Questions: To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow. Temperature 50°C 60°C 70°C 34 29 22 25 30 29 36 34 28 39 22 29 31 30 32 Construct an analysis of variance table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.) Source of Variation Sum of Squares Degrees of Freedom Mean Square F

To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow.

Temperature
50°C 60°C 70°C
34 29 22
25 30 29
36 34 28
39 22 29
31 30 32

Construct an analysis of variance table. (Round your values for MSE and F to two decimal places, and your p-value to four decimal places.)

Source of Variation Sum of Squares Degrees of Freedom Mean Square F

Transcript text: To study the effect of temperature on yield in a chemical process, five batches were produced at each of three temperature levels. The results follow. \begin{tabular}{|c|c|c|} \hline \multicolumn{3}{|c|}{ Temperature } \\ \hline $\mathbf{5 0}^{\circ} \mathbf{C}$ & $\mathbf{6 0}{ }^{\circ} \mathbf{C}$ & $\mathbf{7 0}^{\circ} \mathbf{C}$ \\ \hline 34 & 29 & 22 \\ \hline 25 & 30 & 29 \\ \hline 36 & 34 & 28 \\ \hline 39 & 22 & 29 \\ \hline 31 & 30 & 32 \\ \hline \end{tabular} Construct an analysis of variance table. (Round your values for MSE and $F$ to two decimal places, and your $p$-value to four decimal places.) \begin{tabular}{|c|l|l|l|l|l|l|} \hline \begin{tabular}{c} Source \\ of Variation \end{tabular} & \begin{tabular}{c} Sum \\ of Squares \end{tabular} & \begin{tabular}{c} Degrees \\ of Freedom \end{tabular} & \begin{tabular}{c} Mean \\ Square \end{tabular} & & $\boldsymbol{F}$ \end{tabular}

Solution

Solution Steps

Step 1: Calculate Sum of Squares

The sum of squares between groups ($SS_{between}$) is calculated as:

\[ SS_{between} = \sum_{i=1}^k n_i (\bar{X}_i - \bar{X})^2 = 70.0 \]

The sum of squares within groups ($SS_{within}$) is calculated as:

\[ SS_{within} = \sum_{i=1}^k \sum_{j=1}^{n_i} (X_{ij} - \bar{X}_i)^2 = 244.0 \]

Step 2: Calculate Degrees of Freedom

The degrees of freedom between groups ($df_{between}$) is:

\[ df_{between} = k - 1 = 3 - 1 = 2 \]

The degrees of freedom within groups ($df_{within}$) is:

\[ df_{within} = N - k = 15 - 3 = 12 \]

Step 3: Calculate Mean Squares

The mean square between groups ($MS_{between}$) is:

\[ MS_{between} = \frac{SS_{between}}{df_{between}} = \frac{70.0}{2} = 35.0 \]

The mean square within groups ($MS_{within}$) is:

\[ MS_{within} = \frac{SS_{within}}{df_{within}} = \frac{244.0}{12} = 20.3333 \]

Step 4: Calculate F-statistic

The F-statistic is calculated as:

\[ F = \frac{MS_{between}}{MS_{within}} = \frac{35.0}{20.3333} = 1.7213 \]

Step 5: Calculate p-value

The p-value is calculated using the F-distribution with $df_{between} = 2$ and $df_{within} = 12$:

\[ P = 1 - F(F_{observed}; df_{between}, df_{within}) = 1 - F(1.7213; 2, 12) = 0.2202 \]

Final Answer

The analysis of variance table is summarized as follows:

Degrees of Freedom Between Groups: $2$
Degrees of Freedom Within Groups: $12$
Mean Square Between Groups (MSB): $35.0$
Mean Square Within Groups (MSW): $20.3333$
F-statistic: $1.7213$
p-value: $0.2202$

Since the p-value $0.2202$ is greater than the significance level $\alpha = 0.05$, we do not reject the null hypothesis. There is no significant effect of temperature on yield in the chemical process.

\[ \boxed{\text{No significant effect of temperature on yield.}} \]

Was this solution helpful?

Unhelpful

Helpful

Questions asked by other users

< Previous Next >

Thank you for your feedback.

Copied!