A downtown hotel runs a special promotion to try to fill rooms that aren\'t usua

Question

A downtown hotel runs a special promotion to try to fill rooms that aren't usually occupied on weekends. The long-run average response is 71 rooms per weekend. There is considerable variation due to weather, competing attractions, and unknown cases. The standard deviation of responses is about 15 rooms. a) The hotel schedules adequate staff to handle a response of 80 rooms. If more guests arrive, additional stuff must be brought in at overtime rates. What is the probability that additional staff will be needed on the particular weekend? b) The promotion manager reviews response rates for blocks of 30 weekends, which is regarded as a random sample. Completely describe the sampling distribution of the sample mean. c) If the average demand over the 30-week sample exceeds 80, the manager will increase the scheduled staff. What is the probability that a 30-week sample mean will exceed 80? d) Answering (a) and (c) do you have to make any assumptions about population. Explain why or why not. e) If the average demand over the 30-week sample is 90 rooms, Is this is unusual? What conclusion might you draw?

Explanation / Answer

a) since it is a long average this is a normal distribution

mu = 71, sigma = 15

x be the responses, then p (X > 80) = P ( Z > (80-71)/15) = P (Z > 0.6) = 1-P(Z < 0.6)

= 1-0.7257 = 27.4%

b) The manager reviews response rates of 30 weekends. That is n = 30. Usually for large samples, n>25, the distribution will be normally distributed.

Also if such samples of 30 weekends are repeated and everytime the sample mean is recorded, then by central limit theorem the distribution of the sample mean (30 weekend mean) will follow normal distribution.

c) Ideally the answer is same as that of part a.

We might need standard deviation of the sample to make any observation about the sample.

(x bar - mu)/ sqrt( sd/ sqrt(n)) =

d) When we say long-average we are taking it to be more than 30, because a year has 52 weekends. So definitely the mean responses would have been calculated over more than 30 weekends.

In part c) it is explictly mentioned as 30. So in both the cases the population follows normal distribution

e) t-statistic = (90-71)/[15/sqrt(30)] = 6.937

p-value = almost 0

since p-value < 0.05, reject the null hypothesis that the sample mean is equal to population mean

we can conclude that this 30 weekend mean (90) is different from the population mean(71) and is significantly different and very unlikely.

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