(iii) How large would the sample size need to be so that the margin of error for your 95% confidence interval is at most 5? (iv) Unfortunately, Juanita has not taken statistics in college, and treats her sample variance as the population variance, and carries out a test accordingly. Find out what her p-value is, and whether she accepted or rejected Ho at the same level of significance as in part (ii) above,

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### Hypothesis Testing of Costa Rican Three-wattled Bellbird Loudness

**Context:**
Males of the Costa Rican three-wattled bellbird are considered the loudest birds globally. Their call, reaching an audience over half a mile away, is claimed to reach 100 dB (decibels), comparable to a live rock concert. Juanita, skeptical of such claims, aims to verify this through statistical analysis using a sample of 9 bellbirds.

**Assumptions:**
- The loudness of bellbird songs is approximately normally distributed with mean μ (in dB) and unknown variance σ² (in dB²).

**Tasks:**

1. **Hypotheses Formulation (i):**
   Juanita needs to select the appropriate null (H₀) and alternative hypotheses (Hₐ) from the following options:
   - H₀: μ = 100, Hₐ: μ ≠ 100
   - H₀: μ = 100, Hₐ: μ < 100
   - H₀: μ = 100, Hₐ: μ > 100

2. **Conducting the Test (ii):**
   - Sample results: Average loudness = 94.619 dB, Variance = 49.0001 dB².
   - Conduct a hypothesis test to choose between the hypotheses, reporting the p-value and decision at a 2% significance level (α = 0.02).

3. **95% Confidence Interval (iii):**
   - Construct a 95% confidence interval for μ and provide an interpretation.

4. **Sample Size Calculation (iii, second part):**
   - Determine necessary sample size for a margin of error not exceeding 5 dB within the 95% confidence interval.

5. **P-value Interpretation (iv):**
   - Despite lacking statistical training, Juanita adopts population variance as sample variance. Determine her p-value and decide whether she accepts or rejects H₀ at the established significance level (2%).

This educational exercise demonstrates hypothesis testing and statistical interpretation through analyzing the bellbird's loudness claims.
Transcribed Image Text:### Hypothesis Testing of Costa Rican Three-wattled Bellbird Loudness **Context:** Males of the Costa Rican three-wattled bellbird are considered the loudest birds globally. Their call, reaching an audience over half a mile away, is claimed to reach 100 dB (decibels), comparable to a live rock concert. Juanita, skeptical of such claims, aims to verify this through statistical analysis using a sample of 9 bellbirds. **Assumptions:** - The loudness of bellbird songs is approximately normally distributed with mean μ (in dB) and unknown variance σ² (in dB²). **Tasks:** 1. **Hypotheses Formulation (i):** Juanita needs to select the appropriate null (H₀) and alternative hypotheses (Hₐ) from the following options: - H₀: μ = 100, Hₐ: μ ≠ 100 - H₀: μ = 100, Hₐ: μ < 100 - H₀: μ = 100, Hₐ: μ > 100 2. **Conducting the Test (ii):** - Sample results: Average loudness = 94.619 dB, Variance = 49.0001 dB². - Conduct a hypothesis test to choose between the hypotheses, reporting the p-value and decision at a 2% significance level (α = 0.02). 3. **95% Confidence Interval (iii):** - Construct a 95% confidence interval for μ and provide an interpretation. 4. **Sample Size Calculation (iii, second part):** - Determine necessary sample size for a margin of error not exceeding 5 dB within the 95% confidence interval. 5. **P-value Interpretation (iv):** - Despite lacking statistical training, Juanita adopts population variance as sample variance. Determine her p-value and decide whether she accepts or rejects H₀ at the established significance level (2%). This educational exercise demonstrates hypothesis testing and statistical interpretation through analyzing the bellbird's loudness claims.
Expert Solution
Step 1

Given: Sample size, n = 9

          Sample mean, m = 94.619

          SAmple variance, s2 = 49.0001  

 

Part (iv):  Since the population variance is unknown, the margin of error "E" to be at most 5 implies

                                               E =5  t(n-1, α/2)×s2n  = 5   2.306×49.0001n =5                                n =  2.306×49.000152                                n = 10.4 ~ 10           

 

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