A procurement specialist has purchased 23 resistors from vendor 1 and 30 resistors from vendor 2. Let represent the vendor 1 observed resistances, which are assumed to be normally and independently distributed with mean 120 ohms and standard deviation 1.7 ohms. Similarly, let represent the vendor 2 observed resistances, which are assumed to be normally and independently distributed with mean 125 ohms and standard deviation of 2.0 ohms. What is the sampling distribution of ? What is the standard error of ? The sampling distribution of is What is thesampling distribution of X1 − X2? What is the standard errorof X1 − X2?

Respuesta :

Answer:

The standard error( X1 − X2 ) = 0.547

Step-by-step explanation:

Step:-(1)

Given a procurement specialist has purchased 23 resistors

Given normally and independently distributed with mean 120 ohms and standard deviation 1.7

mean of the Population of the vendor 1 is μ₁ =  120 ohms

Standard deviation of the Population the vendor 1 is σ₁ = 1.7 ohms

similarly represent the vendor 2 observed resistances, which are assumed to be normally and independently distributed with mean 125 ohms and standard deviation of 2.0

mean of the Population of the vendor 2 is μ₂ =  120 ohms

Standard deviation of the Population the vendor 2 is σ₂ = 1.7 ohms

The standard error of the  difference of two means

 Se( X1 − X2) = [tex]\sqrt{\frac{σ^2_{1} }{n_{1} } +\frac{σ^2_{2} }{n_{1} } }[/tex]

 Here σ₁ = 1.7 ohms and σ₂ = 2 ohms and n₁=n₂ =n = 23 resistors

se(X1 − X2) = [tex]\sqrt{\frac{1.7^2}{23 } +\frac{2^2 }{23} }[/tex]

se(X1 − X2) = √0.2995

                   = 0.547

Conclusion:-

The standard error of X1 − X2 = 0.547

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