Friction Stir Welding leaves a hole in the weld that must be closed with a Plug

Question

Friction Stir Welding leaves a hole in the weld that must be closed with a Plug Welding Process. This process must achieve a weld strength of 50 ksi + or - .5 ksi in order to be viable.  

The plug welding tests will use AL2195 Confidence Panels. Aluminum-Lithium Alloy (Al- Li) targeted as advanced materials for aerospace technology and to reduce the weight of U.S. Department of Defense systems primarily because of their low density, high specific modulus, and excellent fatigue and cryogenic toughness properties.

Friction Stir Welding leaves a hole in the weld that must be closed with a Plug Welding Process. This process must achieve a weld strength of 50 ksi + or - .5 ksi in order to be viable.  

How do you calculate the mean, standard deviation, and 95% confidence Interval for the mean, median, standard deviation for the AL2195 confidence panel Length, Width, and Thickness? Basic Statistics Data Length Width Thickness 10.67 7.51 0.551 10.74 7.57 0.546 10.68 7.55 0.546 10.72 7.53 0.554 10.66 7.53 0.546 10.69 7.56 0.542 10.70 7.52 0.545 10.72 7.58 0.538 10.69 7.55 0.552 10.68 7.53 0.547 10.72 7.56 0.546 10.70 7.58 0.545 10.70 7.55 0.546 10.73 7.55 0.548 10.75 7.54 0.546 10.73 7.57 0.543 10.69 7.54 0.548 10.68 7.55 0.545 10.70 7.55 0.553 10.77 7.56 0.539 10.72 7.54 0.549 10.69 7.56 0.541 10.66 7.56 0.543 10.69 7.55 0.545 10.63 7.54 0.549 10.68 7.55 0.546 10.73 7.56 0.545 10.74 7.57 0.540 10.65 7.54 0.541 10.69 7.55 0.545 10.70 7.54 0.547 10.72 7.55 0.543 10.75 7.53 0.545 10.71 7.54 0.546 10.72 7.55 0.547 Upper Spec 11 7.66 0.56 Lower Spec 10.5 7.45 0.54 Target 10.75 7.55 0.55 Drill Hole in Panel Configure Plug elder Settings Cut Panel horizontally (1 nmargiSend plug ean Welder SettinesGirind off Kecess nd plug)sample to pull Repeat process Plug test lab for next test and remove plug weld

Explanation / Answer

Mean of the random variable is,

Mean = sum(Xi) / N

where, Xi is value of variable X , i = 1, 2, 3............., N

N = total no of observation

Median = (N/2)th value of ordered data

i.e. middle value of data but condition is data required in ascending or descending order

Standard Deviation = sqrt [ sum(Xi - Mean(X))^2 / N ]

Confidence Interval:

Lower Confidence interval = mean - t * ( SD / sqrt(n) )

Upper Confidence interval = mean - t * ( SD / sqrt(n) )

t is test statistics of t for single variable

SD = Standard Deviation

But using R we can compute directly,

> Mean_Length = mean(Length)

> Mean_Length

[1] 10.70286

> Median_Length = median(Length)

> Median_Length

[1] 10.7

> Standard_Deviation_Length = sd(Length)

> Standard_Deviation_Length

[1] 0.03054133

> 95 percent confidence interval:

(10.69237, 10.71335)

> Mean_Width = mean(Width)

> Mean_Width

[1] 7.548857

> Median_Width = median(Width)

> Median_Width

[1] 7.55

> Standard_Deviation_Width = sd(Width)

> Standard_Deviation_Width

[1] 0.01567528

> 95 percent confidence interval:

( 7.543472, 7.554242 )

> Mean_Thickness = mean(Thickness)

> Mean_Thickness

[1] 0.5456571

> Median_Thickness = median(Thickness)

> Median_Thickness

[1] 0.546

> Standard_Deviation_Thickness = sd(Thickness)

> Standard_Deviation_Thickness

[1] 0.003629478

> 95 percent confidence interval:

( 0.5444104, 0.5469039)

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