J.D. Power And Associates Publishes Annual Results

2 Each Year Jd Power And Associates Publishes The Results Of Its No

Each year J.D. Power and Associates publishes the results of its North American Hotel Guest Satisfaction Index Study. For 2009, the study revealed that 66% of hotel guests were aware of the hotel’s “green” conservation program. Among these guests, 72% actually participated in the program by reusing towels and bed linens.

In a random sample of 15 hotel guests, consider the number (x) of guests who were aware of and participated in the hotel’s conservation efforts.

Sample Paper For Above instruction

The analysis of hotel guest participation in green conservation efforts offers valuable insights into consumer behavior and supports hotel sustainability initiatives. This paper explores various statistical aspects of hotel guest participation, including the nature of binomial random variables, probability calculations, and the application of normal approximation techniques to binomial distributions.

Understanding the Binomial Random Variable

Part (a) of the problem asks to explain why the variable x, representing the number of guests aware of and participating in the conservation program out of 15, is a binomial random variable. A binomial random variable arises from a sequence of independent Bernoulli trials, each with two possible outcomes: success or failure.

In this context, each hotel guest can be viewed as a trial, where “success” indicates that a guest is aware of and participates in the program. The three key conditions for x to be binomial are: a fixed number of trials (n=15), independence between trials, and a constant probability of success (p) for each trial. Given that each guest's behavior is independent and the probability of success is constant across guests, x follows an approximately binomial distribution.

Determining the Probability of Success, p

Part (b) involves calculating the value of p, which represents the probability that a randomly selected guest both is aware of and participates in the conservation program. The problem provides that 66% of guests are aware and 72% of those aware participate. To find p:

• First, find the probability that a guest both is aware and participates: p = P(guest is aware and participates) = P(aware) × P(participates | aware).
• Given P(aware) = 0.66 and P(participates | aware) = 0.72, then p = 0.66 × 0.72 = 0.4752.

Thus, the probability p for a guest to both be aware and participate in the program is approximately 0.475.

Probability that at Least 10 Guests Participate

Part (c) assumes p=0.45 and asks for the probability that at least 10 out of 15 guests participate in the conservation efforts. This is a binomial probability problem with parameters n=15, p=0.45, and the event of interest x ≥ 10.

Calculating exact binomial probabilities directly is cumbersome; hence, the normal approximation to the binomial distribution is used. The binomial distribution B(n, p) can be approximated by a normal distribution with mean μ = np and standard deviation σ = √(np(1-p)).

With p=0.45:

• μ = 15 × 0.45 = 6.75
• σ = √(15 × 0.45 × 0.55) ≈ √(3.7125) ≈ 1.927

Applying the continuity correction for x ≥ 10, we compute the probability for z ≥ (9.5 - 6.75) / 1.927 ≈ 2.35.

Using standard normal tables or calculator:

P(z ≥ 2.35) ≈ 0.0094.

Therefore, the probability that at least 10 guests out of 15 participate, given p=0.45, is approximately 0.94%.

Testing the Large Hotel Chain’s Claim

Part (d) addresses whether the claim that more than 110 guests out of 200 participated in the conservation program is plausible, based on a sample of 200 guests. The chain's director claims that participation exceeds 55% (since 110/200=0.55).

This is a hypothesis testing problem for a population proportion. Null hypothesis H0: p ≤ 0.55; alternative hypothesis Ha: p > 0.55.

Given that 200 guests are sampled, and assuming the true participation rate equals the claimed rate (p=0.55), the expected number of participants is 200 × 0.55 = 110, with standard deviation σ = √(np(1-p)) ≈ √(200 × 0.55 × 0.45) ≈ √(49.5) ≈ 7.035.

Using the normal approximation, the test statistic z is calculated based on the observed count (x = 111 or more). For maximum variability, we test x=111:

z = (111 - 110) / 7.035 ≈ 0.142.

This z-value corresponds to a p-value well above the typical significance level of 0.05, indicating that the observed data does not provide sufficient evidence to reject the null hypothesis. Therefore, based on this sample, the claim that more than 55% of guests participate is not strongly supported.

In conclusion, statistical analysis using binomial distribution and normal approximation suggests that the hotel chain's claim is plausible, but further data could provide more definitive evidence.

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