Step 1

The data on cube compressive strength (MPa) of concrete specimens is given.

a.

A probability plot is constructed by plotting X- strength against its corresponding Z-score.

\[Z = \frac{{X - \bar X}}{{{s_X}}}\]

Where,

$\bar X$ is the sample mean.

${S_{\bar X}}$ is sample standard deviation.

The sample mean $\bar X$ for the given data is obtained as:

\[\begin{array}{c} \bar X = \frac{{\sum\limits_{i = 1}^{10} {{X_i}} }}{{10}}\\ = \frac{{964.2}}{{10}}\\ = 96.42 \end{array}\]

The sample standard deviation ${S_{\bar X}}$ is obtained as:

\[\begin{array}{c} {S_{\bar X}} = \sqrt {\frac{{\sum\limits_{i = 1}^{10} {{X_i} - \bar X} }}{{n - 1}}} \\ = \sqrt {\frac{{613}}{9}} \\ = 8.258 \end{array}\]
 
Step 2

Data is tabulated as shown below:

X	Z
112.3	1.922851
97	0.07023
92.7	-0.45044
86	-1.26172
102	0.675662
99.2	0.33662
95.8	-0.07507
103.5	0.857291
89	-0.89846
86.7	-1.17696

Normal probability plot is constructed by plotting X observation in vertical axis and its corresponding z score on horizontal axis as shown:

This plot shows straight line relationship between the sample points.

Therefore, it is plausible that the compressive strength is normally distributed.

 
Step 3

b.

Let the true average strength be denoted by $\mu$.

The hypothesis are formulated as:

The null hypothesis:

\[{H_0}:\mu = 100\]

That is, average strength is 100MPa.

VS

The alternative hypothesis:

\[{H_1}:\mu < 100\]

That is, average strength is less than 100MPa.

 
Step 4

The test statistics used is one sample T,

\[t = \frac{{\bar x - {\mu _0}}}{{\frac{s}{{\sqrt n }}}}\]

Here, ${\mu _0}$ is the value of strength considered in null hypothesis.

Substituting the values and obtaining the value of test statistics,

\[\begin{array}{c} t = \frac{{96.42 - 100}}{{\frac{{8.258}}{{\sqrt {10} }}}}\\ = \frac{{ - 3.58}}{{2.611}}\\ = - 1.37 \end{array}\]

The P-value for the two tailed test (using t-table) is obtained as:

\[\begin{array}{c} p - value = 2P\left( {t \ge |{t_{cal}}|} \right)\\ = 2P\left( {t \ge 1.37} \right)\\ = 2 \times \left( {1 - P\left( {t < 1.37} \right)} \right)\\ = 0.102 \end{array}\]

The p-value is 0.102.

 
Step 5

For, $\alpha = 0.05$,

\[P - value\left( {0.102} \right) > \alpha \left( {0.05} \right)\]

That is, the p-value is greater than the level of significance.

Therefore, by rejection rule, fail to reject the null hypothesis and it can be concluded that true average strength is 100 MPa.