A farmer uses a greenhouse to grow his fruits and vegetables year round. He recently purchased...

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A farmer uses a greenhouse to grow his fruits and vegetablesyear round. He recently purchased a machine that uses artificialintelligence to adjust the lighting to optimize plant growth. Hewants to cut down on his energy costs so he started experimentingwith two different light bulbs (HID & LED). The farmer wouldnote down the daily kW usage when either is installed that day,each independent of the other. He randomly selected usage data andcreated the table below. Due to some technical issue, he lost thefiles containing some values for the LED lighting. Does the dataprovide sufficient evidence, at ? = 0.10, to indicate that averagekW 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 & Explanation Solved by verified expert

4.0 Ratings (497 Votes)

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: ?1 = ?2

H_a: ?1 < ?2

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.

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