Question

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...

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

  1. [8] Using the 0.05 level of significance and the critical value approach, is there evidence that the average EER is different from 9.0?

[2] Hypotheses:

[1] Decision Rule:

[3] Calculations:

[2] Conclusions:

  1. [2] Find the p-value and interpret its meaning.

  1. [1] What kind of error might you have committed in part (a)?

  1. [4] Construct a 95% confidence interval for the mean.

Solutions

Expert Solution

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.


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