Question

Example\(7.11\)gave data on the modulus of elasticity obtained\(1\)minute after loading in a certain configuration. The cited article also gave the values of modulus of elasticity obtained\(4\)weeks after loading for the same lumber specimens. The data is presented here. \(\begin{array}{l} Type\\ \begin{array}{*{20}{c}}{}&1&2&3&4&5&6\\{ M: }&{82.6}&{87.1}&{89.5}&{88.8}&{94.3}&{80.0}\\{ LD: }&{86.9}&{87.3}&{92.0}&{89.3}&{91.4}&{85.9}\\{}&7&8&9&{10}&{11}&{12}\\{ M: }&{86.7}&{92.5}&{97.8}&{90.4}&{94.6}&{91.6}\\{ LD: }&{89.4}&{91.8}&{94.3}&{92.0}&{93.1}&{91.3}\\{}&{}&{}&{}&{}&{}&{}\end{array}\end{array}\)

a. Estimate the difference in true average strength under the two drying conditions in a way that conveys information about reliability and precision, and interpret the estimate. What does the estimate suggest about how true average strength under moist drying conditions compares to that under laboratory drying conditions?

b. Check the plausibility of any assumptions that underlie your analysis of (a).

Short answer

(a) \(( - 2.5174,1.0508)\)

(b) Plausibly satisfied

Step-by-step solution

a)Step 1: Find the difference

Given:

\(n = 12\)

I will calculate a\(95\% \)confidence interval. Other confidence bounds can be obtained similarly.

\(c = 95\% = 0.95\)

Sample1	Sample 2	Difference D
\(82.6\)	\(86.9\)	\( - 4.3\)
\(87.1\)	\(87.3\)	\( - 0.2\)
\(89.5\)	\(92\)	\( - 2.5\)
\(88.8\)	\(89.3\)	\( - 0.5\)
\(94.3\)	\(91.4\)	\(2.9\)
\(80\)	\(85.9\)	\( - 5.9\)
\(86.7\)	\(89.4\)	\( - 2.7\)
\(92.5\)	\(91.8\)	\(0.7\)
\(97.8\)	\(94.3\)	\(3.5\)
\(90.4\)	\(92\)	\( - 1.6\)
\(94.6\)	\(93.1\)	\(1.5\)
\(91.6\)	\(91.3\)	\(0.3\)
Mean		\( - 0.7333\)
Sd		\(2.8079\)

Find the boundaries of confidence interval

Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{ - 4.3 - 0.2 - 2.5 + \ldots - 1.6 + 1.5 + 0.3}}{{12}} \approx - 0.7333\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{( - 4.3 - ( - 0.7333))}^2} + \ldots + {{(0.3 - ( - 0.7333))}^2}}}{{12 - 1}}} \approx 2.8079\)

Determine the\({t_{\alpha /2}}\) using the Student's T distribution table in the appendix with\(df = n - 1 = 12 - 1 = 11\):

\({t_{\alpha /2}} = {t_{1 - c/2}} = {t_{0.025}} = 2.201\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 2.201 \cdot \frac{{2.8079}}{{\sqrt {12} }} \approx 1.7841\)

The boundaries of the confidence interval for \({\mu _d}\) are then: \(\begin{array}{l}\bar d + E = - 0.7333 - 1.7841 = - 2.5174\\\bar d + E = - 0.7333 + 1.7841 = 1.0508\end{array}\)

b)Step 3: Plot the normal probability

(b) 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\(\frac{{j - 0.5}}{n}\)(or the closest area) with\(j \in \{ 1,2,3, \ldots .,n\} \)(these z-scores are given in the given table).

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 and does not contain strong curvature, then it is appropriate to assume that the population distribution is approximately normal.

The plot does not contain strong curvature and is roughly linear, thus it is plausible that the differences originate from a normal distribution.

Thus the assumptions underlying the analysis in part (a) appear to be satisfied.