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3. It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Company 2. If independent random samples of size 13 of the two companies? work yield s1 = 0.062 mil and s2 = 0.035 mil, test the null hypothesis ?sigma?1squared = ?sigma?2squared against the alternative hypothesis ?sigma?1squared >?sigma?2squared at the 0.05 level of significance.
4. Apples can be stored in various ways. Four different kinds of storage condition were tested, and the time before spoilage in (months) was recorded as in the table below.
a) By looking at the data, make a judgment about the effect of storage conditions on the response variable.
b) Perform an analysis of variance, and test at the 0.10 level for a significant difference associated with storage conditions.
Storage Replicate
Condition 1 2 3
A 7 8 7
B 3 3 5
C 3 4 3
D 9 7 10
3. It is desired to determine whether there is less variability in the silver plating done by Company 1 than in that done by Company 2. If independent random samples of size 13 of the two companies' work yield s1 = 0.062 mil and s2 = 0.035 mil, test the null hypothesis 'sigma'1squared = 'sigma'2squared against the alternative hypothesis 'sigma'1squared >'sigma'2squared at the 0.05 level of significance.
Answer
H0:
Claim: H1:
Test Statistic : F=
Rejection criteria: Reject the null hypothesis, if the calculated value of F is greater than the critical value.
Significance level ? = 0.05
Critical value F(12,12) =2.68664
Calculated value of F = 3.1380
Conclusion: Reject the null hypothesis. Sample provides evidence to support the claim.
4. Apples can be stored in various ways. Four different kinds of storage condition were tested, and the time before spoilage in (months) was recorded as in the table below.
a) By looking at the data, make a judgment about the effect of storage conditions on the response variable.
b) Perform an analysis of variance, and test at the 0.10 level for a significant difference associated with storage conditions.
Storage Replicate
Condition 1 2 3
A 7 8 7
B 3 3 5
C 3 4 3
D 9 7 10
Answer
SUMMARY
Condition Count Sum Average Variance
A 3 22 7.333333 0.333333
B 3 11 3.666667 1.333333
C 3 10 3.333333 0.333333
D 3 26 8.666667 2.333333
It is clear that the last condition D have high mean time before spoilage and the condition C have the minimum mean time before spoilage
Analysis of variance
H0: There is no significant difference in the mean time before spoilage
H1: There is significant difference in the mean time before spoilage
Test Statistics: F test
Rejection criteria: Reject the null hypothesis, if the calculated value of F is greater than the critical value of F
Significance level ? = 0.10
ANOVA
Source of Variation SS df MS F P-value F crit
Between Groups 63.58333 3 21.19444 19.5641 0.000484 2.923796
Within Groups 8.666667 8 1.083333
Total 72.25 11
Conclusion: Since calculated value of F is greater than the critical value of F , we reject the null hypothesis. Thus the mean times before spoilage are significantly different.
please see the attached file.