Assignment 1 FINAL

# 5 significance level there is not sufficient evidence

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5% significance level, there is not sufficient evidence to infer that there is a relationship between annual profit and the number of rooms. QUESTION 2:

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A) Estimate with 95% confidence the mean profit for hotels with 200 rooms. *Because we want to estimate mean profit given an average value of 200 rooms, we will use the confidence interval estimator equation. *Significance level is 5%, divided by 2 is 2.5% or 0.025 ŷ +/- t (α/2, v) x s ε square root [ 1/n + (x g – x) ̄ 2 /(n-1)s 2 x ] *where ŷ = b 0 + b 1 x g and v= n-2 and x g = the interested value which is 200 t (0.025, 8) = 2.306 ŷ= 13.1356 + 0.03860(200)= 20.8556 20.8556 +/- 2.306 x 4.0851 square root [ 1/10 + (200- 242.6) 2 /(10-1)825.3778 ] 20.8556 +/- 5.5275 Lower limit = 15.3281 or \$15328100 (in millions). Upper limit = 26.3831 or \$26383100 (in millions). B) Suppose that a hotel has 200 rooms. Predict with 90% confidence the profit in that year. *Since we are predicting profit for one year given a specific value of rooms which is 200, we will use the prediction interval estimator. *Significance level is 10%, divided by 2 is 5% or 0.05 ŷ +/- t (α/2, v) x s ε square root [ 1+ 1/n + (x g – x) ̄ 2 /(n-1)s 2 x ] *where ŷ = b 0 + b 1 x g and v= n-2 and x g = the interested value which is 200 t (0.05, 8) = 1.860 ŷ= 13.1356 + 0.03860(200)= 20.8556 20.8556 +/- 1.860 x 4.0851 square root [ 1 + 1/10 + (200- 242.6) 2 /(10-1)825.3778 ] 20.8556 +/- 8.8098 Lower limit = 12.0458 or \$ 12045800 (in millions). Upper limit = 29.6654 or \$ 29665400 (in millions). C) What can you say about the suitability of this model and the available data on the quality of the estimations made in parts (a) and (b) in Question 2? - We can conduct a variety of tests to assess this model and determine whether a relationship exists between annual profit and the number of rooms. First we will look at the standard error of estimate. If sε / x 100 is less than 15%, then this is an excellent model. ӯ SSE= (n-1)[s2y-sxy2/s2x] = 9 [16.0644 – (31.86672/825.3778)]= 133.5066 sε= square root of SSE/n-2 = square root of 133.5066/8 = 4.0851 sε / x 100 = 4.0851/22.5 x 100 = 18.16% ӯ Since it is slightly above 15%, it can be argued that the model is good. Second we look at the coefficient of Determination
R 2 = sxy 2 / s 2 x s 2 y = 31.86672/ (825.3778)(16.0644) = 0.0765. R 2 ranges from the values 0 to 1, where the closer the value is to 1, the more impact the number of rooms has on profit. If the value is closer to zero, the less impact the number of rooms has on profit. R 2 has a value of 0.0765 which means that 7.65% of variation of profits is explained by variation in the number of rooms. The remaining 92.35% is unexplained due to error.
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• Spring '12
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