Question

An amusement park studied methods for decreasing the waiting time (minutes) for rides by loading and unloading riders more efficiently. Two alternative loading/ unloading methods have been proposed. To account for potential differences due to the type of ride and the possible interaction between the method of loading and unloading and the type of ride, a factorial experiment was designed. Use the following data to test for any significant effect due to the loading and unloading method, the type of ride, and interaction. Use α = .05. Use both p-Value and Critical-Value approaches.
Type of Ride
Roller Coaster Screaming Demon Log Flume
Method 1 41 52 50
43 44 46
49 46 48
Method 2
49 50 48
51 46 44
47 48 46
1. The p-value for Factor A is:__________.
a. less than .01
b. between .01 and .025
c. between .025 and .05
d. between .05 and .10
e. greater than .10Item 18
2. What is your conclusion with respect to Factor A?
a. Factor A is significant
b. Factor A is not significant
3. The p-value for Factor B is:_______.
a. less than .01
b. between .01 and .025
c. between .025 and .05
d. between .05 and .10
e. greater than .10
3. What is your conclusion with respect to Factor B?
a. Factor B is significant
b. Factor B is not significant
4. The p-value for the interaction of factors A and B is:_______.
a. less than .01
b. between .01 and .025
c. between .025 and .05
d. between .05 and .10
e. greater than .10
5. What is your recommendation to the amusement park?

Answer 1

Answer:

1. The p-value for Factor A is:__0.4194________. e. greater than0 .10

2. What is your conclusion with respect to Factor A?b. Factor A is not significant

3. What is your conclusion with respect to Factor B?b. Factor B is not significant

4. The p-value for the interaction of factors A and B is:__0.1621_____.e. greater than .10

5. There is no effect of decreasing the waiting time and changing the loading and unloading methods  between the rides.

Step-by-step explanation:

Performing  2 way ANOVA

Observation A       B               C                

1 41, 43, 49 52, 44, 46 50, 46, 48

2 49, 50, 48 50, 46, 48 48, 44, 46          

Row and column sums

                        A      B        C          Row total (xa)

1                 133     142   144                    419

2                 147     144   138                    429

Col total (xb) 280   286 282                            848

∑x²=412+432+492+...+482+442+462=40088→(A)

∑x²b/ra =1/3⋅2(280²+286²+282²)

=1/6(78400+81796+79524)

=1/6(239720)

=39953.3333→(B)

∑x²a/rb=1/3⋅3(419²+429²)

=1/9(175561+184041)

=1/9(359602)

=39955.7778→(C)

∑∑x²ab/r=  

1/3(133²+142²+144²+147²+144²+138²)

=1/3(17689+20164+20736+21609+20736+19044)

=1/3(119978)

=39992.6667→(C)

(∑x)²/rab

=(848)²/3⋅2⋅3

=719104/18

=39950.2222→(D)

Sum of squares total

SST=∑x²-(∑x)²/n

=(A)-(D)

=40088-39950.2222

=137.7778

Sum of squares between rows

SSA=∑x²a/rb-(∑x)²/n

=(C)-(D)

=39955.7778-39950.2222

=5.5556

Sum of squares between columns

SSB=∑x²b/ra- (∑x)²/n

=(B)-(D)

=39953.3333-39950.2222

=3.1111

Sum of squares between columns

SSAB=∑∑x²ab/r-(∑x)²/n-SSA-SSB

=(B)-(D)-SSA-SSB

=39992.6667-39950.2222-5.5556-3.1111

=33.7778

Sum of squares Error (residual)

SSE=SST-SSA-SSB-SSAB

=137.7778-5.5556-3.1111-33.7778

=95.3333

ANOVA table

Source             Sums                Degrees       Mean Squares

of Variation          of Squares     of freedom

                             SS              DF          MS              F         p-value

A          SSA=5.5556         a-1=1 MSR=5.5556    0.6993   0.4194

B            SSB=3.1111        b-1=2 MSC=1.5556    0.1958 0.8247

AB       SSAB=33.7778 2            MSAB=16.8889   2.1259 0.1621

Error (residual)SSE=95.33      12         MSE= 7.9444                      

Total SST=137.7778          17  

Conclusion:

1. F for between columns

The critical value F(1,2) at 0.05 level of significance is 4.7472

As calculated FR=0.6993<4.7472

H0 is accepted,  there is no significant difference between rows.

P value is 0.4194 > 0.05 we fail to reject H0

2. F for between columns

The critical value F(2,2) at 0.05 level of significance is 3.8853

As calculated FC=0.1958<3.8853

H0 is accepted  there is no significant difference between columns.

P value is 0.8247 > 0.05 we fail to reject H0

3. Interaction between the Factors AB

There is not sufficient evidence to support the claim of interaction between factors.

P value is 0.1621 > 0.05 we fail to reject H0