---
title: "Test: electrification effect on barley tillers"
subject: "Probability and Statistics"
chapter: 6
key: "semester_5/probability_and_statistics:barley-pots-electrification-tillers"
source: https://www.bctengineer.com/q/semester_5/probability_and_statistics/barley-pots-electrification-tillers
site: BCT Engineering
---

# Test: electrification effect on barley tillers

Probability and Statistics · Chapter 6: Inference concerning mean

- Asked in: 2068 Chaitra, 6 marks
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/probability_and_statistics/barley-pots-electrification-tillers
- Whole chapter: https://www.bctengineer.com/md/semester_5/probability_and_statistics/chapter-6

## Question

Eight pots growing barley plants each were exposed to a high tension discharge, while nine similar pots were enclosed in an earthed wire cage. The number of tillers in each pot were as follows:
| Caged | 17 | 27 | 18 | 25 | 27 | 29 | 27 | 23 | 17 |
|---|---|---|---|---|---|---|---|---|---|
| Electrified | 16 | 16 | 20 | 16 | 20 | 17 | 15 | 21 | |
Test the hypothesis whether electrification exercises have any real effect on the tillering at α = 0.05 level of significance.

## Short answer

Pooled two-sample t-test: t = (23.333−17.625)/1.869 = 3.055, df=15, critical t=±2.131. Reject H0; electrification significantly affects tillering. Caged pots had higher mean tillers (23.33 vs 17.62).

## Answer

The samples are small and independent, with unknown variances assumed equal, so a pooled two-sample t-test is used.

1. H0: μ₁ = μ₂ (electrification has no effect on tillering)
   H1: μ₁ ≠ μ₂ (two-tailed)
2. Level of significance: α = 0.05
3. Sample statistics:

```text
Caged:       n₁ = 9, Σx = 210, Σx² = 5084
             x̄₁ = 210/9 = 23.333,  s₁² = 23.000
Electrified: n₂ = 8, Σx = 141, Σx² = 2523
             x̄₂ = 141/8 = 17.625,  s₂² = 5.411

Sp² = [(n₁−1)s₁² + (n₂−1)s₂²]/(n₁+n₂−2)
    = [8×23.000 + 7×5.411]/15 = 14.792,  Sp = 3.846
SE  = Sp·√(1/n₁ + 1/n₂) = 3.846 × √(1/9 + 1/8) = 1.869
t   = (x̄₁ − x̄₂)/SE = (23.333 − 17.625)/1.869 = 3.055
```

4. Degrees of freedom = n₁ + n₂ − 2 = 15. Critical value: t(0.025; 15) = ±2.131. Reject H0 if |t| > 2.131.
5. Decision: |t| = 3.055 > 2.131 (p-value = 0.008), so **reject H0**.

**Conclusion:** electrification has a real (significant) effect on tillering at the 5% level; the caged pots produced more tillers on average (23.33 against 17.62).

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which test fits the barley data (n = 9 and 8, unknown equal variances)?
   A. Two-sample z-test
   B. Pooled two-sample t-test
   C. Chi-square test of independence
   D. Paired t-test
   Answer: B. Small independent samples with unknown but assumed equal variances call for a pooled t-test. (Key point: Choice of test)

2. How is the pooled variance Sp² formed?
   A. Simple average of x̄1 and x̄2
   B. (s1² + s2²)/(n1 + n2)
   C. [(n1−1)s1² + (n2−1)s2²]/(n1+n2−2)
   D. s1² × s2²
   Answer: C. Sp² weights each sample variance by its degrees of freedom: here 14.792. (Key point: Pooled variance)

3. How many degrees of freedom does the t-test have?
   A. 8
   B. 7
   C. 17
   D. 15
   Answer: D. d.f. = n1 + n2 − 2 = 9 + 8 − 2 = 15, giving t(0.025;15) = ±2.131. (Key point: Degrees of freedom)

4. t = 3.055 against ±2.131. What is the conclusion?
   A. Reject H0; electrification has a real effect on tillering
   B. Electrification has no effect
   C. The test needs a z critical value
   D. Accept H0 because p = 0.008 < 0.05
   Answer: A. |t| > 2.131 so H0 is rejected: caged pots averaged more tillers (23.33 vs 17.62). (Key point: Decision)

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