Question

38 minutes, 02 seconds. Question Completion Status: INTERPRET THE FOLLOWING SPSS OUTPUT: Jefi the new oumer of a very upscale restaurant betieves thet people uho are most Hlely to patronize his restaurant will very strongly prefer (a value of s on his survey) each of the following (a) woitsh with tuxedos (b) unusual desserts ic) lorge variety of entrees (d) unusual entrees (ej elegant dlécor, and h jaztz combo music. The following hypothesis test mas compiered in sss-interpret the reults for et n spss One Sample Statistics Mean Sad Deiation Std Emor Mean Prefer Formal Waistat 468 72 470 055 Wearning Tux edos Preter Unusual Dessert 72 Prefer Large Vanety af 4 51 503 059 Prefer Unusual Entrees 4 50 Preter Elegant Decor 4 50 72 504 260 Preler Jazz Come 72 One Sample Test 95% Contdence Interval of the Preter Fomal Waitstat -21 5773 Prefer Unuual Dessert 71 Prefer Large Vaniety of 417 Prefer Unuual Enbees 7 121 71 71 Prefer Elegant Decor 4426 Preler Jazz Combe Arial course assessment id-_1239242 1ccourse jid 1694215 18content,id 39728494. 1ästep null Remaining Time: 1 hour, 58 minutes, 02 seconds. e Question Completion Status INTERPRET THE FOLLOWING SPSS OUTPUT Je the new our ero averyu sale restaurant be et es that peoplewho are most lilely to patro ne his restaurant will ery strong y prefer (a ual eo Sonhs si neoch o the fol nga yaon with tusedos (bj unusual desserts ic) lorge voriety of entrees. id) unusual entrees iti elegant decor and (th jare combo music. The following hypothesis test iwas complened in SPSs intergret the resuits for et One Sample Statistics Prefer Fomal Watstat 72 4 68 470 055 Wearing Tuwredes Prefor Unupual Desserts 72 4 54 502 Prefer Large Vanety of 059 503 Preter Unusual Entrees 72 4 58 Preler Elegant Dacor 72 Pter 2 Comb 72 One Sample Test Profer Formal Wat 5773 71 weanng Tugon 71 Prefer Unvoual Desserts -7.751 Prefer Large Vanety af 195 53 7.12 Prefer Utunual Entrees Preter Elegant Decor 8 426 2 403 Prer Jazz Combe

Answer 1

the result shows results of 1-sample t-test to compare 5 population with mu = 5

we observe that sig(2-tailed) otherwise popularly known as p-value = 0.000 is all 5 cases

As p-value < alpha
we reject the null hypothesis
we conclude that mean is different from 5 in each 5 cases