In a study comparing various methods of gold plating, 7 printed circuit edge connectors were gold-plated with control-immersion tip plating. The average gold thickness was 1.5 μm, with a standard deviation of 0.25 μm. Five connectors were masked and then plated with total immersion plating. The av- erage gold thickness was 1.0 μm, with a standard deviation of 0.15 μm. Find a 99% confidence interval for the difference between the mean thicknesses produced by the two methods.

The 99% of confidence intervals for difference between the mean thicknesses produced by the two methods.

( 0.17971  , 0.82028)

Step-by-step explanation:

Step:-(i)

Given data the average gold thickness was 1.5 μm, with a standard deviation of 0.25 μ m.

Given sample size n₁ = 7

mean of first sample x₁⁻ =1.5 μ m.

Standard deviation of first sample S₁ = 0.25 μ m

Given data  Five connectors were masked and then plated with total immersion plating. The average gold thickness was 1.0 μ m, with a standard deviation of 0.15 μ m.

Given second sample size n₂ = 5

The mean of second sample x⁻₂ =  1.0 μ m

Standard deviation of first sample S₂ = 0.15 μ m

Level of significance ∝ =0.01 or 99%

Degrees of freedom γ = n₁+ n₂ -2 = 7+5-2=10

tabulated value t = 2.764

Step(ii):-

The 99% of confidence intervals for μ₁- μ₂ is determined by

(x₁⁻ - x⁻₂)  - z₀.₉₉ Se((x₁⁻ - x⁻₂) ,  (x₁⁻ - x⁻₂)  + z₀.₉₉ Se((x₁⁻ - x⁻₂)

where        

[1.5-1.0 -  2.764 (0.115879) , (1.5+1.0) + 2.764(0.115879 ]

(0.5-0.32029,0.5+0.32029

( 0.17971  , 0.82028)

Conclusion:-

The 99% of confidence intervals for μ₁- μ₂ is determined by

( 0.17971  , 0.82028)

too hard the other guy did it

step-by-step explanation: