Question

Hotel guest satisfaction. Refer to the 2015 North American Hotel Guest Satisfaction Index Study, Exercise 4.49 (p. 239). You determined that the probability that a hotel guest was delighted with his or her stay and would recommend the hotel is .12. Suppose a large hotel chain randomly samples 200 of its guests. The chain’s national director claims that more than 50 of these guests were delighted with their stay and would recommend the hotel.

a. Under what scenario is the claim likely to be false?

b. Under what scenario is the claim likely to be true?

Short answer

a. The guests were delighted with the stay and would recommend the hotel is 0.12. It will not say that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel since the obtained probability value is extremely small; under this scenario claim is likely to be false.

b. The probability that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel must be higher than 0.12, under this scenario, is the claim likely to be true.

Step-by-step solution

Given information

Referring to the 2015 North American Hotel Guest Satisfaction Index Study, Exercise 4.49, the hotel guest was delighted with their stay and would recommend the hotel is 0.12. A large hotel chain randomly samples 200 guests. More than 50 out of 200 guests were delighted with the stay and would recommend the hotel.

Step 2:(a) Identify the scenario in the case the claim is likely to be false

The probability that 50 out of 200 guests were delighted with the stay and would recommend the hotel is,

The mean is,

$\begin{array}{rcl}\mu & =& np\\ & =& 200\times 0.12\\ & =& 24\\ & & \end{array}$

The variance is,

$\begin{array}{rcl}{\sigma }^2& =& np\left(1-p\right)\\ & =& 200\times 0.12\times \left(1-0.12\right)\\ & =& 200\times 0.12\times 0.88\\ & =& 21.12\\ & & \end{array}$

The standard deviation is,

$\begin{array}{rcl}\sigma & =& \sqrt{21.12}\\ & =& 4.5957\\ & & \end{array}$

For,$x=50$

The z-score is,

$\begin{array}{rcl}z& =& \frac{x-\mu }{\sigma }\\ & =& \frac{50-24}{4.5957}\\ & =& 5.6575\\ & & \end{array}$

$\begin{array}{rcl}P\left(x>50\right)& =& 1-P\left(z⩽5.6575\right)\\ & =& 1-1 \left[From S\tan dard Normal Table\right]\\ & =& 0\\ & & \end{array}$

The probability value is 0

Therefore, the guests were delighted with the stay and would recommend the hotel is 0.12. Then it will unlike to say that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel since the obtained probability value is extremely small,under this scenario is claimed likely to be false.

Step 3:(b) Identify the scenario in the case the claim is likely to be true

If the claim is likely to be true, then the probability that more than 50 out of 200 guests were delighted with the stay and would recommend the hotel must be higher than 0.12; under this scenario is, the claim likely to be true.