4.98 Hotels\' use of ecolabels. Refer to the Journal of Vacation Marketing (Janu

Question

4.98 Hotels' use of ecolabels. Refer to the Journal of Vacation Marketing (January 2016) study of travelers' familiarity with ecolabels used by hotels, Exercise 2.64 (p. 80). Recall that adult travelers were shown a list of 6 different ecolabels, and asked, "Suppose the response is measured on a continuous scale from 10 (not familiar at all) to 50 (very familiar)." The mean and standard deviation for the Energy Star ecolabel are 44 and 1.5, respectively. Assume the distribution of the responses is approximately normally distributed a. Find the probability that a response to Energy Star ex- ceeds 43 between 42 and 45 think it is likely that t b. Find the probability that a respo nse to Energy Star falls c. If you observe a response of 35 to an ecolabel, do vou he ecolabel was Energy Star? Explain.

Explanation / Answer

Data given is:

Mean m = 44

Standard deviation S = 1.5

(a)

At X = 43, we have:

z = (X-m)/S = (43-44)/1.5 = -0.67

The p-value for this z-value is:

P(X > 43) = P(z > -0.67) = 0.7486

(b)

At X = 42, we have:

z = (X-m)/S = (42-44)/1.5 = -1.33

The p-value for this z-value is:

P(X < 42) = P(z < -1.33) = 0.0918

At X = 45, we have:

z = (X-m)/S = (45-44)/1.5 = 0.67

The p-value for this z-value is:

P(X < 45) = P(z < 0.67) = 0.7486

So,

P(42 < X < 45) = P(X < 45)-P(X < 42) = 0.7486-0.0918 = 0.6568

(c)

At X = 35, we have:

z = (X-m)/S = (35-44)/1.5 = -6

The p-value for this z-value is:

P(X < 35) = P(z < -6) = approximately 0

This means that it is highly unlikely to get a response of 35 for Energy Star ecolabel.

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