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:
Ho: ?1 = ?2
Ha: ?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.
