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Problem 5 A firm A suggested for market a new coffee packing machine. The firm claims that the new model has higher precision t statement of the producer. Six independent trials gave the following results: han the old one. An old model has precision σ. 5g. Independent expert was invited to test the 103 97 100 97 (a) At 5% significance level test the statement of the machine A producer (b) Find 5% upper bound for the precision σ of the new model A. (c) Find 10% lower bound for the precision σ of the new model A. (d) Another firm B suggested its own model of the packing machine. It claims that model B is more precise than the new model of firm A. The expert tested model B as well, 8 trials gave the following results: model B, weight (g) 100 96 97 98 101 97 Is there enough statistical evidence to support the statement of the firm B?
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Answer #1

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

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