Question 1193796: 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)
Find the expected frequencies of hits given by a Poison distribution having the same mean and total as the observed distribution.
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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