SOLUTION: The following data on annual rates of return were collected from eleven randomly selected stocks listed on the New York Stock Exchange (“the big board”) and twelve randomly sel

Question 1199985: The following data on annual rates of return were collected from eleven randomly selected stocks listed on the New York Stock Exchange (“the big board”) and twelve randomly selected stocks listed on NASDAQ. Assume the population standard deviations are the same. At the 0.10 significance level, can we conclude that the annual rates of return are higher on the big board?
NYSE: 15.0, 10.7, 20.2, 18.6, 19.1, 8.7, 17.8, 13.8, 22.7, 14.0, 26.1
NASDAQ: 8.8, 6.0, 14.4, 19.1, 17.6, 17.8, 15.9, 17.9, 21.6, 6.0, 11.9, 23.4

Answer by textot(100) (Show Source): You can put this solution on YOUR website!
**1. Set up Hypotheses**
* **Null Hypothesis (H0):**
* μ₁ - μ₂ ≤ 0
* The mean annual rate of return for NYSE stocks is less than or equal to the mean annual rate of return for NASDAQ stocks.
* **Alternative Hypothesis (H1):**
* μ₁ - μ₂ > 0
* The mean annual rate of return for NYSE stocks is greater than the mean annual rate of return for NASDAQ stocks.
**2. Calculate Sample Statistics**
* **NYSE:**
* Sample size (n1) = 11
* Sample mean (x̄₁): 16.77
* Sample standard deviation (s1): 5.29
* **NASDAQ:**
* Sample size (n2) = 12
* Sample mean (x̄₂) = 14.59
* Sample standard deviation (s2): 5.96
**3. Calculate Pooled Standard Deviation**
* Pooled standard deviation (sp) = √[((n1 - 1) * s1^2 + (n2 - 1) * s2^2) / (n1 + n2 - 2)]
* sp = √[((11 - 1) * 5.29^2 + (12 - 1) * 5.96^2) / (11 + 12 - 2)]
* sp ≈ 5.64
**4. Calculate t-Statistic**
* t = (x̄₁ - x̄₂) / [sp * √(1/n1 + 1/n2)]
* t = (16.77 - 14.59) / [5.64 * √(1/11 + 1/12)]
* t ≈ 0.67
**5. Determine Degrees of Freedom**
* Degrees of freedom (df) = n1 + n2 - 2 = 11 + 12 - 2 = 21
**6. Find the Critical Value**
* Since this is a one-tailed test with α = 0.10 and df = 21, the critical value from the t-distribution table is approximately 1.323.
**7. Compare t-Statistic to Critical Value**
* Calculated t-statistic (0.67) is less than the critical value (1.323).
**8. Make a Decision**
* Since the t-statistic does not fall in the rejection region, we **fail to reject the null hypothesis**.
**9. Conclusion**
* At the 0.10 significance level, there is **not enough evidence** to conclude that the mean annual rate of return for stocks listed on the NYSE is higher than the mean annual rate of return for stocks listed on NASDAQ.

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