Question

1. A farmer uses a greenhouse to grow his fruits and vegetables year round. He recently purchased a machine that uses artificial intelligence to adjust the lighting to optimize plant growth. He wants to cut down on his energy costs so he started experimenting with two different light bulbs (HID & LED). The farmer would note down the daily kW usage when either is installed that day, each independent of the other. He randomly selected usage data and created the table below. Due to some technical issue, he lost the files containing some values for the LED lighting. Does the data provide sufficient evidence, at α = 0.10, to indicate that average kW usage is lower for the LED bulbs?

LED (kW usage)	516	487	502	501	498
	515
HID (kW usage)	517	506	513	554	550
	499	535	490	510	529

Answer 1

Here we have data:

LED	HID
516	517
487	506
502	513
501	554
498	550
515	499
	535
	490
	510
	529

Here we use Excel to calculate:

Excel oputput:

t-Test: Two-Sample Assuming Unequal Variances
	Variable 1	Variable 2
Mean	503.1666667	520.3
Variance	119.7666667	450.67778
Observations	6	10
Hypothesized Mean Difference	0
df	14
t Stat	-2.12465636
P(T<=t) one-tail	0.025953184
t Critical one-tail	1.345030374
P(T<=t) two-tail	0.051906368
t Critical two-tail	1.761310136

Hypothesis:

H_o: μ1 = μ2

H_a: μ1 < μ2

Test statistics:

t = - 2.12

P-value = p(t>|-2.12|)

= 0.0259

Critical value:

t-critical = - 1.345

Conclusion: Reject the null hypothesis.

Here we have sufficient evidence to reject the null hypothesis, because t-observed value (-2.12) is less than t-critical value (-1.345) so it is in the rejection region.

we can support the Claim that average kW usage is lower for the LED bulbs.