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**\[\begin{array}{|c|c|c|}\hline50^\circ C & 60^\circ C & 70^\circ C \\\hline33 & 28 & 25 \\23 & 29 & 30 \\35 & 32 & 30 \\38 & 21 & 32 \\31 & 25 & 33 \\\hline\end{array}\]Construct an analysis of variance table (to 2 decimals but *p*-value to 4 decimals, if necessary):| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | *p*-value ||---------------------|----------------|--------------------|-------------|---|-----------|| Treatments | | | | | || Error | | | | | || Total | | | | | |Use a 0.05 level of significance to test whether the temperature level has an effect on the mean yield of the process.Calculate the value of the test statistic (to 2 decimals).- [Input Box for Test Statistic]The *p*-value is - [Dropdown for Selecting Answer]What is your conclusion?- [Dropdown for Selecting Answer]

Obtain the values in the ANOVA table. Use EXCEL Procedure to obtain the values in the ANOVA table. Follow the instruction to obtain the values in the ANOVA table. Open EXCEL Go to Data>Data Analysis. Choose ANOVA: Single Factor. Enter the input range as $A$1:$C$6. Select Grouped By as Columns. Check the Labels in the first row. Set the alpha as 0.05. Click OK. EXCEL output: From EXCEL output, the values in the ANOVA table are as follows:State the hypotheses. Determine the value of test statistic. From the EXCEL output, the value of test statistic is 1.61. Thus, value of test statistic is 1.61. Determine the P-value. From EXCEL output, the P-value is 0.2402 Thus, the P-value is 0.2402 The conclusion is obtained as follows: Use the level of significance as 0.05. The P-value is greater than the level of significance. Therefore by the rejection rule do not reject the null hypothesis. That is, there is no evidence to conclude that temperature has an effect on the mean yield of the process.