Question

Part A and B are done, kindly do part C

2) The results of a survey comparing the costs of staying one night in a full-service hotel [including food, beverages, and telephone calls, but not taxes or gratuities) for several major cities are given in the following table. An Excel version of this data set is available on Blackboard (File: Hotel Prices) Los Angeles 240 250 230 245 235 New York 300 320 325 350 275 Atlanta HoustonPhoenix 190 198 185 195 182 238 240 236 248 228 190 200 192 198 Use the Excel file to construct the ANOVA table. Attach it to this sheet or email it to me. a. Solution Anova: Single Factor SUMMARY Gro New York Los Angel Atlanta Houston Phoenix CountSumAverage Variance 314 792.5 62.5 44.5 17 52 5 1570 5 1200 950 975 51190 190 195 238 ANOVA ce of Varic SS Between 49496 Within Gr 3874 P-value F crit 4 12374 63.88229 4.17E-11 2.866081 20 193.7 53370 24 otal b. Use a 0.05 significance level to test the claim that the one night in a full-service hotel costs the same in all five major cities. 1. esi Based on the information provided, the significance level is, and the degree of freedom and dfı-4 and df2 - 4, therefore, the rejection region for this F-test is R-fF: F> Fc2.8661) 3.Test Statistics F 63.8823 Level of significance is D.F(4, 20) a 0.05 4. P-value P-value at F = 63.8823, α= 0.05 is; P-value 0.000 5. Conclusion Since, p-value« α-0.05, so reject the null hypothesis Ho, Hence, there is a significance difference in the average cost of one night in a full-service hotel of five cities. Assumption: All the sample are drawn independently of each other Each group sample is drawn from a normally distributed All population have a common yarance If there is a difference in means, perform the Bonferroni method with EER = .05. needed. vou may round α and α/2 to four decimal places.) Summarize your results with the underline approach and also state which means can be concluded larger than which. (Continue on back if needed.) c.

Answer 1

We will use R software to answer this question

c)

The R output is

> data-(300,320,325,350,275,240,250,230,245,235,190,198,185,195,182, 195,190,200,192,198,238,240,236,248,228) > city-as.factor (rep (seq(1,5),each-5)) > pairwise.t.test(data,city,p. adj bnferroni" Pairwise comparisons using t tests with pooled SD data: data and city 1 2 5.4e-07 3 7.6e-11 0.00015 4 1.6e-10 0.00053 1.00000 5 3.5e-07 1.00000 0.00024 0. 00090 4 P value adjustment method: bonferroni

according to the above p-value table we can say the city2 and city5 having same mean and city3 and city4 has same mean.

conclusion: city1 is significantly higher mean than other cities. Then comes city2 but city5 has statistically same mean as city2.