Answered

a. Given the following information obtained from four normally distributed populations, construct an ANOVA table. (Round "SS" to 2 decimal places, "MS" to 4 decimal places, and " [tex]$P$[/tex] " to 3 decimal places.)

[tex]\[
SST = 76.83; \quad SST_R = 11.41; \quad C = 4; \quad n_1 = n_2 = n_3 = n_4 = 15
\][/tex]

\begin{tabular}{|l|l|l|l|l|l|}
\hline
\multicolumn{6}{|c|}{ANOVA} \\
\hline
Source of Variation & SS & df & MS & [tex]$F$[/tex] & [tex]$p$[/tex]-value \\
\hline
Between Groups & & & & & 0.028 \\
\hline
Within Groups & & & & & \\
\hline
Total & 0.00 & 0 & & & \\
\hline
\end{tabular}

b. At the [tex]$5\%$[/tex] significance level, what is the conclusion to the ANOVA test of mean differences?

A. Reject [tex]$H_0$[/tex]: we can conclude that some means differ.

B. Do not reject [tex]$H_0$[/tex]: we cannot conclude that some means differ.

C. Do not reject [tex]$H_0$[/tex]: we can conclude that some means differ.

D. Reject [tex]$H_0$[/tex]: we cannot conclude that some means differ.



Answer :

GinnyAnswer
Alright, let's walk through each step in constructing the ANOVA table based on the given information. We'll fill out the ANOVA table and then interpret the results.

### Step-by-Step Solution:

#### Given Information:
- Total Sum of Squares (SST) = 76.83
- Sum of Squares for Treatment (SSTR) = 11.41
- Number of groups (C) = 4
- Number of observations per group ([tex]\( n_1 = n_2 = n_3 = n_4 \)[/tex]) = 15

#### Calculations:

1. Total Number of Observations [tex]\( (N) \)[/tex]
[tex]\[ N = n_1 + n_2 + n_3 + n_4 = 15 + 15 + 15 + 15 = 60 \][/tex]

2. Degrees of Freedom (df) Calculation:
- Total Degrees of Freedom [tex]\( (df_{\text{total}}) \)[/tex]
[tex]\[ df_{\text{total}} = N - 1 = 60 - 1 = 59 \][/tex]
- Between Groups Degrees of Freedom [tex]\( (df_{\text{between}}) \)[/tex]
[tex]\[ df_{\text{between}} = C - 1 = 4 - 1 = 3 \][/tex]
- Within Groups Degrees of Freedom [tex]\( (df_{\text{within}}) \)[/tex]
[tex]\[ df_{\text{within}} = df_{\text{total}} - df_{\text{between}} = 59 - 3 = 56 \][/tex]

3. Mean Squares Calculation:
- Mean Square Between Groups [tex]\( (MS_{\text{between}}) \)[/tex]
[tex]\[ MS_{\text{between}} = \frac{SSTR}{df_{\text{between}}} = \frac{11.41}{3} = 3.8033 \][/tex]
- Mean Square Within Groups [tex]\( (MS_{\text{within}}) \)[/tex]
[tex]\[ MS_{\text{within}} = \frac{SST - SSTR}{df_{\text{within}}} = \frac{76.83 - 11.41}{56} = \frac{65.42}{56} = 1.1682 \][/tex]

4. F-statistic Calculation:
[tex]\[ F = \frac{MS_{\text{between}}}{MS_{\text{within}}} = \frac{3.8033}{1.1682} = 3.2557 \][/tex]

5. Conclusion based on the p-value:
- Given p-value: 0.028
- Significance Level [tex]\( (\alpha) \)[/tex]: 0.05

Since the p-value (0.028) is less than the significance level (0.05), we reject the null hypothesis [tex]\( H_0 \)[/tex].

### ANOVA Table:

[tex]\[ \begin{array}{|l|l|l|l|l|l|} \hline \text{Source of Variation} & \text{SS} & \text{df} & \text{MS} & F & p \text{-value} \\ \hline \text{Between Groups} & 11.41 & 3 & 3.8033 & 3.2557 & 0.028 \\ \hline \text{Within Groups} & 65.42 & 56 & 1.1682 & & \\ \hline \text{Total} & 76.83 & 59 & & & \\ \hline \end{array} \][/tex]

### Conclusion:
Based on the ANOVA table and the p-value:
- Reject [tex]\( H_0 \)[/tex]: we can conclude that some means differ.