Source dF MS F A 2 45 4.5 B 1 70 7.0 AB 2 170 17...
|
Source |
dF |
MS |
F |
|
A |
2 |
45 |
4.5 |
|
B |
1 |
70 |
7.0 |
|
AB |
2 |
170 |
17 |
|
Within |
60 |
10 |
For which source of variation is the null hypothesis rejected at the .01 level of significance?
a. A
b. B
c. AB
D. All of the above
Homework Answers
Complete the following ANOVA table to answer the following questions. Source df MS F FCRIT City_Size...
Complete the following ANOVA table to answer the following questions. Source df MS F FCRIT City_Size (A) 2 380 Gender (B) 1 20 A x B 700 Error 100 Total 59 For which source (s) is/are the null hypothesis rejected at the .05 level of significance? How many cells are there in the above design? What is the total number of subjects in this design? In the ANOVA table, SS A x B would equal to what? The partial eta-squared...
Refer to the following partial ANOVA results from EXCEL. source of variation ss df MS F-sta...
Refer to the following partial ANOVA results from EXCEL. source of variation ss df MS F-sta among-group 172.740 43.185 within-group 508.972 total total # of observation: 58 e. the critical value , F x at x=0.05 (3 decimals) f. at x= 0.05 your conclusion for testing the null hypothesis is that all the groups have the same mean is:
Source of Variation SS df MS F P-value F crit Rows 4.80 4 c f j...
Source of Variation SS df MS F P-value F crit Rows 4.80 4 c f j h Columns a b d g k i Error 15.09 12 e Total 114.07 19 In this Two Way ANOVA, the conclusion you arrive at is that mean blocks are not all the same. True False
A randomized block design yielded the following ANOVA table. Source df SS MS F Treatments 4...
A randomized block design yielded the following ANOVA table. Source df SS MS F Treatments 4 501 125.25 9.109 Blocks 2 1 225 112.50 8.182 Error 8 110 13.75 Total 14 836 A. How many blocks and treatments were used in the experiment? B. How many observations were collected in the experiment? C. Specify the null and alternative hypothesis you may use to compare the treatment means. D. Conduct the test of hypothesis and comment on the result. Use a...
Source Between treatments Within treatments Sum of Squares (Ss) df Mean Square (MS) 2 310,050.00 2,650.00 In some ANOVA summary tables you will see, the labels in the first (source) column are Treatm...
Source Between treatments Within treatments Sum of Squares (Ss) df Mean Square (MS) 2 310,050.00 2,650.00 In some ANOVA summary tables you will see, the labels in the first (source) column are Treatment, Error, and Total Which of the following reasons best explains why the within-treatments sum of squares is sometimes referred to as the "error sum of squares"? O Differences among members of the sample who received the same treatment occur when the researcher O Differences among members of...
In a test of weight loss programs, 120 subjects were divided such that 40 subjects followed...
In a test of weight loss programs, 120 subjects were divided such that 40 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation SS df MS F P-value Fcrit Between Groups 91.990 2 45.99487 1.3743 0.257070 3.073763 Within Groups 3915.738 117...
ANOVA Source of Variation SS df MS F p-value Factor A 30,865.45 3 10,288.48 Factor B...
ANOVA Source of Variation SS df MS F p-value Factor A 30,865.45 3 10,288.48 Factor B 22,557.30 2 11,278.65 Interaction 119,155.58 6 19,859.26 Error 90,553.57 36 2,515.38 Total 263,131.90 47 (a) What kind of ANOVA is this? One-factor ANOVA Two-factor ANOVA with replication Two-factor ANOVA without replication (b) Calculate each F test statistic and the p-value for each F test using Excel's function =F.DIST.RT(F,DF1,DF2). (Round your Fcalcvalues to 3 decimal places and p-values to 4 decimal places.) Source of Variation...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response f...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response for each treatment a. The MINITAB ANOVA printout is shown here. Test for interaction at the a = 0.05 b. level of significance. Analysis of variance for response SS MS df Source 0.11851 0.53777 0.53777 0.55391 5.02708 2.51334 2 1.48678 13.49334 6.74667 2 AB 27.22667 4.53778 6 Error 46.28486 11 Total Does the result warrant...
Let's say that a researcher conducts a study with 3 groups, each with 6 participants. Fill...
Let's say that a researcher conducts a study with 3 groups, each with 6 participants. Fill in the degrees of the following ANOVA table. freedom in df Source Between Within Total MS 34.43 6.95 F 4.95 D 17 Use the following Distributions tool to find the boundary for the critical region at a = .05 and - .01. To use the tool to find the critical F value, set both the numerator and the denominator degrees of freedom; this will...
Consider the data set that is summarized in below. -------------------------------------------------------------------------------------------------- Source DF Adj SS Adj MS...
Consider the data set that is summarized in below. -------------------------------------------------------------------------------------------------- Source DF Adj SS Adj MS F-Value P-Value C2 2 78.73 39.36 2.80 0.075 Error 34 477.90 14.06 Total 36 556.62 S = 3.749 R-Sq = 14.14% R-Sq(adj) = 9.09% Level N Mean StDev 1 10 7.699 3.808 2 10 11.656 3.859 3 17 9.461 3.651 Pooled StDev = 3.749 Fisher Individual Tests for Differences in Means Difference of Levels Difference of Means 93% CI T-Value Adjusted P-Value 2 -...
A randomized block design yielded the following ANOVA table. Source df SS MS F Treatments 4 501 125.25 9.109 Blocks 2 1 225 112.50 8.182 Error 8 110 13.75 Total 14 836 A. How many blocks and treatments were used in the experiment? B. How many observations were collected in the experiment? C. Specify the null and alternative hypothesis you may use to compare the treatment means. D. Conduct the test of hypothesis and comment on the result. Use a...
Source Between treatments Within treatments Sum of Squares (Ss) df Mean Square (MS) 2 310,050.00 2,650.00 In some ANOVA summary tables you will see, the labels in the first (source) column are Treatment, Error, and Total Which of the following reasons best explains why the within-treatments sum of squares is sometimes referred to as the "error sum of squares"? O Differences among members of the sample who received the same treatment occur when the researcher O Differences among members of...
In a test of weight loss programs, 120 subjects were divided such that 40 subjects followed each of 3 diets. Each was weighed a year after starting the diet and the results are in the ANOVA table below. Use a 0.05 significance level to test the claim that the mean weight loss is the same for the different diets. Source of Variation SS df MS F P-value Fcrit Between Groups 91.990 2 45.99487 1.3743 0.257070 3.073763 Within Groups 3915.738 117...
17) FACTOR B Level 1 2 3 FACTOR A 1 4.1, 4.1 5.0,5.2 6.3, 6.1 8.8,9.0 2 5.8, 5.6 5.0, 5.4 Calculate the mean response for each treatment a. The MINITAB ANOVA printout is shown here. Test for interaction at the a = 0.05 b. level of significance. Analysis of variance for response SS MS df Source 0.11851 0.53777 0.53777 0.55391 5.02708 2.51334 2 1.48678 13.49334 6.74667 2 AB 27.22667 4.53778 6 Error 46.28486 11 Total Does the result warrant...
Let's say that a researcher conducts a study with 3 groups, each with 6 participants. Fill in the degrees of the following ANOVA table. freedom in df Source Between Within Total MS 34.43 6.95 F 4.95 D 17 Use the following Distributions tool to find the boundary for the critical region at a = .05 and - .01. To use the tool to find the critical F value, set both the numerator and the denominator degrees of freedom; this will...
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