Answer to Question #248938 in Statistics and Probability for Hailey

Question

A milk processing company test implemented a plant-wide energy conservation program with a goal of reducing the mean daily consumption rate of at least 1,000 kWh from its normal operating plants. The conservation program was implemented in Plant B. The following data was collected on weekdays where consumption level is at its 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 are not equal.

 Plant A (kWh) 3,952.80 3,276.00 3,636.00 3,636.00 3,636.00 3,636.00 4,068.00 4,068.00 4,362.00 4,362.00 4,362.00 4,362.00 3,882.00 3,808.80 3,808.80

Plant B (kWh) 4,036.00 4,036.00 4,036.00 3,264.00 864.00 1,368.00 2,196.00 4,392.00 5,220.00 3,600.00 3,960.00 4,428.00 756.00 612.00 684.00

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 H₀ if $t ≥ t_{0.05, 14}$

So, calculated t is less than 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.