a)
Ho : σ = 5
Ha : σ > 5
Level of Significance , α = 0.05
sample Std dev , s = 2.5033
Sample Size , n = 6
Chi-Square Statistic X² = (n-1)s²/σ² =
1.2533
degree of freedom, DF=n-1 = 5
one tail test
Upper Critical Value = 11.070
p-Value = 0.9397
since, p- value >alpha=0.05 , Do not reject the null
hypothesis
so, there is no enough evidence to support claim of A
b)
alpha=0.05
Upper Chi-Square Value= | χ²α = | 11.0705 |
confidence interval for variance is | ||
upper bound= | (n-1)s²/χ²1-α/2 = | =5*2.5033²/11.0705 = 27.353 |
c)
alpha=0.10
Lower Chi-Square Value= χ²1-α = 1.6103
confidence interval for variance is
lower bound=(n-1)s²/χ²1-α = 5*2.2033² / 1.6103 =
3.3923
d)
X |
100 |
96 |
97 |
98 |
101 |
99 |
99 |
97 |
mean = ΣX/n = 98.375
sample variance = [ Σ(X - X̄)²/(n-1)] = 2.839
Ho:σ B = σ A
H1: σ B > σA
Level of Significance 0.05
Sample B
Sample Size 8
Sample Variance 2.839
Sample A
Sample Size 6
Sample Variance 6.267
Intermediate Calculations
F Test Statistic = s²B/s²A = 0.4531
Population 1 Sample Degrees of Freedom = 7
Population 2 Sample Degrees of Freedom =5
Upper-Tail Test
Upper Critical Value =4.8759
p-Value = 0.8345
Do not reject the null hypothesis
so, there is no enough evidence to support B
Problem 5 A firm A suggested for market a new coffee packing machine. The firm claims...
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