Ejercicios minitab
Enviado por doras123 • 3 de Febrero de 2023 • Trabajos • 1.268 Palabras (6 Páginas) • 40 Visitas
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EJERCICIOS
-M I N I T A B-
DOCENTE:
Fecha de Entrega:
09 de DICIEMBRE de 2022
One-way ANOVA: metodo A, metodo B, metodo C, metodo D
Method
Null hypothesis | All means are equal |
Alternative hypothesis | Not all means are equal |
Significance level | α = 0.05 |
Equal variances were assumed for the analysis.
Factor Information
Factor | Levels | Values |
Factor | 4 | metodo A, metodo B, metodo C, metodo D |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Factor | 3 | 69.50 | 23.167 | 9.42 | 0.002 |
Error | 12 | 29.50 | 2.458 |
|
|
Total | 15 | 99.00 |
|
|
|
Model Summary
S | R-sq | R-sq(adj) | R-sq(pred) |
1.56791 | 70.20% | 62.75% | 47.03% |
Means
Factor | N | Mean | StDev | 95% CI |
metodo A | 4 | 7.250 | 0.957 | (5.542, 8.958) |
metodo B | 4 | 8.500 | 1.291 | (6.792, 10.208) |
metodo C | 4 | 12.75 | 2.36 | (11.04, 14.46) |
metodo D | 4 | 10.500 | 1.291 | (8.792, 12.208) |
Pooled StDev = 1.56791
Tukey Pairwise Comparisons
Grouping Information Using the Tukey Method and 95% Confidence
Factor | N | Mean | Grouping | |
metodo C | 4 | 12.75 | A |
|
metodo D | 4 | 10.500 | A | B |
metodo B | 4 | 8.500 |
| B |
metodo A | 4 | 7.250 |
| B |
Means that do not share a letter are significantly different.
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Fisher Pairwise Comparisons
Grouping Information Using the Fisher LSD Method and 95% Confidence
Factor | N | Mean | Grouping | ||
metodo C | 4 | 12.75 | A |
|
|
metodo D | 4 | 10.500 | A | B |
|
metodo B | 4 | 8.500 |
| B | C |
metodo A | 4 | 7.250 |
|
| C |
Means that do not share a letter are significantly different.
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WORKSHEET 1
GRAFICA DE SESIDUOS
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WORKSHEET 1
GRAFICA DE SESIDUOS
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Conclusión
La hipótesis nula se rechaza ya que es menor que el valor de significancia, está por debajo de la tolerancia.
Podemos observar que existe interacción entre los materiales ya que, el que está afectando de algún modo es el método podemos ver que el método c esta fuera del valor.
One-way ANOVA: 15%, 20%, 25%, 30%, 35%
Method
Null hypothesis | All means are equal |
Alternative hypothesis | Not all means are equal |
Significance level | α = 0.05 |
Rows unused | 1 |
Equal variances were assumed for the analysis.
Factor Information
Factor | Levels | Values |
Factor | 5 | 15%, 20%, 25%, 30%, 35% |
Analysis of Variance
Source | DF | Adj SS | Adj MS | F-Value | P-Value |
Factor | 4 | 516.6 | 129.140 | 15.12 | 0.000 |
Error | 20 | 170.8 | 8.540 |
|
|
Total | 24 | 687.4 |
|
|
|
Model Summary
S | R-sq | R-sq(adj) | R-sq(pred) |
2.92233 | 75.15% | 70.18% | 61.17% |
Means
Factor | N | Mean | StDev | 95% CI |
15% | 5 | 9.80 | 3.35 | (7.07, 12.53) |
20% | 5 | 15.40 | 3.13 | (12.67, 18.13) |
25% | 5 | 17.600 | 2.074 | (14.874, 20.326) |
30% | 5 | 22.20 | 3.03 | (19.47, 24.93) |
35% | 5 | 10.80 | 2.86 | (8.07, 13.53) |
Pooled StDev = 2.92233
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