91Ó°ÊÓ

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. \(1\begin{array}{*{20}{c}}{Observation}&{1 min}&{4 weeks}&{Difference}\\1&{10,490}&{9,110}&{1380}\\2&{16,620}&{13,250}&{3370}\\3&{17,300}&{14,720}&{2580}\\4&{15,480}&{12,740}&{2740}\\5&{12,970}&{10,120}&{2850}\\6&{17,260}&{14,570}&{2690}\\7&{13,400}&{11,220}&{2180}\\8&{13,900}&{11,100}&{2800}\\9&{13,630}&{11,420}&{2210}\\{10}&{13,260}&{10,910}&{2350}\\{11}&{14,370}&{12,110}&{2260}\\{12}&{11,700}&{8,620}&{3080}\\{13}&{15,470}&{12,590}&{2880}\\{14}&{17,840}&{15,090}&{2750}\\{15}&{14,070}&{10,550}&{3520}\\{16}&{14,760}&{12,230}&{2530}\end{array}\)

Calculate and interpret an upper confidence bound for the true average difference between\(1\)-minute modulus and\(4\)-week modulus; first check the plausibility of any necessary assumptions.

Step by step solution

01

Determine the upper confidence bound

Given:

\(n = 16\)

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

\(c = 95\% = 0.95\)Determine the sample mean of the differences. The mean is the sum of all values divided by the number of values.

\(\bar d = \frac{{1380 + 3370 + 2580 + \ldots + 2750 + 3520 + 2530}}{{16}} \approx 2635.625\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{(1380 - 2635.625)}^2} + \ldots + {{(2530 - 2635.625)}^2}}}{{16 - 1}}} \approx 508.6448\)

Determine the\({t_\alpha }\)using the Student's T distribution table in the appendix with\(df = n - 1 = 16 - 1 = 15\):

\({t_\alpha } = {t_{1 - c}} = {t_{0.05}} = 1.753\)

The margin of error is then:

\(E = {t_{\alpha /2}} \cdot \frac{{{s_d}}}{{\sqrt n }} = 1.753 \cdot \frac{{508.6448}}{{\sqrt {16} }} \approx 222.9136\)

The upper confidence bound for\({\mu _d}\)is then:

\(\bar d + E = 2635.625 + 222.9136 = 2858.5386\)

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