Question 1176053: Suppose a random sample of size n = 10 is taken from a normally distributed population with variance sigma^2 = 1.5.
a. Use a Chi-Squared table to approximate the probability that the sample variance is less than 2.446. That is, approximate P(S^2 <= 2.446) where S^2 is the sample variance. (HINT: According to theorem the random variable C = (10-1)S^2/1.5 has a Chi-Squared(9) distribution. This means that the event S^2 <= 2.446 corresponds to the event C = 9S^2/1.5 <= 9(2.446)/1.5 = 14.68. Thus P(S^2 <= 2.446) = P(C <=14.68))
b. Now suppose the sample size is n = 31. Approximate P(S^2 <= 0.83955)
Answer by CPhill(2189)   (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**a. n = 10, S^2 <= 2.446**
1. **Calculate the Chi-Squared Statistic (C):**
* C = (n - 1) * S^2 / σ^2
* C = (10 - 1) * 2.446 / 1.5
* C = 9 * 2.446 / 1.5
* C = 22.014 / 1.5
* C = 14.676 (which rounds to 14.68 as given in the problem)
2. **Degrees of Freedom (df):**
* df = n - 1 = 10 - 1 = 9
3. **Find the Probability Using the Chi-Squared Table:**
* We need to find P(C ≤ 14.68) with 9 degrees of freedom.
* Look up 14.68 in the Chi-Squared distribution table with 9 degrees of freedom.
* Using a Chi-Squared table, we find that 14.68 falls between the values corresponding to cumulative probabilities of 0.10 and 0.05.
* More specifically, for df = 9, the chi-square value for 0.10 is 14.684, and for 0.05 it is 16.919.
* Since 14.676 is very close to 14.684, the probability P(C ≤ 14.68) is approximately 0.10.
**Therefore, P(S^2 ≤ 2.446) ≈ 0.10**
**b. n = 31, S^2 <= 0.83955**
1. **Calculate the Chi-Squared Statistic (C):**
* C = (n - 1) * S^2 / σ^2
* C = (31 - 1) * 0.83955 / 1.5
* C = 30 * 0.83955 / 1.5
* C = 25.1865 / 1.5
* C = 16.791
2. **Degrees of Freedom (df):**
* df = n - 1 = 31 - 1 = 30
3. **Find the Probability Using the Chi-Squared Table or Scipy:**
* We need to find P(C ≤ 16.791) with 30 degrees of freedom.
* Using a Chi-Squared table, we find that the value 16.791 falls between the values corresponding to cumulative probabilities of 0.025 and 0.01.
* More specifically, for df = 30, the chi-square value for 0.025 is 16.791.
* Alternatively, we can use Python's Scipy library, as illustrated in the provided code, which calculates the cumulative distribution function (CDF) of the Chi-Squared distribution.
* Based on the provided code, P(S^2 <= 0.83955) = 0.0250
**Therefore, P(S^2 ≤ 0.83955) ≈ 0.025**
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