Question

Shoveling is not exactly a high-tech activity, but it will continue to be a required task even in our information age. The article "A Shovel with a Perforated Blade Reduces Energy Expenditure Required for Digging Wet Clay" (Human Factors, 2010: 492-502) reported on an experiment in which 13 workers were each provided with both a conventional shovel and a shovel whose blade was perforated with small holes. The authors of the cited article provided the following data on stable energy expenditure ((kcal/kg(subject)//b(clay)):

Worker : 1 2 3 4

Conventional : .0011 .0014 .0018 .0022

Perforated : .0011 .0010 .0019 .0013

Worker: 5 6 7

Conventional : .0010 .0016 .0028

Perforated : .0011 .0017 .0024

Worker 8 9 10

Conventional : .0020 .0015 .0014

Perforated : .0020 .0013 .0013

Worker: 11 12 13

Conventional : .0023 .0017 .0020

Perforated : .0017 .0015 .0013

a. Calculate a confidence interval at the 95 % confidence level for the true average difference between energy expenditure for the conventional shovel and the perforated shovel (the relevant normal probability plot shows a reasonably linear pattern). Based on this interval, does it appear that the shovels differ with respect to true average energy expenditure? Explain.

b. Carry out a test of hypotheses at significance level .05 to see if true average energy expenditure using the conventional shovel exceeds that using the perforated shovel.

Short answer

(a) \((0.00002, 0.00038)\)

Shovels appear to differ with respect to the true average energy expenditure.

(b) There is sufficient evidence to support the claim that the true average energy expenditure using the conventional shovel exceeds that using the perforated shovel.

Step-by-step solution

To Calculate a confidence interval at the 95 % confidence level for the true average difference between energy expenditure for the conventional shovel and the perforated shovel 

(a)

Given:

\(\begin{array}{l}n = 13 \\c = 95\% = 0.95\end{array}\)

Determine the difference in value of each pair.

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

\(\bar d = \frac{{0 + 0.0004 - 0.0001 + \ldots + 0.0006 + 0.0002 + 0.0007}}{{13}} \approx 0.0002\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{(0 - 0.0002)}^2} + \ldots + {{(0.0007 - 0.0002)}^2}}}{{13 - 1}}} \approx 0.0003\)

Determine the\({t_{\alpha /2}}\) using the Student's T distribution table in the appendix with\(df = n - 1 = 13 - 1 = 12\) :

\({t_{0.025}} = 2.179\)

The margin of error is then:

\(E = {t_{\alpha /2}} \times \frac{{{s_d}}}{{\sqrt n }} = 2.179 \times \frac{{0.0003}}{{\sqrt {13} }} \approx 0.00018\)

The endpoints of the confidence interval for\({\mu _d}\) are:

\(\begin{array}{l}\bar d - E = 0.0002 - 0.00018 = 0.00002\\\bar d + E = 0.0002 + 0.00018 = 0.00038\end{array}\)

Shovels appear to differ with respect to the true average energy expenditure, because 0 does not lie in the confidence interval.

To find test of hypotheses at significance level .05 to see if true average energy expenditure using the conventional shovel exceeds that using the perforated shovel.

(b)

Given:

\(\begin{array}{l}n = 13 \\\alpha = 0.05\end{array}\)

Given claim: exceeds

The claim is either the null hypothesis or the alternative hypothesis. The null hypothesis and the alternative hypothesis state the opposite of each other. The null hypothesis needs to contain the value mentioned in the claim.

\(\begin{array}{l}{H_0}:{\mu _d} = 0\\{H_a}:{\mu _d} > 0\end{array}\)

Determine the difference in value of each pair.

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

\(\bar d = \frac{{0 + 0.0004 - 0.0001 + \cdots + 0.0006 + 0.0002 + 0.0007}}{{13}} \approx 0.0002\)

Determine the sample standard deviation of the differences:

\({s_d} = \sqrt {\frac{{{{(0 - 0.0002)}^2} + \ldots + {{(0.0007 - 0.0002)}^2}}}{{13 - 1}}} \approx 0.0003\)

Determine the value of the test statistic:

\(t = \frac{{\bar d}}{{{s_d}/\sqrt n }} = \frac{{0.0002}}{{0.0003/\sqrt {13} }} \approx 2.404\)

The P-value is the probability of obtaining the value of the test statistic, or a value more extreme, assuming that the null hypothesis is true. The P-value is the number (or interval) in the column title of the Student's T distribution in the appendix containing the t-value in the row\(df = n - 1 = 13 - 1 = 12\) :

\(0.01 < P < 0.025\)

If the P-value is less than the significance level, reject the null hypothesis.

\(P < 0.05 \Rightarrow Reject {H_0}\)

There is sufficient evidence to support the claim that the true average energy expenditure using the conventional shovel exceeds that using the perforated shovel.