Question 1193797
**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.