Question

13.19 It is suspected that the environmental temper- ature at which batteries are activated affects their life. Thirty homogeneous batteries were tested, six at each of five temperatures, and the data are shown below (activated life in seconds). Analyze and interpret the data. (From C. R. Hicks, Fundamental Concepts in Design of Experiments, Holt, Rinehart and Winston, New York, 1973.) Temperature (°C) 0 25 50 75 100 55 61 72 72 57 60 72 72 54 60 68 7064 54 60 77 68 56 60 77 69 65 Perform ANOVA and interpret overall F-Test and show Ho conclusion from the test. Perform Tukey's and Duncan's Test

Answer 1

I used MINITAB to solve this question.

Enter data in MINITAB. Then goto Stat menu ----> ANOVA ---> One way ANOVA .

Minitab - Untitled Eile Edit Data HO Calc Session Stat Graph Editor Tools Window Help Assistant Basic Statistics O@EISO 90 O SODE Regression VIA ORTOON ANOVA One-Way... DOE # Analysis of Means... Control Charts Balanced ANOVA... Quality Tools General Linear Model Reliability/Survival Fully Nested ANOVA... Multivariate General MANOVA... Time Series Tables Test for Equal Variances... Nonparametrics 111 Interval Plot... Equivalence Tests Main Effects Plot... Power and Sample Size * Interaction Plot... Worksheet 1 *** + 01 O 06 07 08 09 10 02 25 60 61 60 03 50 70 72 72 68 77 04 75 72 72 72 70 68 05 100 65 66 1 60 77

New Window will pop up on screen. Given input range. Go to comparison option and select Tuckey's and Dunnet

Minitab - Untitled Eile Edit Data H 1% Calc Stat Graph Editor Tools Window Help Assistant INR OO. SOT > SODEX - - A + VE X IQOTOO LUM 2 Session One-Way Analysis of Variance One-Way Analysis of Variance: Comparisons C1 0 Response data are in a separate column for each factor level Error rate for comparisons: 6 8888 c350 75 Responses: '0'-100 Comparison procedures assuming equal variances Tukey Fisher Dunnett Control group level: 0 Hsu MCB Best: Largest mean is best Worksh 015 016 14 Graphs... 1 Results Interval plot for differences of means ✓ Grouping information Tests Options... Results... Comparisons... | Storage... ] Select Help OK Cancel Help OK Cancell

Minitab output:

One-way ANOVA: 0, 25, 50, 75, 100

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 5 0, 25, 50, 75, 100

Analysis of Variance

Source DF Adj SS Adj MS F-Value P-Value
Factor 4 1252.1 313.033 69.05 0.000
Error 25 113.3 4.533
Total 29 1365.5

Model Summary

S R-sq R-sq(adj) R-sq(pred)
2.12916 91.70% 90.37% 88.05%

Means

Factor N Mean StDev 95% CI
0 6 55.167 1.169 (53.376, 56.957)
25 6 60.167 0.408 (58.376, 61.957)
50 6 72.67 3.67 ( 70.88, 74.46)
75 6 70.500 1.761 (68.710, 72.290)
100 6 64.167 2.137 (62.376, 65.957)

Pooled StDev = 2.12916

Tukey Pairwise Comparisons

Grouping Information Using the Tukey Method and 95% Confidence

Factor N Mean Grouping
50 6 72.67 A
75 6 70.500 A
100 6 64.167 B
25 6 60.167 C
0 6 55.167 D

Means that do not share a letter are significantly different.

Tukey Simultaneous 95% CIs

Dunnett Multiple Comparisons with a Control

Grouping Information Using the Dunnett Method and 95% Confidence

Factor N Mean Grouping
0 (control) 6 55.167 A
50 6 72.67
75 6 70.500
100 6 64.167
25 6 60.167

Means not labeled with the letter A are significantly different from the control level mean.

P value for F test is 0.000 which is less than 0.05, hence we reject null hypothesis and conclude that at least one mean is significantly different than other.

From Tuckey's we see that mean of the temperature 50 and 75 are same but these are different than remaining temperature means.

From Dunnets test we see that mean of temperature zero is different than all other temperature means.