SE (with t_crit negative). For two-sided, you had both lower and upper critical bounds.
CI equivalence: A left-tailed α test corresponds to a one-sided 95% CI (upper bound). We also show a two-sided 95% CI for context.
Power: You can still approximate with normal (plug-in s) or use noncentral t if available in Excel.

Quick Q&A (with answers)

Why t instead of z? σ is unknown and n=23 ⇒ use t with df=22.
Why left-tailed? The research question is “faster than 10” (μ < 10).
What is SE here? SE = s/√n using the sample standard deviation.
What’s the decision rule in x̄? Reject if x̄ < μ₀ + t_crit·SE with t_crit=T.INV(α,df).
How do I get the p-value? Use T.DIST(t, df, TRUE) for the left-tail.
Why does “one-tailed” help here? All α is placed in the hypothesized direction, giving more power in that direction than a two-tailed test.
Do I still report a 95% CI? Yes; we show the two-sided 95% CI. A one-sided 95% CI (upper bound) also matches the left-tailed α=0.05 test.
How is power computed? Approximate with normal using s (or use Excel’s T.DIST.NC for exact noncentral-t power if available).
What changed vs the z-test page? Tail (one vs two), distribution (t vs z), and critical/p-value formulas.
As n grows, what happens? t ≈ z and results converge to the z-methods.