using mintab>stat>basic stat>ANOVA
we have
One-way ANOVA: Imperial, Parkey, Blue Bonnet, Chiffon, Mazola, Fieischmann
Method
Null hypothesis All means are equal
Alternative hypothesis At least one mean is different
Significance level α = 0.05
Equal variances were assumed for the analysis.
Factor Information
Factor Levels Values
Factor 6 Imperial, Parkey, Blue Bonnet, Chiffon, Mazola,
Fieischmann
Analysis of Variance
Source DF Adj SS Adj MS F-Value P-Value
Factor 5 107.492 21.4984 74.01 0.000
Error 20 5.809 0.2905
Total 25 113.302
Model Summary
S R-sq R-sq(adj) R-sq(pred)
0.538957 94.87% 93.59% 91.30%
Means
Factor N Mean StDev 95% CI
Imperial 4 14.100 0.356 (13.538, 14.662)
Parkey 5 12.800 0.430 (12.297, 13.303)
Blue Bonnet 4 13.825 0.443 (13.263, 14.387)
Chiffon 4 13.150 0.686 (12.588, 13.712)
Mazola 5 17.140 0.598 (16.637, 17.643)
Fieischmann 4 18.100 0.648 (17.538, 18.662)
Pooled StDev = 0.538957
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor N Mean Grouping
Fieischmann 4 18.100 A
Mazola 5 17.140 A
Imperial 4 14.100 B
Blue Bonnet 4 13.825 B C
Chiffon 4 13.150 B C
Parkey 5 12.800 C
Means that do not share a letter are significantly different.
Tukey Simultaneous Tests for Differences of Means
Difference SE of Adjusted
Difference of Levels of Means Difference 95% CI T-Value
P-Value
Parkey - Imperial -1.300 0.362 (-2.438, -0.162) -3.60 0.019
Blue Bonnet - Imperial -0.275 0.381 (-1.474, 0.924) -0.72
0.977
Chiffon - Imperial -0.950 0.381 (-2.149, 0.249) -2.49 0.173
Mazola - Imperial 3.040 0.362 ( 1.902, 4.178) 8.41 0.000
Fieischmann - Imperial 4.000 0.381 ( 2.801, 5.199) 10.50
0.000
Blue Bonnet - Parkey 1.025 0.362 (-0.113, 2.163) 2.84 0.092
Chiffon - Parkey 0.350 0.362 (-0.788, 1.488) 0.97 0.923
Mazola - Parkey 4.340 0.341 ( 3.267, 5.413) 12.73 0.000
Fieischmann - Parkey 5.300 0.362 ( 4.162, 6.438) 14.66 0.000
Chiffon - Blue Bonnet -0.675 0.381 (-1.874, 0.524) -1.77
0.505
Mazola - Blue Bonnet 3.315 0.362 ( 2.177, 4.453) 9.17 0.000
Fieischmann - Blue Bonnet 4.275 0.381 ( 3.076, 5.474) 11.22
0.000
Mazola - Chiffon 3.990 0.362 ( 2.852, 5.128) 11.04 0.000
Fieischmann - Chiffon 4.950 0.381 ( 3.751, 6.149) 12.99 0.000
Fieischmann - Mazola 0.960 0.362 (-0.178, 2.098) 2.66 0.129
Individual confidence level = 99.49%
a ) the value of F stat = 74.01
p value <0.01
Reject Ho . there is convincing evidence that the mean PAPUFA percentages for six brands are not all equal .
lower limit | upper limit | ||
Imperial and Parkay | 0.162 | 2.438 | |
Imperial and Blue bonnet | -0.924 | 1.474 | |
Imperial and chiffon | -0.249 | 2.149 | |
Imperial and Mazoia | -4.178 | -1.902 | |
Imperial and Fleischmann | -5.199 | -2.801 | |
Parkey and Blue Bonnet | -2.163 | 0.113 | |
Parkay and Chiffon | -1.488 | 0.788 | |
Parkay and Mazola | -5.413 | -3.267 | |
Parkay and Fleischmann | -6.438 | -4.162 | |
Blue Bonnet and Chiffon | -0.524 | 1.874 | |
Blue Bonnet and Mazola | -4.453 | -3.076 | |
Blue Bonnet and Flieschmann | -5.474 | -3.076 | |
Chiffon and Mazola | -5.128 | -2.852 | |
Chiffon and Fieschmann | -6.149 | -3.751 | |
Mazola and Fieichmann | -2.098 | 0.178 | |
Suppose samples of six different brands of diet or imitation margarine were analyzed to determine the...
Suppose samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. Imperial 14.1 13.6 14.4 14.3 Parkay 12.8 12.5 13.4 13.0 12.3 Blue Bonnet 13.5 13.4 14.1 14.3 Chiffon 13.2 12.7 12.6 14.1 Mazola 16.8 17.2 16.4 17.3 18.0 17.2 18.7 18.4 Fleischmann's 18.1 (a) Test for differences among the true average PAPUFA percentages for the...
Suppose samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. Imperial 14.1 13.6 14.4 14.3 Parkay 12.8 12.5 13.4 13.0 12.3 Blue Bonnet 13.5 13.4 14.1 14.3 Chiffon 13.2 12.7 12.6 13.9 Mazola 16.8 17.3 16.4 17.3 18.0 Fleischmann's 18.1 17.2 18.7 18.4 (a) Test for differences among the true average PAPUFA percentages for the...
Suppose samples of six different brands of diet or imitation margarine were analyzed to determine the level of physiologically active polyunsaturated fatty acids (PAPUFA, in percent), resulting in the data shown in the accompanying table. Imperial 14.1 13.6 14.5 14.3 Parkay 12.8 12.5 13.4 13.0 12.3 Blue Bonnet 13.5 13.4 14.1 14.4 Chiffon 13.2 12.7 12.6 14.1 Mazola 16.8 17.2 16.4 17.3 18.0 Fleischmann's 18.1 17.2 18.7 18.4 Use the T-K procedure to compute 95% simultaneous confidence intervals for all...
Suppose the accompanying summary statistics for a measure of social marginality for samples of youths, young adults, adults, and seniors appeared in a research paper. The social marginality score measured actual and perceived social rejection, with higher scores indicating greater social rejection. Age Group Youths Young Adults Adults Seniors Sample Size X 107 2.00 255 3.30 3 13 3.09 32 2.83 1.57 1.68 1.69 1.86 For purposes of this exercise, assume that it is reasonable to regard the four samples...
An experiment was carried out to compare electrical resistivity for six different low-permeability concrete bridge deck mixtures. There were 26 measurements on concrete cylinders for each mixture; these were obtained 28 days after casting. The entries in the accompanying ANOVA table are based on information in an article. Fill in the remaining entries. (Round your answer for f to two decimal places.) Source df Sum of Squares Mean Square f Mixture Error 13.989 Total 5664.495 Test appropriate hypotheses at level...
1.An experiment was carried out to compare electrical resistivity for six different low-permeability concrete bridge deck mixtures. There were 26 measurements on concrete cylinders for each mixture; these were obtained 28 days after casting. The entries in the accompanying ANOVA table are based on information in an article. Fill in the remaining entries. (Round your answer for f to two decimal places.) Source df sum of squares mean square f Mixture : _____ _____________ __________ ____ Error : ______ _____________...
1. The lumen output was determined for each of I = 3 different brands of lightbulbs having the same wattage, with J = 8 bulbs of each brand tested. The sums of squares were computed as SSE = 4775.1 and SSTr = 590.8. State the hypotheses of interest (including word definitions of parameters). A) μi = true average lumen output for brand i bulbs H0: μ1 = μ2 = μ3 Ha: at least two μi's are unequal B) μi =...
Problem 3 A particular brand of diet margarine was analyzed to determine the level of polyunsaturated fatty acid (in percentages). A sample of six packages resulted in the following data: 16.8, 17.2, 17.4, 16.9, 16.5, 17.1. Question 7 0.5 pts For problem 3 what is the p-value? Almost zero Between 0.01 and 0.025 Between 0.025 and 0.05 Between 0.4 and 0.5 Question8 0.5 pts For problem 3, at a significance level of a-0.05, the company should fail to reject the...
1. In an experiment to compare the tensile strengths of 1 - 6 different types of copper wire, ) = 5 samples of each type were used. The between-samples and within samples estimates of a were computed as MSTR = 2649.3 and MSE - 1169.2, respectively. Use the F test at level 0.05 to test Ho: H1 12 - ... - versus Ha: at least two wi's are unequal Calculate the test statistic (Round your answer to two decimal places.)...
s samples of each of our types of cereal grain grown in a certain region were analyzed to determine am n content, resulting in the following data ㎍/g Note: If you want to use R this data has been set up so you may copy ıt d ec y into R Wheat- 5.1, 4.4, 6.1, 6.2, 6.6, 5.7) Sarley- 6.5, 7.9, 6.0, 7.5, 5.8, 5.7) Maize- 5.8, 4.7, 6.5, 5.0, 6.1, 5.3) Oats- 8.2, 6.0, 7.7, 6.9, 5.4, 7.2 In...