Step 1

Values of work of adhesion measurements are provided for a sample of 5 UHPC specimens adhered to steel. It is required to compute the 95% confidence interval for the true average work of adhesion, 95% prediction interval to predict a single work of adhesion value and a tolerance interval that includes at least 95% of all UHPC specimens adhered to steel.

 
Step 2

(a)

To verify if the given data comes from a normal population, a normal probability plot is plotted as shown below:

As most of the data points follow a straight line, it can be said that the data follows a normal distribution.

 
Step 3

(b)

Here, the sample size (n) is equal to 5.

The confidence level is equal to 0.95.

Thus,

\[\begin{array}{c} 1 - \alpha = 0.95\\ \alpha = 1 - 0.95\\ = 0.05 \end{array}\]

The sample mean $\left( {\bar x} \right)$ is calculated as follows:

\[\begin{array}{c} \bar x = \frac{{107.1 + 109.5 + 107.4 + 106.8 + 108.1}}{5}\\ = 107.78 \end{array}\]

The sample standard deviation is calculated as follows:

\[\begin{array}{c} s = \sqrt {\frac{{{{\left( {107.1 - 107.78} \right)}^2} + {{\left( {109.5 - 107.78} \right)}^2} + {{\left( {107.4 - 107.78} \right)}^2} + {{\left( {106.8 - 107.78} \right)}^2} + {{\left( {108.1 - 107.78} \right)}^2}}}{{5 - 1}}} \\ = 1.08 \end{array}\]

To compute the 95% confidence interval for the population mean work of adhesion $\left( \mu \right)$ the following formula is used:

\[CI = \bar x \pm {t_{\frac{\alpha }{2},n - 1}}\frac{s}{{\sqrt n }}\]

For the t- multiplier, the level of significance is calculated as follows:

\[\begin{array}{c} \alpha = 0.05\\ \frac{\alpha }{2} = \frac{{0.05}}{2}\\ = 0.025 \end{array}\]

For the t- multiplier, the degrees of freedom are calculated as follows:

\[\begin{array}{c} df = n - 1\\ = 5 - 1\\ = 4 \end{array}\]

Corresponding to 0.025 column and 4 row, the one-sided t-critical from the t-distribution is equal to 2.7764.

The 95% confidence interval is computed as follows:

\[\begin{array}{c} CI = \bar x \pm {t_{\frac{\alpha }{2},n - 1}}\frac{s}{{\sqrt n }}\\ = 107.78 \pm 2.7764 \times \frac{{1.08}}{{\sqrt 5 }}\\ = 107.78 \pm 2.7764\left( {0.481041} \right)\\ \approx \left( {106.4,109.1} \right) \end{array}\]

Therefore, the 95% confidence interval for the true mean work of adhesion is equal to (106.4,109.1).

As the value 107 lies in the confidence interval, it can be said that it is a plausible value of the true average work of adhesion.

As 110 does not lie in the confidence interval, it can be said that it is not a plausible value of the true average work of adhesion.

 
Step 4

(c)

The 95% prediction interval to predict the value of work of adhesion is computed using the given formula:

\[PI = \bar x \pm {t_{\frac{\alpha }{2},n - 1}}\left( {s \times \sqrt {1 + \frac{1}{n}} } \right)\]

The following calculations are done to compute the prediction interval:

\[\begin{array}{c} PI = \bar x \pm {t_{\frac{\alpha }{2},n - 1}}\left( {s \times \sqrt {1 + \frac{1}{n}} } \right)\\ = 107.78 \pm {t_{0.025,4}}\left( {1.08 \times \sqrt {1 + \frac{1}{5}} } \right)\\ = 107.78 \pm 2.7764 \times \left( {1.18} \right)\\ \approx \left( {104.5,111.1} \right) \end{array}\]

Therefore, the 95% prediction interval for predicting a single value of the work of adhesion is equal to (104.5,111.1).

It is observed that the width of the prediction interval is approximately 2.5 times that of the width of the confidence interval.

 
Step 5

(d)

To consist of at least 95% of the work of adhesion values with a 95% confidence level, a tolerance interval is constructed using the following formula:

\[TI = \bar x \pm to{l_{k,n}}\left( s \right)\]

Let, k=0.95 and n=5.

The tolerance critical value is looked up in the tolerance critical value table under the column corresponding to k and the row corresponding to n.

Here,

\[to{l_{0.95,5}} = 5.079\]

The following calculations are done to compute the tolerance interval:

\[\begin{array}{c} TI = \bar x \pm to{l_{k,n}}\left( s \right)\\ = 107.78 \pm 5.079\left( {1.08} \right)\\ \approx \left( {102.3,113.3} \right) \end{array}\]

Therefore, the tolerance interval that included 95% of the work of adhesion values with a 95% confidence is equal to (102.3,113.3).