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Consider a sensor system for which we know that the sensor noise is normally distributed (and...

Consider a sensor system for which we know that the sensor noise is normally distributed (and thus that an actual reading is taken from a normal distribution). Given an existing sensor with a known mean reading of µ = 3.75 and a standard deviation of 0.4, we want to compare a new sensor to it. For the new sensor we take 10 measurements: {3, 4.5, 4, 3.6, 3.9, 4.15, 3.2, 4.1, 3.2, 3.35}. Given this data we want to show that the sensors generate different data and that the new sensor is more reliable (i.e. has noise with a lower variance).

Evaluate whether the mean of the data sample from the new sensor is significantly different from that of data samples obtained from the original sensor. Include your calculations and the significance scores.

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Answer #1

Ho :   σ² =   0.16
Ha :   σ² <   0.16

Level of Significance ,    α =    0.05
sample Variance,   s² =    0.2506
Sample Size ,   n =    10
      
Chi-Square Statistic   X² = (n-1)s²/σ² =    14.094
      
degree of freedom,   DF=n-1 =    9
      
one tail test       
lower critical value   =   3.33
test stat > critical value,   Do not reject the null hypothesis      

there is no evidence that it has lower variance

======================

Ho :   µ =   3.75                  
Ha :   µ ╪   3.75       (Two tail test)          
                          
Level of Significance ,    α =    0.05                  
sample std dev ,    s = √(Σ(X- x̅ )²/(n-1) ) =   0.5006                  
Sample Size ,   n =    10                  
Sample Mean,    x̅ = ΣX/n =    3.7000                  
                          
degree of freedom=   DF=n-1=   9                  
                          
Standard Error , SE = s/√n =   0.5006   / √    10   =   0.1583      
t-test statistic= (x̅ - µ )/SE = (   3.700   -   3.75   ) /    0.1583   =   -0.316
                          
  
p-Value   =   0.759298   [Excel formula =t.dist.2t(t-stat,df) ]              
Decision:   p-value>α, Do not reject null hypothesis                       
Conclusion: There is not enough evidence that mean of the data sample from the new sensor is significantly different from that of data samples obtained from the original sensor

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