Let's answer each part step-by-step using the information from the TI-84 Plus ANOVA output. ### a. **State the null hypothesis.** The null hypothesis (H₀) for a one-way ANOVA is: > **H₀:** All group means are equal. > Mathematically, μ₁ = μ₂ = μ₃ = μ₄ --- ### b. **How many levels were there for the factor?** From the output: - **df (factor) = 3** So, **number of levels = df + 1 = 3 + 1 = 4** --- ### c. **Assume the design was balanced. What was the sample size for each factor?** From the output: - **df (error) = 28** - **df (factor) = 3** - **k = 4 groups (from part b)** - **Total n = df(factor) + df(error) + 1 = 3 + 28 + 1 = 32** If design is balanced: - Sample size per group = Total n / Number of groups = 32 / 4 = **8** --- ### d. **What are the values of SStr, SSE, MSTr, and MSE?** From the output: - **SStr (Sum of Squares for treatment/factor):** 1070.5 - **SSE (Sum of Squares for error):** 15711.5 - **MSTr (Mean Square for treatment):** 356.83333 - **MSE (Mean Square for error):** 561.125 --- ### e. **What is the value of the test statistic?** From the output: - **F = 0.6359248533** --- ### f. **What is the P-value?** From the output: - **p = 0.5981345619** --- ### g. **State your conclusion regarding H₀. Use the α = 0.05 significance level.** - **P-value = 0.5981 > 0.05** - Since the P-value is much greater than α = 0.05, **we fail to reject the null hypothesis**. **Conclusion:** There is not enough evidence to conclude that at least one group mean is different. All group means appear to be equal. --- ## **Final Summary Table** | Part | Answer | |------|------------------------------------------------------------------------| | a | H₀: μ₁ = μ₂ = μ₃ = μ₄ (All group means are equal) | | b | 4 levels | | c | Sample size per group = 8 | | d | SStr = 1070.5, SSE = 15711.5, MSTr = 356.83333, MSE = 561.125 | | e | F = 0.6359 | | f | P-value = 0.5981 | | g | Fail to reject H₀; not enough evidence to show group means differ | Let me know if you want a breakdown of the calculations!

Let's go through each part step-by-step, using the information provided in the ANOVA output from the TI-84 Plus. --- ### **Given Data from the TI-84 Output:** | Statistic | Value | |----------------------------|-------------------| | F-statistic | 0.6359248533 | | df (Factor) | 3 | | df (Error) | 28 | | P-value | 0.5981346619 | | SSTr (Sum of Squares for treatment) | 1070.5 | | SSE (Sum of Squares for error) | 15711.5 | | MSTr (Mean Square for treatment) | 356.833333 | | MSE (Mean Square for error) | 561.125 | --- ### **a. State the null hypothesis** **Null hypothesis (H₀):** All group means are equal. \[ H_0: \mu_1 = \mu_2 = \mu_3 = \mu_4 \] **Alternative hypothesis (H₁):** At least one group mean differs. --- ### **b. How many levels were there for the factor?** - The degrees of freedom for the factor (df) = 3 - Number of levels = df + 1 = 3 + 1 = **4** --- ### **c. Assume the design was balanced. What was the sample size for each factor?** - The degrees of freedom for error = 28 - Total number of observations (N): \[ N = \text{df}_{\text{factor}} + \text{df}_{\text{error}} + 1 = 3 + 28 + 1 = 32 \] - Since there are 4 groups (levels), and the design is balanced: \[ \text{Sample size per group} = \frac{N}{\text{number of groups}} = \frac{32}{4} = \boxed{8} \] --- ### **d. What are the values of SSTr, SSE, MSTr, and MSE?** - **SSTr (Sum of Squares for treatment):** 1070.5 (given) - **SSE (Sum of Squares for error):** 15711.5 (given) - **MSTr (Mean Square for treatment):** \[ MSTr = \frac{SSTr}{df_{\text{factor}}} = \frac{1070.5}{3} \approx 356.833 \] - **MSE (Mean Square for error):** \[ MSE = \frac{SSE}{df_{\text{error}}} = \frac{15711.5}{28} \approx 561.125 \] --- ### **e. What is the value of the test statistic?** - The F-statistic is directly given: \[ F \approx \boxed{0.6359} \] --- ### **f. What is the P-value?** - The P-value from the output: \[ p \approx \boxed{0.5981} \] --- ### **g. Conclusion regarding H₀ at α = 0.05** - Since **p ≈ 0.5981 > 0.05**, **we fail to reject the null hypothesis**. - **Interpretation:** There is not enough evidence to conclude that the group means differ. --- ## **Summary of Calculations:** | Calculation | Value | Explanation | |----------------------------------------------|----------------------------------|----------------------------------------------------------| | Number of levels | 4 | df = 3 ⇒ levels = df + 1 | | Total observations | 32 | df_error + df_factor + 1 = 28 + 3 + 1 = 32 | | Sample size per group | 8 | 32 / 4 = 8 | | MSTr | 356.833 | SSTr / df_factor | | MSE | 561.125 | SSE / df_error | | F statistic | 0.6359 | Directly from output | | P-value | 0.5981 | From output | --- **If you'd like further clarification or calculations, feel free to ask!**