Question

PRIMARY REASON FOR NOT RETURNING	Golden Palm	Palm Royal	Palm Princess
price	32	27	37
location	39	13	18
Accommodations	25	15	14
Other	23	8	8

I wanted to know what Kind of Anova do I run for those two tables. and how my anova would look like.

Choose hotel Again	Golden Palm	Palm Royal	Palm Princess
yes	133	201	179
No	32	89	66

Answer 1

1)

Null and Alternative Hypothesis:

H₀: µ_{Golden Palm} = µ_{Palm Royal} = µ_{Palm Princess}

H₁: Not all Means are equal

Alpha = 0.05

N = 16

n = 4

Degress of Freedom:

df_Between = a – 1 = 3-1 =2

df_Within = N-a = 16-3 = 13

df_Total = N-1 = 16-1 = 15

Critical Values:

Time (df_Between, df_Within): (2,13) = 4.26

Decision Rule:

If F is greater than 4.26, reject the null hypothesis

Test Statistics:

SS_Betwen = ∑(∑a_i)²/n - T²/N = 424.67

SS_Within = ∑(Y)² - ∑(∑a_i)²/n = 824.25

SS_Total = SS_{Betwen ­­}+ SS_Within = 1248.92

MS = SS/df               

F = MS_effect / MS_error

Hence,

F = 212.33/91.58 = 2.32

Result:

Our F = 2.32, we fail to reject the null hypothesis

Conclusion:

All means are equal.

2)

For Second Question, Chi Square Test would be appropriate.

Null and Alternative Hypothesis:

H₀: The two variables (Choice and Hotel) are independent.

H_a: The two variables are associated.

Given,

Observed Values:

Expected Values:

Expected Values are calculated as:

E_ij = (T_i * T_j)/N

where,

T_i = Total in ith row

T_j = Total in jth column

N = table grand total

Alpha = 0.05

df = (r-1)*(c-1) = (2-1)*(3-1) = 1*2 = 2

Chi Square _Critical = 5.99

Decision Rule:

If Chi Square> Chi Square _Critical reject the null hypothesis

Test Statistic:

Chi Square = ∑(O_ij – E_ij)²/E_ij = (133-120.92)²/120.92 + ……………….. + (66-65.45)²/65.45 = 6.86

Result:

Since, Chi Square> Chi Square _Critical we reject the null hypothesis.

Conclusion:

Choice and Hotel are associated.