A doughnut shop wants to determine if there is a difference in donut sales at different times of the day and for different types of doughnuts. They are open in the morning, afternoon, and night, and offer the following flavors: vanilla, chocolate, red velvet, and marbled. There were a total of 48 sales recorded. The shop conducted a two-way ANOVA test and found an F test statistic for Flavor of 14.87. What would be the numerator degree of freedom for the F test statistic to determine if the factor flavor was significant

Solution :

Let

= number of levels for the factors 'flavors' = 4

      (4 levels vanilla, chocolate, red velvet and marbled)

The degree of freedom for the factor 'flavors' = - 1

                                                                            = 4 - 1

                                                                            = 3

Now defining the F test statistics for testing the significance of the factors, 'flavors' :

F test statics =          ,   Ms = mean square

where F = , error df.

Thus the numerator degrees of the freedom for the F test statistics to determine if the factor flavor was significant is = - 1

                                                                            = 4 - 1

                                                                            = 3                                                                              

step-by-step explanation:

the options are kind of confusing. i can tell you what i see going on here when i look at the graph.

the class and exam medians are closely related, and it looks like the low outlier on the exam did pull the median lower. [tex][/tex]