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A sample of 14 joint specimens of a particular type gave a sample mean proportional limit stress of 8.48 MPa and a sample standard deviation of 0.79 MPa ("Characterization of Bearing Strength Factors in Pegged Timber Connections," J. of Structural Engr., 1997: 326332). Calculate and interpret a 95% lower confidence bound for the true average proportional limit stress of all such joints. What, if any, assumptions did you make about the distribution of proportional limit stress?

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Answer:

95% CI = ( 8.024 , 8.936 )

Step-by-step explanation:

Developing the exercise for a normal distribution then we have to calculate the degrees of freedom,

[tex]df = n-1 = 14-1=13[/tex]

Where n, is our sample.

So in the table,

t = 2.1604

[tex]95\% CI = (8.48 - \frac{2.1604*0.79}{\sqrt{14}} ,8.48 + \frac{2.1604*0.79}{\sqrt{14}} )[/tex]

[tex]95\% CI = ( 8.024 , 8.936 )[/tex]

**Assumptions:

Population is normal distributed

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