Three brands of batteries are under study. It is suspected that the lives (in we
ID: 3132566 • Letter: T
Question
Three brands of batteries are under study. It is suspected that the lives (in weeks) of the three brands are different. Five randomly selected batteries of each brand are tested with the following results: Are the lives of these brands of batteries different? Analyze the residuals from this experiment. Construct a 95 percent confidence interval estimate on the mean life of battery brand 2. Construct a 99 percent confidence interval estimate on the mean difference between the lives of battery brands 2 and 3. Which brand would you select for use? If the manufacturer will replace without chaise any battery that fails in less than 85 weeks, what percentage would the company expect to replace?Explanation / Answer
One-way ANOVA: Lives versus Brand
Source DF SS MS F P
Brand 2 1196.1 598.1 38.34 0.000
Error 12 187.2 15.6
Total 14 1383.3
S = 3.950 R-Sq = 86.47% R-Sq(adj) = 84.21%
Conclusion: The calculated p-value is 0.000 and less than 0.05 level of significant. The null hypothesis is not accepted and accepts the alternative hypothesis. Hence, there is a different in the lives of batteries among the three brands of batteries.
Comment: i) The residuals are scatter from the straight line in normal probability plot. Hence, the normal distribution assumption on the residual cannot accept.
ii) The histogram of the residual shows a positive skew. Hence, the normal assumption on the residual is violated.
iii) The scatter plot of the residual versus order does not show a random pattern and the variance is high at the two end sides of graph. Hence, our assumption on the residual is violated.
Test hypothesis mu = 0 vs not = 0
Variable N Mean StDev SE Mean 95% CI T P
Live 5 79.40 3.85 1.72 (74.62, 84.18) 46.15 0.000
Comment: The calculated p-value is 0.000 and less than 0.05 level of significant. Hence, the mean live spent of battery bran 2 is not equal to zero and the 95% confidence interval is (74.62, 84.18).
Second part
Paired T for Live 2 - Live 3
N Mean StDev SE Mean
Live 2 5 79.40 3.85 1.72
Live 3 5 100.40 4.56 2.04
Difference 5 -21.00 6.71 3.00
99% CI for mean difference: (-34.81, -7.19)
T-Test of mean difference = 0 (vs not = 0): T-Value = -7.00 P-Value = 0.002
Comment: The calculated p-value is 0.002 and less than 0.09 level of significant. Hence, There is significant different among the mean lives of battery of bran 2 and 3. The 99% confidence interval is (-34.81, -7.19).
Individual 95% CIs For Mean Based on
Pooled StDev
Level N Mean StDev ------+---------+---------+---------+---
1 5 95.20 3.35 (----*----)
2 5 79.40 3.85 (----*----)
3 5 100.40 4.56 (----*---)
------+---------+---------+---------+---
80.0 88.0 96.0 104.0
Pooled StDev = 3.95
Grouping Information Using Tukey Method
Brand N Mean Grouping
3 5 100.400 A
1 5 95.200 A
2 5 79.400 B
Means that do not share a letter are significantly different.
Tukey 95% Simultaneous Confidence Intervals
All Pairwise Comparisons among Levels of Brand
Individual confidence level = 97.94%
Brand = 1 subtracted from:
Brand Lower Center Upper --------+---------+---------+---------+-
2 -22.459 -15.800 -9.141 (---*----)
3 -1.459 5.200 11.859 (---*----)
--------+---------+---------+---------+-
-15 0 15 30
Brand = 2 subtracted from:
Brand Lower Center Upper --------+---------+---------+---------+-
3 14.341 21.000 27.659 (---*---)
--------+---------+---------+---------+-
-15 0 15 30
Comment: The Brand 3 has the longest mean live, So, we should select the Brand 3.
Second Part
One-Sample T: Lives
Test of mu = 85 vs > 85
95% Lower
Variable N Mean StDev SE Mean Bound T P
Lives 15 91.67 9.94 2.57 87.15 2.60 0.011
Comment: The calculated p-value is 0.011 and less than 0.05 level of significant. Hence, the mean lives of batteries are more than 85 weeks. The percentage of the replacement is 1%.
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