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Magazine Chapter 7: Q38E (page 303)
Ultra high performance concrete (UHPC) is a rela- tively new construction material that is characterized by strong adhesive properties with other materials. The article 鈥淎dhesive Power of Ultra High Performance Concrete from a Thermodynamic Point of View鈥� described an investigation of the intermolecular forces for UHPC connected to various substrates. The following work of adhesion measurements (in mJ/m2) for UHPC specimens adhered to steel appeared in the article:
\({\rm{107}}{\rm{.1\;109}}{\rm{.5\;107}}{\rm{.4\;106}}{\rm{.8\;108}}{\rm{.1}}\)
a. Is it plausible that the given sample observations were selected from a normal distribution?
b. Calculate a two-sided \({\rm{95\% }}\) confidence interval for the true average work of adhesion for UHPC adhered to steel. Does the interval suggest that \({\rm{107}}\) is a plausible value for the true average work of adhesion for UHPC adhered to steel? What about \({\rm{110}}\)?
c. Predict the resulting work of adhesion value resulting from a single future replication of the experiment by calculating a \({\rm{95\% }}\)prediction interval, and compare the width of this interval to the width of the CI from (b).
d. Calculate an interval for which you can have a high degree of confidence that at least \({\rm{95\% }}\)of all UHPC specimens adhered to steel will have work of adhesion values between the limits of the interval.
Short Answer
a) It is plausible that the given sample observations were selected from a normal distribution.
b)\((106.4447,109.1153)\)
\(107\): Plausible
\(110\): Not plausible
c)\((104.5091,111.0509)\)
The prediction interval is wider than the confidence interval.
d) The boundaries of the prediction interval then become:
\((102.3553,113.2047)\)
Step by step solution
Normal probability plot
The data values are on the horizontal axis and the standardized normal scores are on the vertical axis.
If the data contains n data values, then the standardized normal scores are the z-scores in the normal probability table of the appendix corresponding to an area of \(\frac{{{\rm{j - 0}}{\rm{.5}}}}{{\rm{n}}}\) (or the closest area) with \({\rm{j脦\{ 1,2,3, \ldots ,n\} }}\).
The smallest standardized score corresponds with the smallest data value, the second smallest standardized score corresponds with the second smallest data value, and so on.
If the pattern in the normal probability plot is roughly linear, then it is plausible that the observations originate from a normal distribution.
The normal probability plot does not contain strong curvature and is roughly linear,
Thus it is plausible that the given sample observations were selected from a normal distribution.
To Calculate a two-sided \({\rm{95\% }}\) confidence interval
(b)
Given:
\(\begin{aligned}{\rm{n = 5 }}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{aligned}\)
\(107.1109.5107.4108.6108.1\)
The mean is the sum of all values divided by the number of values:
\({\rm{\bar x = }}\frac{{{\rm{107}}{\rm{.1 + 109}}{\rm{.5 + 107}}{\rm{.4 + 106}}{\rm{.8 + 108}}{\rm{.1}}}}{{\rm{5}}}{\rm{ = }}\frac{{{\rm{538}}{\rm{.9}}}}{{\rm{5}}}{\rm{ = 107}}{\rm{.78}}\)
The variance is the sum of squared deviations from the mean divided by \({\rm{n - 1}}\).The standard deviation is the square root of the variance:
\({\rm{s = }}\sqrt {\frac{{{{{\rm{(107}}{\rm{.1 - 107}}{\rm{.78)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{.(108}}{\rm{.1 - 107}}{\rm{.78}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{5 - 1}}}}} {\rm{\gg 1}}{\rm{.0756}}\)
Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 5 - 1 = 4}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = 0}}{\rm{.025}}\)in the table of the critical values for t distributions in the appendix:
\({{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.776}}\)
The margin of error is then:
\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times }}\frac{{\rm{s}}}{{\sqrt {\rm{n}} }}{\rm{ = 2}}{\rm{.776 \times }}\frac{{{\rm{1}}{\rm{.0756}}}}{{\sqrt {\rm{5}} }}{\rm{\gg 1}}{\rm{.3353}}\)
The boundaries of the confidence interval then become:
\(\begin{aligned}{\rm{\bar x - E = 107}}{\rm{.78 - 1}}{\rm{.3353 = 106}}{\rm{.4447}}\\{\rm{\bar x + E = 107}}{\rm{.78 + 1}}{\rm{.3353 = 109}}{\rm{.1153}}\end{aligned}\)
07 is a plausible value for the true average work of adhesion for UHPC adhered to steel, because 107 lies between the boundaries of the confidence interval.
10 is not a plausible value for the true average work of adhesion for UHPC adhered to steel, because 110 does not lie between the boundaries of the confidence interval.
Hence \((106.4447,109.1153)\)
\(107\): Plausible
\(110\): Not plausible
To predict the resulting work of adhesion value
(c)
Given:
\(\begin{aligned}{\rm{n = 5 }}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{aligned}\)
Result part (b): \(\left( {106.4447,{\rm{ }}109.1153} \right)\)
\(107.1109.5107.4108.6108.1\)
The mean is the sum of all values divided by the number of values:
\({\rm{\bar x = }}\frac{{{\rm{107}}{\rm{.1 + 109}}{\rm{.5 + 107}}{\rm{.4 + 106}}{\rm{.8 + 108}}{\rm{.1}}}}{{\rm{5}}}{\rm{ = }}\frac{{{\rm{538}}{\rm{.9}}}}{{\rm{5}}}{\rm{ = 107}}{\rm{.78}}\)
The variance is the sum of squared deviations from the mean divided by \({\rm{n - 1}}\) The standard deviation is the square root of the variance:
\({\rm{s = }}\sqrt {\frac{{{{{\rm{(107}}{\rm{.1 - 107}}{\rm{.78)}}}^{\rm{2}}}{\rm{ + \ldots }}{\rm{. + (108}}{\rm{.1 - 107}}{\rm{.78}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{5 - 1}}}}} {\rm{\gg 1}}{\rm{.0756}}\)
Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 5 - 1 = 4}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = 0}}{\rm{.025}}\)in the table of the critical values for $t$ distributions in the appendix:
\({{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 2}}{\rm{.776}}\)
The margin of error is then:
\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 2}}{\rm{.776 \times 1}}{\rm{.0756}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{5}}}} {\rm{\gg 3}}{\rm{.2709}}\)
The boundaries of the prediction interval then become:
\(\begin{aligned}{\rm{\bar x - E = 107}}{\rm{.78 - 3}}{\rm{.2709 = 104}}{\rm{.5091}}\\{\rm{\bar x + E = 107}}{\rm{.78 + 3}}{\rm{.2709 = 111}}{\rm{.0509}}\end{aligned}\)
We note that the prediction interval is wider than the confidence interval.
Hence \((104.5091,111.0509)\)
The prediction interval is wider than the confidence interval.
To Calculate an interval
(d)
Given:
\(\begin{aligned}{\rm{n = 5 }}\\{\rm{c = 95\% = 0}}{\rm{.95}}\end{aligned}\)
\(107.1109.5107.4108.6108.1\)
Multiplication rule for independent events:
\({\rm{P(A and B) = P(A) \times P(B)}}\)
We want the probability of five confidence intervals to be \(95\% \)each:
\({\rm{P(5 confidence intervals ) = 95\% = 0}}{\rm{.95}}\)
Use the multiplication rule:
\({{\rm{(P( each confidence interval ))}}^{\rm{5}}}{\rm{ = 0}}{\rm{.95}}\)
Take the \(5\)th root of each side:
\({\rm{P( each confidence interval ) = }}\sqrt({\rm{5}}){{{\rm{0}}{\rm{.95}}}}{\rm{\gg 0}}{\rm{.99}}\)
Thus we will determine a \({\rm{c = 0}}{\rm{.99 = 99\% }}\)confidence interval.
The mean is the sum of all values divided by the number of values:
\({\rm{\bar x = }}\frac{{{\rm{107}}{\rm{.1 + 109}}{\rm{.5 + 107}}{\rm{.4 + 106}}{\rm{.8 + 108}}{\rm{.1}}}}{{\rm{5}}}{\rm{ = }}\frac{{{\rm{538}}{\rm{.9}}}}{{\rm{5}}}{\rm{ = 107}}{\rm{.78}}\)
The variance is the sum of squared deviations from the mean divided by\({\rm{n - 1}}\)The standard deviation is the square root of the variance:
\({\rm{s = }}\sqrt {\frac{{{{{\rm{(107}}{\rm{.1 - 107}}{\rm{.78)}}}^{\rm{2}}}{\rm{ + \ldots + (108}}{\rm{.1 - 107}}{\rm{.78}}{{\rm{)}}^{\rm{2}}}}}{{{\rm{5 - 1}}}}} {\rm{\gg 1}}{\rm{.0756}}\)
Determine the t-value by looking in the row starting with degrees of freedom \({\rm{df = n - 1 = 5 - 1 = 4}}\) and in the column with \({\rm{\alpha = (1 - c)/2 = 0}}{\rm{.025}}\) in the table of the critical values for t distributions in the appendix:
\({{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ = 4}}{\rm{.604}}\)
The margin of error is then:
\({\rm{E = }}{{\rm{t}}_{{\rm{\alpha /2}}}}{\rm{ \times s}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{n}}}} {\rm{ = 4}}{\rm{.604 \times 1}}{\rm{.0756}}\sqrt {{\rm{1 + }}\frac{{\rm{1}}}{{\rm{5}}}} {\rm{\gg 5}}{\rm{.4247}}\)
The boundaries of the prediction interval then become:
\(\begin{aligned}{\rm{\bar x - E = 107}}{\rm{.78 - 5}}{\rm{.4247 = 102}}{\rm{.3553}}\\{\rm{\bar x + E = 107}}{\rm{.78 + 5}}{\rm{.4247 = 113}}{\rm{.2047}}\end{aligned}\)
Thus we predict that all five observations in a sample of \(5\) are between \({\rm{102}}{\rm{.3553\;mJ/}}{{\rm{m}}^{\rm{2}}}{\rm{ and 113}}{\rm{.2047\;mJ/}}{{\rm{m}}^{\rm{2}}}\).
Hence \((102.3553,113.2047)\)
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