Question 1200717: Arbitron Media Research Inc. conducted a study of the iPod listening habits of men and women. One facet of the study involved the mean listening time. It was discovered that the mean listening time for a sample of 10 men was 42 minutes per day. The standard deviation was 16 minutes per day. The mean listening time for a sample of 13 women was also 42 minutes, but the standard deviation of the sample was 15 minutes. At the 0.02 significance level, can we conclude that there is a difference in the variation in the listening times for men and women?
Compute the p-value. (Round your answer to 4 decimal places.?
Answer by textot(100)   (Show Source):
You can put this solution on YOUR website! **1. Set up Hypotheses**
* **Null Hypothesis (H0):** σ1² = σ2² (The variances of listening times for men and women are equal)
* **Alternative Hypothesis (H1):** σ1² ≠ σ2² (The variances of listening times for men and women are not equal)
**2. Calculate Test Statistic (F-statistic)**
* F = s1² / s2²
* F = 16² / 15²
* F = 256 / 225
* F = 1.138
**3. Determine Degrees of Freedom**
* Degrees of freedom for the numerator (df1) = n1 - 1 = 10 - 1 = 9
* Degrees of freedom for the denominator (df2) = n2 - 1 = 13 - 1 = 12
**4. Find the P-value**
* Using an F-distribution table or statistical software (like R or Python), find the p-value associated with the calculated F-statistic (1.138), df1 = 9, and df2 = 12.
* **P-value ≈ 0.7724**
**5. Make a Decision**
* **Significance Level (α) = 0.02**
* **Since the p-value (0.7724) is greater than α (0.02), we fail to reject the null hypothesis.**
**Conclusion**
* At the 0.02 significance level, there is **not enough evidence** to conclude that there is a difference in the variation in listening times for men and women.
**Note:**
* This analysis assumes that the listening times for both men and women are normally distributed.
* If the normality assumption is not met, other tests like the Levene's test or Bartlett's test might be more appropriate.
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