Question 1193797: The table below gives the distribution of the number of hits by flying bombs in 450 equally sized areas in City X during World War II.
Number of hit (x) (0) (1) (2) (3) (4) (5) (6 or more)
Frequency (f) (180) (173) (69) (20) (6) (2) (0)
Use χ^2 distribution and a 10% level of significance to test the adequacy of the Poisson distribution as a model for these data.
Answer by proyaop(69)   (Show Source):
You can put this solution on YOUR website! **1. Calculate the Mean Number of Hits**
* Mean number of hits (λ) = (Σf * x) / Σf
* Where f is the frequency and x is the number of hits.
* λ = (0*180 + 1*173 + 2*69 + 3*20 + 4*6 + 5*2) / 450
* λ = 0.5
**2. Calculate Expected Frequencies under Poisson Distribution**
* Use the Poisson probability mass function:
* P(X = x) = (e^(-λ) * λ^x) / x!
* Where:
* X is the number of hits
* λ is the mean number of hits (0.5)
* e is the base of the natural logarithm (approximately 2.71828)
* x! is the factorial of x
* Calculate the expected frequency (E) for each number of hits:
* E(X = 0) = 450 * P(X = 0) = 450 * (e^(-0.5) * 0.5^0) / 0! = 270.27
* E(X = 1) = 450 * P(X = 1) = 450 * (e^(-0.5) * 0.5^1) / 1! = 135.14
* E(X = 2) = 450 * P(X = 2) = 450 * (e^(-0.5) * 0.5^2) / 2! = 33.78
* E(X = 3) = 450 * P(X = 3) = 450 * (e^(-0.5) * 0.5^3) / 3! = 5.63
* E(X = 4) = 450 * P(X = 4) = 450 * (e^(-0.5) * 0.5^4) / 4! = 0.70
* E(X = 5) = 450 * P(X = 5) = 450 * (e^(-0.5) * 0.5^5) / 5! = 0.07
* E(X ≥ 6) = 450 * (1 - [P(X = 0) + P(X = 1) + P(X = 2) + P(X = 3) + P(X = 4) + P(X = 5)]) = 0.01
**3. Calculate the Chi-Square Test Statistic**
* **Chi-Square (χ²) = Σ [(O - E)² / E]**
* Where:
* O = Observed frequency
* E = Expected frequency
* **Calculate for each category:**
* (180 - 270.27)² / 270.27 = 32.18
* (173 - 135.14)² / 135.14 = 12.19
* (69 - 33.78)² / 33.78 = 36.03
* (20 - 5.63)² / 5.63 = 38.84
* (6 - 0.70)² / 0.70 = 42.57
* (2 - 0.07)² / 0.07 = 50.43
* (0 - 0.01)² / 0.01 = 0.01
* **Sum the values:** χ² = 32.18 + 12.19 + 36.03 + 38.84 + 42.57 + 50.43 + 0.01 = 212.25
**4. Determine Degrees of Freedom**
* Degrees of Freedom (df) = Number of categories - Number of parameters estimated - 1
* Number of categories = 7 (0 hits, 1 hit, ..., 5 hits, 6 or more hits)
* Number of parameters estimated = 1 (the mean, λ)
* df = 7 - 1 - 1 = 5
**5. Find the Critical Value**
* Using a chi-square distribution table, find the critical value for α = 0.10 and df = 5.
* The critical value is approximately 9.24.
**6. Make a Decision**
* **Compare the calculated chi-square statistic to the critical value:**
* 212.25 > 9.24
* **Decision:** Since the calculated chi-square statistic (212.25) is greater than the critical value (9.24), we **reject the null hypothesis**.
**Conclusion**
* There is sufficient evidence at the 0.10 level of significance to conclude that the Poisson distribution is not an adequate fit for the observed data.
* The observed frequencies of bomb hits deviate significantly from what would be expected under a Poisson distribution.
**Note:**
* This analysis assumes that the expected frequencies in each category are sufficiently large (generally, expected frequencies should be greater than 5).
* For categories with expected frequencies less than 5, you may need to combine them with adjacent categories to meet this assumption.
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