Question

Please answer showing all the work by hand ie not using excel , minitab,...

(For (a) please show all the work on how you got the values for the table)

2.) Suppose customer satisfaction tests are given to random samples of customers from five different restaurant chains. The samples are chosen so that each chain provides data from three major cities (Montreal, Ottawa, Toronto). Perform an analysis using the data below, treating the city as blocks. Restaurant 52 79 (a) Fill in the following table (4 points) Cit Montreal 73 Ottawa Toronto 82 82 86 91 34 42 45 49 80 Source of Variation MS otal (b) what do you conclude based on the results? Use alpha of 5% in the right tail. (1 point) (c) Based on your answer to (b), would it be helpful to test for the difference in means using the Tukey method? If not, why? If so, complete the calculation (with alpha 0.05) and draw your conclusions. (1.5 points)

Answer 1

Treating city's as groups

One-Way ANOVA Table Sum of Squares Degrees of Freedom Mean Square Source F-Stat P-Value DF MS Right tail of Between Groups k -1 SSB MSB = SSB / (k-1) F=MSB / MSw F(k-1,N-k) Within Groups N-k SSw Total: N-1

Between Groups Degrees of Freedom: DF k- 1, where k is the number of groups Within Groups Degrees of Freedom: DF N-k, where N is the total number of subjects Total Degrees of Freedom: DF N -1 Sum' of Squares Between Groups: SSB =Σ ini (xī-x).where ni is the number of subjects n the ith group Sum of Squares with nGroups SSw -(n; 1)Sj. here Sisthe standard de ation of the ith group Total Sum of Squares: Ss SsB+SSw lk Mean Square Between Groups: MSB SSB / (k -1) Mean Square Within Groups: MSw SSw (N -k) F-Statistic (or F-ratio): FMSB / MSw

The fratio value is 1.87236. The p-value is.19601. The result is not significant at p<.05.