Question

. The article "Wind-Uplift Capacity of Residential Wood Roof-Sheathing Panels Retrofitted with Insulating Foam Adhesive" (P. Datin, D. Prevatt, and W. Pang, Journal of Architectural Engineering, 2011:144–154) presents a study of the failure pressures of roof panels. Following are the failure pressures, in kPa, for five panels constructed with 6d smooth shank nails. These data are consistent with means and standard deviations presented in the article. 3.32 2.53 3.45 2.38 3.01 Find a 95% confidence interval for the mean failure pressure for this type of roof panel.

Answer 1

The 95% confidence interval is
`2.354 <   3.526`

Explanation:

From the question we are told that

The sample size is n = 5

The sample data is 3.32 2.53 3.45 2.38 3.01

Gnerally the sample mean is mathematically represented as

`\= x  =  (\sum x_i)/(n)`

=>
`\= x  =  (3.32+  2.53 +3.45+ 2.38+ 3.01 )/(5)`

=>
`\= x  =   2.94`

Generally the standard deviation is mathematically represented as

`s =  \sqrt{ ( \sum (x_i -  \= x )^2)/(n-1) }`

=>
`s =  \sqrt{ (  ( 3.32 -  2.94  )^2 +  ( 2.53 -  2.94  )^2+ \cdots + ( 3.01  -  2.94  )^2)/(5-1) }`

=>
`s =   0.472`

Generally the degree of freedom is mathematically represented as

`df =  n-1`

=>
`df =  5-1`

=>
`df =  4`

From the question we are told the confidence level is 95% , hence the level of significance is

`\alpha = (100 - 95 ) \%`

=>
`\alpha = 0.05`

Generally from the t distribution table the critical value of
`(\alpha )/(2)` at a degree of freedom of
`df =  4` is

`t_{(\alpha )/(2) ,df } =  2.77`

Generally the margin of error is mathematically represented as

`E =t_{(\alpha )/(2) ,df }  *  (\sigma )/(√(n) )`

`E = 2.77 *  (0.472 )/(√(5) )`

`E = 0.586`

Generally 95% confidence interval is mathematically represented as

`\= x -E <  \mu <  \=x  +E`

`2.94-0.586 <  2.94  + 0.586`

`2.354 <   3.526`