Question

Amilkprocessing company testimplementedaplant-wide energy conservation programwitha goal of reducing themeandaily consumptionrateof atleast 1,000kWhfromitsnormaloperatingplants.The conservationprogramwas implementedinPlant B.The following data was collected on weekdays where consumption level is atits peak.Were the conservation efforts effective in achieving its goal? Compare the results with Plant A’s data where the program is not implemented using 𝛼 = 0.05. Assume that the population variances arenot equal. Plant A (kWh) Plant B (kWh) 3,952.80 4,036.00 3,276.00 4,036.00 3,636.00 4,036.00 3,636.00 3,264.00 3,636.00 864.00 3,636.00 1,368.00 4,068.00 2,196.00 4,068.00 4,392.00 4,362.00 5,220.00 4,362.00 3,600.00 4,362.00 3,960.00 4,362.00 4,428.00 3,882.00 756.00 3,808.80 612.00 3,808.80 684.0

Answer 1

The solution is based on Welch’s Test

X = energy consumption rate for Plant A

Y = energy consumption rate for Plant B

$H_0: \mu_1- \mu_2 \leq 1000 \\ H_1: \mu_1- \mu_2 > 1000$

Test-statistic:

$t= \frac{(\bar{X} - \bar{Y})-1000}{\sqrt{\frac{s^2_1}{n_1}+\frac{s^2_2}{n_2}}} \\ \bar{X}=\frac{3952.80 + 3276.00+...+3808.80 + 3808.80}{15} = 3923.633 \\ \bar{Y} = \frac{4036.00+4036.00+...+612.00+684.00}{15} = 2896.80 \\ s^2_1= \frac{(3952.80-3923.633)^2+(3276.00-3923.633)^2+...+(3808.80-3923.633)^2+(3808.80-3923.633)^2}{15-1}= 113572.4452 \\ s^2_2 = \frac{(4036.00-2896.8)^2+(4036.00-2896.8)^2+...+(612.00-2896.8)^2+(684.00-2896.8)^2}{15-1} = 2663639.3143 \\ t= \frac{(3923.633 -2896.8)-1000}{\sqrt{\frac{113572.4452}{15}+\frac{2663639.3143}{15}}} \\ = \frac{26.833}{\sqrt{185147.44}} \\ = 0.0623 \\ α=0.05 \\ df= 15-1= 14 \\ t_{0.05, 14} = 1.7613$

One-tailed test. Reject $H_0\; if \;t ≥ t_{0.05, 14}$

So, calculated t is less than the critical t-value. So, we accept the null hypothesis.

There is sufficient evidence to suggest that the claim is NOT valid at 0.05 level of significance. Hence, we conclude that energy conservation program does not reduce the mean consumption rate at least by 1000 kWh.