The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal arc welds used in LNG ship containment tanks. The resulting sample standard deviation was s = 2.84 mils. Assuming normality, derive a 95% CI for ² and for o. (Round your answers to two decimal places.) CI for ² mils² CI for o mils

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A: Solution

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A: Given information and calculation is shown below

Q: The amount of lateral expansion (mils) was determined for a sample of n = 5 pulsed-power gas metal…

A: Given:

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Q: Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores…

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Q: M1= 37 SS1=360 n1=11 M2= 43 SS2=576 n2=10 Assume two failed test and alpha= .05 What is the t value?

A: We have given that M1= 37, SS1= 360, n1=11, M2 = 43, SS2 = 576 n2=10

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A: Sample size n =20 Sample mean=1428 Standard deviation =58

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A: Given Data: X~ Normal (266, 16)   P(X>285)=0.1175 P(X<285)=0.8825

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A: Given that μ=2,σ=0.05 , n =16

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A: Note: As per our Honor code policy, we are authorized to answer only three sub parts at a time. So,…

Q: The amount of lateral expansion (mils) was determined for a sample of n = 7 pulsed-power gas metal…

A:

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A: Given, Standard deviation = 5.7 Error = 0.91 confidence level =92%

Q: The amount of lateral expansion (mils) was determined for a sample of n = 8 pulsed-power gas metal…

A: Solution : Given information: n= 8 Sample size  s= 2.89 Sample standard deviations  α=0.05 Level of…

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A: As per our guidelines we are supposed to answer only one question per post so i am solving first…

Q: Find the indicated IQ score. The graph to the right depicts IQ scores of​ adults, and those scores…

A: From the provided information, the respective IQ score is to be obtained, for which the area to the…