✔ 最佳答案
ANSWER: To significance level of 99%, more than 0.15 (15%) damaged trees in Hopkins Forest
POPULATION PROPORTION HYPOTHESIS TESTING, NORMAL DISTRIBUTION, 7-Step Procedure
1. PARAMETER OF INTEREST: p = POPULATION PROPORTION
2. HYPOTHESES :
NULL HYPOTHESIS H0: p = 0.15 (15% DAMAGED TREES)
ALTERNATIVE HYPOTHESIS H1: p > 0.15 (exceeding 15% DAMAGED TREES)
3. COMPUTATION OF POPULATION PARAMETER:
STANDARD DEVIATION σ = sqrt[ p * ( 1 - p) / n ]
n = sample size [100]
σ = sqrt [0.15 * (1 - 0.15)/100] = 0.035707142
4. COMPUTATION OF TEST STATISTIC ("critical value of z")
z = (p_hat - p) / σ
p_hat = SAMPLE PROPORTION [0.25] (25/100)
z = (0.25-0.15)/ 0.035707142 = 2.8
Valid when both conditions true:
n * p > 10 [100 *0.15 = 5 > 10] TRUE
n (1 − p) > 10 [100 * (1 - 0.15) = 85 > 10] TRUE
5. "P-value" "Look-up" value of Normal Distribution "area under the curve" "to the right" of z = 2.8
"P-value" = 0.0026
6. TEST of P-value
P-value ≤ α (significance level) [0.0026 ≤ 0.01]; reject NULL HYPOTHESIS stating proportion of damaged trees = 0.15 (15%) for any reasonable significance level.
"When P-value is low, let it go" (let go the NULL HYPOTHESIS) P-value ≤ α
"When P-value is high, let it fly" (let fly the NULL HYPOTHESIS) P-value > α
7. CONCLUSION:
For signficance level α = 0.01, NULL HYPOTHESIS H0: p = 0.15 (15% damaged trees) should be rejected. More than 0.15 (15%) damaged trees.