In: Statistics and Probability
Energy Efficiency Rating
8.9
9.1
9.2
9.1
8.4
9.5
9
9.6
9.3
9.3
8.9
9.7
8.7
9.4
8.5
8.9
8.4
9.5
9.3
9.3
8.8
9.4
8.9
9.3
9
9.2
9.1
9.8
9.6
9.3
9.2
9.1
9.6
9.8
9.5
10
Problem 1 [15] A consumer’s advocate group would like to evaluate the average energy efficiency rating (EER) of window-mounted large-capacity (i.e. in excess of 7,000 BTU) air-conditioning units. A random sample of 36 such air-conditioning units is selected and tested for a fixed period of time. Their EER records are given the file EER.xlsx
[2] Hypotheses:
[1] Decision Rule:
[3] Calculations:
[2] Conclusions:
a) Using the 0.05 level of significance, is there evidence that the average EER is different from 9.0?
Hypothesis:
H0: μ = 9.0
Ha: μ ≠ 9.0
Formula:
t = (x̄ - μ)/(s/√n)
Decision Rule:
t = ± 2.03; We reject the null hypothesis if t-calc is less than -2.03 or greater than 2.03.
Calculations:
x̄ = Σxi/n = 9.2111; n = 36; s = √(Σx2 - (Σx)2 /n)/n - 1) = 0.384
t = (9.2111 - 9.0)/(0.384/√36) = 3.29
Conclusion:
3.29 is greater than 2.03, therefore we can reject the null hypothesis. There is sufficient evidence to suggest that the average EER is different from 9.0.
b) Find the p-value and interpret its meaning.
P (t > 3.29) = 0.002
0.5 - 0.002 = 0.498
0.498 * 2 = 0.996
There is a 99.6% chance of encountering an average EER that is different from the predicted 9.0.
c) What kind of error might you have committed in part (a)?
We may have committed a Type I error.
Type I error = α = .05
d) Construct a 95% confidence interval for the mean.
x̄ ± t(s/√n)
9.2111 + 3.29(0.384/√36)
= 9.00054 and 9.42166
Therefore, we can say with 95% confidence that the average EER will fall between 9.00054 and 9.42166.