10) The Baldwin Piano Company has long argued that their method of teaching people to play the piano is superior to that of its competitors. To estimate any difference in the length of time required to learn the basic techniques, as the new staff analyst for Baldwin, you select 100 students who used your method and find the mean time was 159 hours with \( s=38.6 \) hours for your method. In a sample of 120 students using a competitor's method, the mean was 184 hours with \( s=58.6 \). a) Construct a \( 99 \% \) confidence interval for the difference in means. b) Test the hypothesis, at the \( 5 \% \) level, that the Baldwin method reduces the time needed to learn the piano.

**Part b: Hypothesis Testing** To test whether the Baldwin method reduces the time needed to learn the piano at the \( 5\% \) significance level, we perform the following steps: 1. **State the Hypotheses:** - **Null Hypothesis (\( H_0 \))**: The Baldwin method does not reduce the learning time. Mathematically, \( \mu_1 - \mu_2 \geq 0 \). - **Alternative Hypothesis (\( H_a \))**: The Baldwin method reduces the learning time. Mathematically, \( \mu_1 - \mu_2 < 0 \). 2. **Calculate the Test Statistic:** - **Difference in Sample Means (\( \bar{x}_1 - \bar{x}_2 \))**: \( 159 - 184 = -25 \) hours. - **Standard Error (SE)**: As calculated in Part a, \( SE \approx 6.595 \) hours. - **Test Statistic (\( z \))**: \[ z = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{SE} = \frac{-25}{6.595} \approx -3.785 \] 3. **Determine the Critical Value:** - For a one-tailed test at the \( 5\% \) significance level, the critical z-value is \( z_{0.05} = -1.645 \). 4. **Decision Rule:** - If \( z \leq -1.645 \), reject the null hypothesis. 5. **Conclusion:** - Since \( -3.785 < -1.645 \), we reject the null hypothesis. - **Interpretation**: There is sufficient evidence at the \( 5\% \) level to conclude that the Baldwin Piano Company’s method reduces the time needed to learn the basic piano techniques compared to the competitor’s method. **Summary:** - **Test Statistic**: \( z \approx -3.785 \) - **Decision**: Reject \( H_0 \) - **Conclusion**: The Baldwin method significantly reduces the learning time. Answer: Problem b Answer We perform a one-tailed test with H₀: μ₁ – μ₂ ≥ 0 versus H₁: μ₁ – μ₂ < 0. The test statistic is (–25)/6.595 ≈ –3.785, which is less than the critical value –1.645 for α = 0.05. Therefore, we reject H₀ and conclude at the 5% significance level that the Baldwin method reduces the time needed to learn the piano.