By using excel we can solve this question easily.
First, enter all data into excel.
Click on Data -------> Data Analysis--------> Anova: Two-Factor Without Replication-----> Input Data -------> Select all data -------> Output Range : Select any empty cell --------> ok
We get
Anova: Two-Factor Without Replication | ||||||
SUMMARY | Count | Sum | Average | Variance | ||
Row 1 | 4 | 30 | 7.5 | 4.333333 | ||
Row 2 | 4 | 26 | 6.5 | 3 | ||
Row 3 | 4 | 32 | 8 | 0.666667 | ||
Column 1 | 3 | 21 | 7 | 1 | ||
Column 2 | 3 | 19 | 6.333333 | 2.333333 | ||
Column 3 | 3 | 27 | 9 | 1 | ||
Column 4 | 3 | 21 | 7 | 4 | ||
ANOVA | ||||||
Source of Variation | SS | df | MS | F | P-value | F crit |
Rows | 4.666667 | 2 | 2.333333 | 1.166667 | 0.373248 | 5.143253 |
Columns | 12 | 3 | 4 | 2 | 0.215553 | 4.757063 |
Error | 12 | 6 | 2 | |||
Total | 28.66667 | 11 |
4) SSE = 12 ( Using above table)
5) SST = 28.67 ( Using above table)
6) Critical F value for treatments = 5.14 ( Using above table)
Data from a randomized block design are shown in the following table. Treatment Levels 2 3...
Topic: ANOVA Topic: ANOVA 1- An experiment was conducted using a randomized block design. The data from the experiment are displayed in the following table. Block Treatment 1 2 3 1 2 3 5 2 8 6 7 3 7 6 5 a) Fill in the missing entries in the ANOVA table. Source df SS MS F Treatment 2 21.5555 Block 2 Error 4 Total 8 30.2222 b) Specify the null use to investigate whether a difference exists among the...
The following data are from a completely randomized design. Treatment Treatment Treatment 32 30 30 26 32 30 35 38 37 38 42 38 6.5 45 45 47 49 46 Sample mean Sample variance At the α-.05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean Squares, Treatment Mean Squares, Error
The following data are from a completely randomized design. Treatment Treatment Treatment A B C 32 47 34 30 46 37 30 47 36 26 49 37 32 51 41 Sample mean 30 48 37 Sample variance 6 4 6.5 At the = .05 level of significance, can we reject the null hypothesis that the means of the three treatments are equal? Compute the values below (to 1 decimal, if necessary). Sum of Squares, Treatment Sum of Squares, Error Mean...
Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table. Treatment 10 98 12 18 21 2 3 4 Blocks Use a - .05 to test for any significant differences. Show entries to 2 decimals, if necessary. Round p-value to four decimal places. If your answer is zero enter "o". Source of Variation Sum of Squares Degrees of Freedom Mean Square Treatments Blocks Error Total
The following data are from a completely randomized design. Treatment 145 145 145 149134 151 140 129 Sample mean Sample variance a. Compute the sum of squares between treatments. 159 310 142 108.8 134 145.2 b. Compute the mean square between treatments. c. Compute the sum of squares due to error d. Compute the mean square due to error (to 1 decimal), e. Set up the ANOVA table for this problem. Round all Sum of Squares to nearest whole numbers....
Consider the experimental results for the following randomized block design. Make the calculations necessary to set up the analysis of variance table Treatment 1 10 2 13 3 19 4 20 Blocks 15 18 7 Use α-.05 to test for any significant differences. Show entries to 2 decimals, if necessary. Round p-value to four decimal places. If your answer is zero enter "O 15 19 Sum of Squares Source of Degrees of Freedom Mean p-value Variation Square Treatments Blocks Error...
The following data are from a completely randomized design. Treatment 164 149 142 157 167 124 145 149 149 137 169 136 156 142 144 141.6 119.6 126 122 133 141 152 130 134 Sample mean Sample variance a. Compute the sum of squares between treatments. Round the intermediate calculations to whole number 1488 b. Compute the mean squ are between treatments. 744 c. Compute the sum of squares due to error. 135.33 d. Compute the mean square due to...
The following data are from a completely randomized design. Treatment 163 145 123 142 157 121 166 129 132 144 145 144 148 138 153 191 138 131 159 142 134 344.8 88.8 152.8 Sample mean Sample variance a. Compute the sum of squares between treatments. Round the intermediate calculations to whole number. b. Compute the mean square between treatments. c. Compute the sum of squares due to error. d. Compute the mean square due to error (to 1 decimal).
STA1502/101/3/2019 QUESTION 18 A randomized block design with 4 treatments and 5 blocks produce the following sum of squares values: SS Total 1951 SST 349 SSE 18 The value of SSB must be: 1. 1414 2. 537 3. 1763 4. 1602 5. 534
5. A randomized block design is used in an experiment. There are 4 treatments and 3 blocks. Use the information below to complete the table. b a b a (*; – x)² = 50 {(xy - x)= 712 (#4 - 3)2 = 98 i=1 j=1 i= Degrees of Freedom Sum of Squares Mean Square f P value Source of Variation Treatments Blocks Error Total