Quiz 3:Statistical Inference(Data Science Specialization):Answers2025

Question 1

In a sample of n=9: mean = 1100 cc, sd = 30 cc. 95% Student’s t CI for the mean?

❌ [1092, 1108]
❌ [1080, 1120]
❌ [1031, 1169]
✅ [1077,1123]

Explanation: SE = 30/√9 = 10. t_{0.975, df=8} ≈ 2.306 → margin ≈ 2.306·10 = 23.06. CI = 1100 ± 23.06 ≈ [1076.94, 1123.06] → [1077,1123].

Question 2

9 subjects, mean difference = −2. For the 95% CI to have upper endpoint = 0, what must sd(diff) be?

❌ 0.30
❌ 2.10
❌ 1.50
✅ 2.60

Explanation: Want −2 + t·(sd/√9) = 0 → sd = 6 / t. With df=8, t≈2.306 → sd ≈ 6/2.306 ≈ 2.60.

Question 3

Crossover with 5 runners, each measured under both treatments. Use which test?

✅ A paired interval
❌ Independent groups, since all subjects were seen under both systems
❌ You could use either
❌ It’s necessary to use both

Explanation: Each subject provides a pair of measurements (treatment vs placebo); differences are paired — use paired t-test / interval.

Question 4

20 nights, 10 per group. New mean = 3 (var=0.60), Old mean = 5 (var=0.68). Assume equal variances. 95% CI for (New − Old)?

❌ [-2,70, -1.29]
❌ [1.25, 2.75]
✅ [-2.75, -1.25]
❌ [1.29, 2.70]

Explanation: Pooled var = 0.64 → sp = 0.8. SE = 0.8·√(1/10+1/10)=0.3578. t_{0.975,df=18}≈2.101 → margin ≈0.75. Difference = 3−5 = −2 → CI ≈ −2 ± 0.75 → [−2.75, −1.25].

Question 5

You make 95% CI, then 90% CI on same data. Relationship?

❌ It is impossible to tell.
❌ The interval will be wider
✅ The interval will be narrower.
❌ The interval will be the same width, but shifted.

Explanation: Lower confidence level → smaller critical value → narrower interval.

Question 6

n=100 per group. New mean 4 (sd=0.5), Old mean 6 (sd=2). Consider hypothesis that new decreases mean. What does 95% unequal-variance CI (use Z) suggest for (old − new)?

❌ When subtracting (old – new) the interval is entirely above zero. The new system does not appear to be effective.
✅ When subtracting (old – new) the interval is entirely above zero. The new system appears to be effective.
❌ When subtracting (old – new) the interval contains 0. There is not evidence suggesting that the new system is effective.
❌ When subtracting (old – new) the interval contains 0. The new system appears to be effective.

Explanation: old−new = 2. SE = √(4/100 + 0.25/100) = √0.0425 ≈0.2062. 95% margin ≈1.96·0.2062 ≈0.403 → CI ≈ [1.597, 2.403] > 0 ⇒ old > new ⇒ new reduces waiting time.

Question 7

18 subjects (9 treated, 9 placebo), treated mean diff = −3 (sd 1.5), placebo mean diff = 1 (sd 1.8). 90% t CI for (Treated − Placebo) = ?

❌ [2.636, 5.364]
✅ [-5.364, -2.636]
❌ [2.469, 5.531]
❌ [-5.531, -2.469]

Explanation: Difference = −3 − 1 = −4. Pooled var = ((8·1.5²)+(8·1.8²))/16 = 2.745 → sp ≈1.6565. SE = sp·√(1/9+1/9) ≈0.7817. For df=16, t_{0.95}≈1.746 → margin ≈1.364. CI = −4 ±1.364 → [−5.364, −2.636].

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