Levelness of concrete slabs. Geotechnical engineers use water-level "manometer" surveys to assess the levelness of newly constructed concrete slabs. Elevations are typically measured at eight points on the slab; the maximum differential between elevations is of interest. The Journal of Performance of Constructed Facilities (February 2005) published an article on the levelness of slabs in California residential developments. Elevation data collected for more than 1,300 concrete slabs before tensioning revealed that the maximum differential, x, has a mean of $\mathbf{\mu }\mathbf{=}\mathbf{0}\mathbf{.53}$an inch and a standard deviation of $\mathit{\sigma }\mathbf{=}\mathbf{0.193}$an inch. Consider a sample of n = 50 slabs selected from those surveyed, and let$\overline{\mathbf{X}}$ represent the sample's mean.

1. Fully describe the sample sampling distribution of $\overline{\mathbf{x}}$.
2. Find$\mathit{P}\mathbf{(}\overline{\mathbf{x}}\mathbf{>}\mathbf{0.58}\mathbf{)}$
3. The study also revealed that the mean maximum differential of concrete slabs measured after tensioning and loading is$\mathit{\mu }\mathbf{=}\mathbf{0}\mathbf{.58}$ an inch. Suppose the sample data yield$\overline{\mathbf{x}}\mathbf{=}\mathbf{0}\mathbf{.59}$ is an inch. Comment on whether the sample measurements were obtained before or after tensioning and loading.

Mean$\mathrm{\mu }=0.53$

Standard deviation$\mathrm{\sigma }=0.193$

Sample size $\mathrm{n}=50$

According to Central Limit Theorem, if large-sized samples are drawn from any population, with mean = $\mu$and standard deviation = $\sigma$, then the sampling distribution of sample means approximates a normal distribution, irrespective of the underlying distribution of population. The greater the sample size, the better the approximation (typically zero). The sample mean follows the normal distribution, with the mean of the sampling mean is:

$\begin{array}{c}{\mathrm{\mu }}_{\overline{\mathrm{x}}}=\mathrm{\mu }\\ =0.53\end{array}$

The standard deviation of sample mean or standard error:

$\begin{array}{c}{\mathrm{\sigma }}_{\overline{\mathrm{x}}}=\frac{\mathrm{\sigma }}{\sqrt{\mathrm{n}}}\\ =\frac{0.193}{\sqrt{50}}\\ \approx 0.0272\end{array}$

Let

$\begin{array}{c}\mathrm{P}(\overline{\mathrm{x}}>0.58)=\mathrm{P}\left(\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}>\frac{0.58-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}\right)\\ =\mathrm{P}\left(\mathrm{z}>\frac{0.58-0.53}{0.193/\sqrt{50}}\right)\\ =\mathrm{P}(\mathrm{z}>1.83)\end{array}$

From the z-score table,

$\mathrm{P}(\mathrm{z}>1.83)=0.0335$

Clearly, 0.59 is closer to the mean after tensioning, and that is what we would believe. In perspective, the probability of getting this extreme mean is lower than 0.0336(obtained in the last part) in the case of "before tensioning." Following is the probability of getting this sample mean in the case of "after."

Let,

$\mathrm{\mu }=0.58,\mathrm{\sigma }=0.193,\mathrm{n}=50$

So,

$\begin{array}{c}\mathrm{P}(\overline{\mathrm{x}}>0.59)=\mathrm{P}\left(\frac{\overline{\mathrm{x}}-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}>\frac{0.59-\mathrm{\mu }}{\mathrm{\sigma }/\sqrt{\mathrm{n}}}\right)\\ =\mathrm{P}\left(\mathrm{z}>\frac{0.59-0.58}{0.193/\sqrt{50}}\right)\\ =\mathrm{P}(\mathrm{z}>0.37)\end{array}$

From the z-score table,

$\mathrm{P}(\mathrm{z}>0.37)=0.3557$