Question

QUESTION 2 The director of a green energy company is interested in comparing two different methods of installing solar light panels. An experiment is conducted with 28 technicians using the old method and the other 45 technicians using the new method. For the 28 technicians using the old method, the average time was 483 minutes with 35.8 standard deviation For the 45 technicians using the new method, the average time was 384 minutes with 42.1 standard deviation. Conduct an analysis to determine whether there is a difference in the average time to complete the installation of the solar panels. Provide the test statistic value.

Answer 1

Solution:

Given:

Sample 1:

Sample size = n₁ = 28

Sample mean =

1=483

Sample standard deviation = s₁ = 35.8

Sample 2:

Sample size = n₂ = 45

Sample mean = $\bar{x}_{2}=384$

Sample standard deviation = s₂ = 42.1

Find Test statistic value:

Formula:

$t = \frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{S_{p}^{2}\times \left ( \frac{1}{n_{1}}+\frac{1}{n_{2}} \right )}}$

$S_{p}^{2}=\frac{(n_{1}-1)\times s_{1}^{2}+(n_{2}-1)\times s_{2}^{2}}{n_{1}+n_{2}-2}$

$S_{p}^{2}=\frac{(28-1)\times 35.8^{2}+(45-1)\times 42.1^{2}}{28+45-2}$

$S_{p}^{2}=\frac{ 27 \times 1281.64+ 44\times 1772.41}{71 }$

$S_{p}^{2}=\frac{ 34604.28+ 77986.04 }{71 }$

$S_{p}^{2}=\frac{112590.32 }{71 }$

$S_{p}^{2}=1585.7792$

Thus t test statistic is:

$t = \frac{\bar{x}_{1}-\bar{x}_{2}}{\sqrt{S_{p}^{2}\times \left ( \frac{1}{n_{1}}+\frac{1}{n_{2}} \right )}}$

$t = \frac{483-384}{\sqrt{1585.7792 \times \left ( \frac{1}{28}+\frac{1}{45} \right )}}$

$t = \frac{99 }{\sqrt{1585.7792 \times \left ( 0.0357143+ 0.0222222 \right )}}$

99 1585.7792 x 0.0579365

$t = \frac{99 }{\sqrt{91.8745 }}$

$t = \frac{99 }{9.585119 }$

t10.3285

= 10.329)