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 = 0.05. factor a is method of loading and unloading; factor b is the type of ride.

type of ride
roller coaster screaming demon long plume
method 1
47 53 52
49 45 48
method 2
46 48 51
48 44 47

set up the anova table (to whole number, but p - value to 4 decimals and f value to 2 decimals, if necessary). do not round intermediate calculations.

source of variation sum of squares degrees of freedom mean square f p - value
factor a *
factor b *
interaction *
error *
total *

the p - value for factor a is select your answer *
what is your conclusion with respect to factor a? select your answer *
the p - value for factor b is select your answer *
what is your conclusion with respect to factor b? select your answer *
the p - value for the interaction of factors a and b is select your answer *
what is your conclusion with respect to the interaction of factors a and b? select your answer *
what is your recommendation to the amusement park? select your answer *

Explanation:

Step1: Organize Data for Factor A (Method)

Factor A has 2 levels (Method 1, Method 2) and 2 replicates (since there are 2 rows per method). For each method, sum the values across rides:
- Method 1: \(47 + 53 + 52 + 46 + 48 + 51 = 297\) (Wait, no, actually, each method has 2 replicates for 3 rides? Wait, the table: Method 1 has two rows? Wait, looking at the table:

Wait, the data is:

Type of Ride: Roller Coaster, Screaming Demon, Long Plume

Method 1: 47, 53, 52; then another row? Wait, the table shows:

Method 1:
Roller Coaster: 47
Screaming Demon: 53
Long Plume: 52

Method 2:
Roller Coaster: 49
Screaming Demon: 45
Long Plume: 48

Then another set? Wait, maybe it's a 2x3 factorial with 2 replicates? Wait, the rows: maybe Method 1 has two observations per ride? Wait, the table has:

Method 1:
Row 1: 47 (RC), 53 (SD), 52 (LP)
Row 2: 46 (RC), 48 (SD), 51 (LP)

Method 2:
Row 1: 49 (RC), 45 (SD), 48 (LP)
Row 2: 48 (RC), 44 (SD), 47 (LP)

Ah, so it's a 2 (Factor A: Method 1, Method 2) x 3 (Factor B: Ride Type) factorial design with \(n = 2\) replicates per cell.

So total observations: \(2 \times 3 \times 2 = 12\).

First, calculate totals for Factor A (Method):

Method 1 (A1): Sum all values for A1. Cells: (47,46), (53,48), (52,51). So per cell: 47+46=93, 53+48=101, 52+51=103. Total for A1: 93 + 101 + 103 = 297.

Method 2 (A2): Cells: (49,48), (45,44), (48,47). Per cell: 49+48=97, 45+44=89, 48+47=95. Total for A2: 97 + 89 + 95 = 281.

Grand Total (G): 297 + 281 = 578.

Step2: Sum of Squares for Factor A (SSA)

Formula: \(SSA = \frac{\sum T_A^2}{n_B \times n} - \frac{G^2}{N}\), where \(n_B\) is number of levels of B (3), \(n\) is replicates per cell (2), \(N = a \times b \times n = 2 \times 3 \times 2 = 12\).

\(\sum T_A^2 = 297^2 + 281^2 = 88209 + 78961 = 167170\)

\(n_B \times n = 3 \times 2 = 6\)

\(\frac{\sum T_A^2}{n_B \times n} = \frac{167170}{6} \approx 27861.6667\)

\(\frac{G^2}{N} = \frac{578^2}{12} = \frac{334084}{12} \approx 27840.3333\)

\(SSA = 27861.6667 - 27840.3333 = 21.3333\) (Wait, but maybe I messed up. Wait, actually, for two-way ANOVA with replication, the formula for SSA is \(\frac{\sum (T_{A_i})^2}{b \times n} - \frac{G^2}{a \times b \times n}\), where \(a\) is levels of A (2), \(b\) levels of B (3), \(n\) replicates (2). So yes, that's correct.

But wait, the problem says "Set up the ANOVA table". Let's proceed step by step.

First, define:

- \(a = 2\) (Factor A levels: Method 1, Method 2)
- \(b = 3\) (Factor B levels: RC, SD, LP)
- \(n = 2\) (replicates per cell)
- \(N = a \times b \times n = 12\)

Sum of Squares for Factor A (SSA):

\(T_{A1} = 47 + 46 + 53 + 48 + 52 + 51 = 297\) (sum of all Method 1 observations)
\(T_{A2} = 49 + 48 + 45 + 44 + 48 + 47 = 281\) (sum of all Method 2 observations)
\(SSA = \frac{T_{A1}^2 + T_{A2}^2}{b \times n} - \frac{G^2}{N}\)
\(G = 297 + 281 = 578\)
\(SSA = \frac{297^2 + 281^2}{3 \times 2} - \frac{578^2}{12}\)
\(= \frac{88209 + 78961}{6} - \frac{334084}{12}\)
\(= \frac{167170}{6} - \frac{334084}{12}\)
\(= 27861.6667 - 27840.3333 = 21.3333\)

Sum of Squares for Factor B (SSB):

Calculate \(T_{B1}\) (RC: 47,46,49,48), \(T_{B2}\) (SD: 53,48,45,44), \(T_{B3}\) (LP: 52,51,48,47)
\(T_{B1} = 47 + 46 + 49 + 48 = 190\)
\(T_{B2} = 53 + 48 + 45 + 44 = 190\)
\(T_{B3} = 52 + 51 + 48 + 47 = 198\)
\(SSB = \frac{T_{B1}^2 + T_{B2}^2 + T_{B3}^2}{a \times n} - \frac{G^2}{N}\)
\(= \frac{190^2 + 190^2 + 198^2}{2 \times 2} - \frac{578^2}{12}\)
\(= \frac{36100 + 36100 + 39204}{4} - 27840.3333\)
\(= \frac{111404}{4} - 27840.3333\)
\(= 27851 - 27840.3333 = 10.6667\)

Sum of Squares for Interaction (SSAB):

First, calculate cell totals:
- Cell (A1,B1): 47 + 46 = 93
- Cell (A1,B2): 53 + 48 = 101
- Cell (A1,B3): 52 + 51 = 103
- Cell (A2,B1): 49 + 48 = 97
- Cell (A2,B2): 45 + 44 = 89
- Cell (A2,B3): 48 + 47 = 95

\(SSAB = \frac{\sum (cell\ totals)^2}{n} - \frac{G^2}{N} - SSA - SSB\)
\(\sum (cell\ totals)^2 = 93^2 + 101^2 + 103^2 + 97^2 + 89^2 + 95^2\)
\(= 8649 + 10201 + 10609 + 9409 + 7921 + 9025 = 55814\)
\(\frac{55814}{2} = 27907\)
\(SSAB = 27907 - 27840.3333 - 21.3333 - 10.6667\)
\(= 27907 - 27872.3333 = 34.6667\)

Sum of Squares for Error (SSE):

Total Sum of Squares (SST) = \(\sum x^2 - \frac{G^2}{N}\)
Calculate \(\sum x^2\):
\(47^2 + 46^2 + 53^2 + 48^2 + 52^2 + 51^2 + 49^2 + 48^2 + 45^2 + 44^2 + 48^2 + 47^2\)
\(= 2209 + 2116 + 2809 + 2304 + 2704 + 2601 + 2401 + 2304 + 2025 + 1936 + 2304 + 2209\)
Let's sum:
2209+2116=4325; +2809=7134; +2304=9438; +2704=12142; +2601=14743; +2401=17144; +2304=19448; +2025=21473; +1936=23409; +2304=25713; +2209=27922.

\(SST = 27922 - \frac{578^2}{12} = 27922 - 27840.3333 = 81.6667\)

Then \(SSE = SST - SSA - SSB - SSAB = 81.6667 - 21.3333 - 10.6667 - 34.6667 = 15\)

Degrees of Freedom:

- Factor A: \(df_A = a - 1 = 2 - 1 = 1\)
- Factor B: \(df_B = b - 1 = 3 - 1 = 2\)
- Interaction: \(df_{AB} = (a - 1)(b - 1) = 1 \times 2 = 2\)
- Error: \(df_E = a \times b \times (n - 1) = 2 \times 3 \times 1 = 6\)
- Total: \(df_T = N - 1 = 12 - 1 = 11\)

Mean Squares:

- \(MSA = \frac{SSA}{df_A} = \frac{21.3333}{1} = 21.3333\)
- \(MSB = \frac{SSB}{df_B} = \frac{10.6667}{2} = 5.3333\)
- \(MSAB = \frac{SSAB}{df_{AB}} = \frac{34.6667}{2} = 17.3333\)
- \(MSE = \frac{SSE}{df_E} = \frac{15}{6} = 2.5\)

F - Values:

- \(F_A = \frac{MSA}{MSE} = \frac{21.3333}{2.5} = 8.5333\)
- \(F_B = \frac{MSB}{MSE} = \frac{5.3333}{2.5} = 2.1333\)
- \(F_{AB} = \frac{MSAB}{MSE} = \frac{17.3333}{2.5} = 6.9333\)

p - Values:

- For \(F_A\) with \(df_1 = 1\), \(df_2 = 6\): \(p - value = P(F > 8.5333) \approx 0.0275\) (using F - distribution table or calculator)
- For \(F_B\) with \(df_1 = 2\), \(df_2 = 6\): \(p - value = P(F > 2.1333) \approx 0.2075\)
- For \(F_{AB}\) with \(df_1 = 2\), \(df_2 = 6\): \(p - value = P(F > 6.9333) \approx 0.0235\)

Now, let's fill the ANOVA table:

| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p - value |
|---------------------|----------------|--------------------|-------------|---|-----------|
| Factor A | 21.33 | 1 | 21.33 | 8.53 | 0.0275 |
| Factor B | 10.67 | 2 | 5.33 | 2.13 | 0.2075 |
| Interaction | 34.67 | 2 | 17.33 | 6.93 | 0.0235 |
| Error | 15.00 | 6 | 2.50 | - | - |
| Total | 81.67 | 11 | - | - | - |

(Note: Rounded to two decimal places for SS, MS, F; four decimals for p - value as per problem request.)

Now, for the conclusions:

- Factor A (Method): p - value ≈ 0.0275 < 0.05, so reject \(H_0\); there is a significant effect of the loading/unloading method.
- Factor B (Ride Type): p - value ≈ 0.2075 > 0.05, so fail to reject \(H_0\); no significant effect of ride type.
- Interaction (A×B): p - value ≈ 0.0235 < 0.05, so reject \(H_0\); there is a significant interaction effect.

For the amusement park, since Factor A (method) and interaction are significant, they should consider the method of loading/unloading and its interaction with ride type to reduce waiting times.

Answer:

ANOVA Table:

Source	SS	df	MS	F	p - value
Factor B	10.67	2	5.33	2.13	0.2075
Interaction	34.67	2	17.33	6.93	0.0235
Error	15.00	6	2.50	-	-
Total	81.67	11	-	-	-

Conclusions:

• Factor A: Reject \(H_0\) (significant effect).
• Factor B: Fail to reject \(H_0\) (no significant effect).
• Interaction: Reject \(H_0\) (significant interaction).

Recommendation:

The park should focus on the loading/unloading method (Factor A) and account for its interaction with ride type to reduce waiting times, as ride type alone does not have a significant effect.