Question

The article “Concrete Pressure on Formwork” (Mag. of Concrete Res., 2009: 407–417) gave the following observations on maximum concrete pressure (kN/m² ):

33.2 41.8 37.3 40.2 36.7 39.1 36.2 41.8

36.0 35.2 36.7 38.9 35.8 35.2 40.1

a.Is it plausible that this sample was selected from a normal population distribution?

b. Calculate an upper confidence bound with confidence level 95% for the population standard deviation of maximum pressure.

Short answer

a) Yes, It is plausible.

b) The upper confidence bound with confidence level of \(95\% \) for \(\sigma \) is \(1.834\).

Step-by-step solution

Step 1:Sample variance and Sample deviation:

The sample variance\({s^2}\)is \({s^2} = \frac{1}{{n - 1}} \cdot {S_{xx}}\)

Where

\(\begin{array}{l}{S_{xx}} = \sum {({x_i} - \overline x } {)^2}\\ = \sum {{x_i}^2} - \frac{1}{n} \cdot {\left( {\sum {{x^i}} } \right)^2}\end{array}\)

The sample standard deviation \(s\)is,\(s = \sqrt {{s^2}} \),

\(s = \sqrt {\frac{1}{{n - 1}} \cdot {S_{xx}}} \)

Solution for part a).

From the normal probability plot, we can say that it is plausible that the sample was taken from a normal population distribution.

Solution for part b).

An upper confidence bound for standard deviation with confidence level \(100(1 - \alpha )\% \) can be computed using formula

\(\sqrt {\frac{{(n - 1){s^2}}}{{{X^2}_{1 - \alpha ,n - 1}}}} \)

Using the standard variance and standard deviation formulas, it is easy to obtain the sample standard deviation, and it is,

\(s = 1.579\)

There are total of \(n = 15\) observations.

In order to obtain \(95\% CI\) for \(\sigma \), value \(\alpha \)is needed and can be obtain as,

\(\begin{array}{l}100(1 - \alpha ) = 95\\\alpha = 0.05\end{array}\)

The only missing value for the upper confidence interval is,

\(\begin{array}{l}{X^2}_{1 - \alpha ,n - 1} = {X^2}_{0.95,14}\\{X^2}_{1 - \alpha ,n - 1} = 6.571\end{array}\)

Therefore, the upper confidence bound with confidence level of \(95\% \)for\(\sigma \) is

\(\begin{array}{l}\sqrt {\frac{{(n - 1){s^2}}}{{{X^2}_{1 - \alpha ,n - 1}}}} = \sqrt {\frac{{14(1.579)}}{{6.571}}} \\ = 1.834\end{array}\)

Hence, The upper confidence bound with confidence level of \(95\% \) for \(\sigma \) is \(1.834\)..