Question

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 each of 13 workers was 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 (measured in kilocalories per kg of subject per pounds of clay):
1 2 3 4 5 6 7 8 9 10 11 12 13 Worker Conventional 0.0011 0.0014 0.0018 0.0022 0.001 0.0016 0.0028 0.002 0.0015 0.0023 0.0017 0.002 0.0014 Perforated 0.0011 0.001 0.0019 0.0013 0.0011 0.0017 0.0024 0.002 0.0013 0.0017 0.002 0.0013 0.0013 Carry out a test of hypotheses at significance level 0.05 to see whether the true average energy expenditure using the conventional shovel exceeds that using the perforated shovel. (hint: this is a paired-design!). Make sure to calculate a p-value!

Answer 1

Answer:

Step-by-step explanation:

Corresponding true average energy expenditure of shovel with conventional blade and true average energy expenditure of shovel with perforated blades form matched pairs.

The data for the test are the differences between the true average energy expenditures of the shovels.

μd = true average energy expenditure of shovel with conventional blade minus true average energy expenditure of shovel with perforated blades.

Conventional perforated diff

0.0011 0.0011 0

0.0014 0.0010 0.0004

0.0018 0.0019 -0.0001

0.0022 0.0013 0.0009

0.0010 0.0011 -0.0001

0.0016 0.0017 -0.0001

0.0028 0.0024 0.0004

0.0020 0.002 0

0.0015 0.0013 0.0002

0.0023 0.0017 0.0006

0.0017 0.002 -0.0003

0.0020 0.0013 0.0007

0.0014 0.0013 0.0001

Sample mean, xd

= (0 + 0.0004 - 0.0001 + 0.0009 - 0.0001 - 0.0001 + 0.0004 + 0 + 0.0002 + 0.0006 - 0.0003 + 0.0007 + 0.0001)/13 = 0.0002077

xd = 0.0002077

Standard deviation = √(summation(x - mean)²/n

n = 13

Summation(x - mean)² = (0 - 0.0002077)^2 + (0.0004 - 0.0002077)^2 + (- 0.0001 - 0.0002077)^2+ (0.0009 - 0.0002077)^2 + (- 0.0001 - 0.0002077)^2 + ( - 0.0001 - 0.0002077)^2 + (0.0004 - 0.0002077)^2 + (0 - 0.0002077)^2 + (0.0002 - 0.0002077)^2 + (0.0006 - 0.0002077)^2 + (- 0.0003 - 0.0002077)^2 + (0.0007 - 0.0002077)^2 + (0.0001 - 0.0002077)^2 = 0.00000158923

Standard deviation = √(0.00000158923/13

sd = 0.00035

For the null hypothesis

H0: μd ≥ 0

For the alternative hypothesis

H1: μd < 0

The distribution is a students t. Therefore, degree of freedom, df = n - 1 = 13 - 1 = 12

The formula for determining the test statistic is

t = (xd - μd)/(sd/√n)

t = (0.0002077 - 0)/(0.00035/√13)

t = 2.14

We would determine the probability value by using the t test calculator.

p = 0.027

Since alpha, 0.05 > than the p value, 0.027, then we would reject the null hypothesis. Therefore, at 5% significance level, we can conclude that the true average energy expenditure using the conventional shovel does not exceed that using the perforated shovel.