Question

a study reported in the accounting review examined the separate and joint effects of two levels of time pressure (low and moderate) and three levels of knowledge (naive, declarative, and procedural) on key word memory behavior in tax research. subjects were given a tax case containing a set of facts, a tax issue, and a key word index consisting of 1386 key words. they were asked to mark the key words they believed would help them in a tax authority search in resolving the tax case. prior to the experiment, a group of tax experts determined that the text contained 10 relevant key words. subjects in the naive group had little or no declarative or procedural knowledge; subjects in the declarative group had significant declarative knowledge but little or no procedural knowledge; and subjects in the procedural group had significant declarative knowledge and procedural knowledge. declarative knowledge consists of knowledge of both the applicable tax rules and the technical terms used to describe such rules. procedural knowledge is knowledge of the rules that guide the tax researcher’s search for relevant key words. subjects in the low time pressure situation were told they had 25 minutes to complete the problem, an amount of time which should be ‘more than adequate’ to complete the case; subjects in the moderate time pressure situation were told they would have ‘only’ 11 minutes to complete the case. suppose 25 subjects were selected for each of the six treatment combinations and the sample means for each treatment combination are as follows (standard deviations are in parentheses):
table: time pressure (low, moderate) vs knowledge (naive, declarative, procedural) with means (sds): low-naive: 1.13 (1.13), low-declarative: 1.56 (1.33), low-procedural: 2.00 (1.54); moderate-naive: 0.48 (0.80), moderate-declarative: 1.66 (1.26), moderate-procedural: 2.66 (1.80)
use the anova procedure to test for any significant differences due to time pressure, knowledge, and interaction. use a 0.05 level of significance. assume that the total sum of squares for this experiment is 127.50. letting time pressure be factor a and knowledge be factor b, show the entries in the anova table (to 2 decimals, if necessary). do not round intermediate calculations. if your answer is zero, enter 0.
anova table: source of variation (factor a, factor b, interaction, error, total), sum of squares, degrees of freedom, mean square, f, p-value (in 4 decimals)
the p-value for factor a is - dropdown
what is your conclusion with respect to factor a? - dropdown
the p-value for factor b is - dropdown
what is your conclusion with respect to factor b? - dropdown
the p-value for the interaction of factors a and b is - dropdown
what is your conclusion with respect to the interaction of factors a and b? - dropdown

Explanation:

Step 1: Determine the number of levels and total subjects

Factor A (Time Pressure) has 2 levels, Factor B (Knowledge) has 3 levels. Each treatment combination has 25 subjects, so total subjects \( N = 2\times3\times25 = 150 \).

Step 2: Calculate Degrees of Freedom (df)

- **Factor A (df_A):** \( k - 1 = 2 - 1 = 1 \)
- **Factor B (df_B):** \( l - 1 = 3 - 1 = 2 \)
- **Interaction (df_AB):** \( (k - 1)(l - 1) = 1\times2 = 2 \)
- **Error (df_E):** \( N - kl = 150 - 6 = 144 \)
- **Total (df_T):** \( N - 1 = 149 \)

Step 3: Calculate Sum of Squares (SS)

First, calculate the grand mean and group means. Let's denote the groups:
- Low - Naive: \( n = 25, \bar{x} = 1.13 \)
- Low - Declarative: \( n = 25, \bar{x} = 1.56 \)
- Low - Procedural: \( n = 25, \bar{x} = 2.00 \)
- Moderate - Naive: \( n = 25, \bar{x} = 0.48 \)
- Moderate - Declarative: \( n = 25, \bar{x} = 1.66 \)
- Moderate - Procedural: \( n = 25, \bar{x} = 2.66 \)

**Factor A (Time Pressure) means:**
- Low: \( \bar{x}_L = \frac{25(1.13 + 1.56 + 2.00)}{75} = \frac{25(4.69)}{75} = \frac{117.25}{75} \approx 1.5633 \)
- Moderate: \( \bar{x}_M = \frac{25(0.48 + 1.66 + 2.66)}{75} = \frac{25(4.80)}{75} = \frac{120}{75} = 1.60 \)

**SS_A (Sum of Squares for Factor A):**
\( SS_A = \sum_{i = 1}^{k} n_i (\bar{x}_i - \bar{x}_{grand})^2 \)
First, grand mean \( \bar{x}_{grand} = \frac{75(1.5633) + 75(1.60)}{150} = \frac{117.2475 + 120}{150} = \frac{237.2475}{150} \approx 1.5817 \)
\( SS_A = 75(1.5633 - 1.5817)^2 + 75(1.60 - 1.5817)^2 \)
\( = 75(-0.0184)^2 + 75(0.0183)^2 \)
\( = 75(0.00033856) + 75(0.00033489) \)
\( \approx 0.025392 + 0.02511675 \approx 0.0505 \) (Wait, this seems low. Maybe better to use another approach. Alternatively, use the formula for two-way ANOVA with equal cell sizes: \( SS_A = \frac{n}{l}\sum_{i = 1}^{k} (\bar{x}_{i\cdot} - \bar{x}_{\cdot\cdot})^2 \), where \( n = 25 \), \( l = 3 \), \( k = 2 \))

\( \bar{x}_{1\cdot} = 1.5633 \), \( \bar{x}_{2\cdot} = 1.60 \), \( \bar{x}_{\cdot\cdot} = \frac{1.5633 + 1.60}{2} = 1.58165 \)

\( SS_A = \frac{25}{3}[(1.5633 - 1.58165)^2 + (1.60 - 1.58165)^2] \times 3 \)? Wait, no, equal cell sizes: \( n = 25 \) per cell, so \( n_{i\cdot} = 25 \times 3 = 75 \) per level of A.

So \( SS_A = 75[(1.5633 - 1.5817)^2 + (1.60 - 1.5817)^2] \)
Wait, maybe I made a mistake in group means. Let's recalculate group means for Factor A:

Low group (3 cells, 25 each): total sum for Low = 25(1.13 + 1.56 + 2.00) = 25(4.69) = 117.25

Moderate group: 25(0.48 + 1.66 + 2.66) = 25(4.80) = 120

Grand total = 117.25 + 120 = 237.25

Grand mean \( \bar{x}_{\cdot\cdot} = \frac{237.25}{150} \approx 1.5817 \)

SS_A = sum over A levels of \( n_{i\cdot}(\bar{x}_{i\cdot} - \bar{x}_{\cdot\cdot})^2 \)
\( n_{1\cdot} = 75 \), \( \bar{x}_{1\cdot} = \frac{117.25}{75} \approx 1.5633 \)
\( n_{2\cdot} = 75 \), \( \bar{x}_{2\cdot} = \frac{120}{75} = 1.60 \)

SS_A = 75(1.5633 - 1.5817)^2 + 75(1.60 - 1.5817)^2
= 75(-0.0184)^2 + 75(0.0183)^2
= 75(0.00033856) + 75(0.00033489)
= 0.025392 + 0.02511675 ≈ 0.0505

**Factor B (Knowledge) means:**
- Naive: \( \bar{x}_{\cdot1} = \frac{25(1.13 + 0.48)}{50} = \frac{25(1.61)}{50} = 0.805 \)
- Declarative: \( \bar{x}_{\cdot2} = \frac{25(1.56 + 1.66)}{50} = \frac{25(3.22)}{50} = 1.61 \)
- Procedural: \( \bar{x}_{\cdot3} = \frac{25(2.00 + 2.66)}{50} = \frac{25(4.66)}{50} = 2.33 \)

SS_B = sum over B levels of \( n_{\cdot j}(\bar{x}_{\cdot j} - \bar{x}_{\cdot\cdot})^2 \)
\( n_{\cdot j} = 50 \) (2 levels of A, 25 each)
SS_B = 50(0.805 - 1.5817)^2 + 50(1.61 - 1.5817)^2 + 50(2.33 - 1.5817)^2
= 50(-0.7767)^2 + 50(0.0283)^2 + 50(0.7483)^2
= 50(0.6033) + 50(0.0008) + 50(0.5600)
= 30.165 + 0.04 + 28.0
= 58.205 (approx, maybe more accurate calculation)

Wait, 0.7767² = 0.6033, 0.0283² = 0.0008, 0.7483² = 0.5600

So 50*0.6033 = 30.165, 50*0.0008 = 0.04, 50*0.5600 = 28.0, total SS_B ≈ 30.165 + 0.04 + 28.0 = 58.205

**Interaction (SS_AB):**
SS_AB = sum over cells of \( n_{ij}(\bar{x}_{ij} - \bar{x}_{i\cdot} - \bar{x}_{\cdot j} + \bar{x}_{\cdot\cdot})^2 \)
For each cell:
- Low - Naive: \( 25(1.13 - 1.5633 - 0.805 + 1.5817)^2 = 25(1.13 - 0.7866)^2 = 25(0.3434)^2 ≈ 25(0.1180) = 2.95 \)
- Low - Declarative: \( 25(1.56 - 1.5633 - 1.61 + 1.5817)^2 = 25(1.56 - 1.5916)^2 = 25(-0.0316)^2 ≈ 25(0.0010) = 0.025 \)
- Low - Procedural: \( 25(2.00 - 1.5633 - 2.33 + 1.5817)^2 = 25(2.00 - 2.3116)^2 = 25(-0.3116)^2 ≈ 25(0.0971) = 2.4275 \)
- Moderate - Naive: \( 25(0.48 - 1.60 - 0.805 + 1.5817)^2 = 25(0.48 - 0.8233)^2 = 25(-0.3433)^2 ≈ 25(0.1189) = 2.9725 \)
- Moderate - Declarative: \( 25(1.66 - 1.60 - 1.61 + 1.5817)^2 = 25(1.66 - 1.6283)^2 = 25(0.0317)^2 ≈ 25(0.0010) = 0.025 \)
- Moderate - Procedural: \( 25(2.66 - 1.60 - 2.33 + 1.5817)^2 = 25(2.66 - 2.3483)^2 = 25(0.3117)^2 ≈ 25(0.0972) = 2.43 \)

Sum these up: 2.95 + 0.025 + 2.4275 + 2.9725 + 0.025 + 2.43 ≈ 10.83

**Error (SS_E):**
Total SS = 127.50 (given)
SS_E = Total SS - SS_A - SS_B - SS_AB = 127.50 - 0.0505 - 58.205 - 10.83 ≈ 58.4145

Step 4: Calculate Mean Squares (MS)

- MS_A = SS_A / df_A = 0.0505 / 1 ≈ 0.0505
- MS_B = SS_B / df_B = 58.205 / 2 ≈ 29.1025
- MS_AB = SS_AB / df_AB = 10.83 / 2 ≈ 5.415
- MS_E = SS_E / df_E = 58.4145 / 144 ≈ 0.4057

Step 5: Calculate F - statistics

- F_A = MS_A / MS_E = 0.0505 / 0.4057 ≈ 0.1245
- F_B = MS_B / MS_E = 29.1025 / 0.4057 ≈ 71.73
- F_AB = MS_AB / MS_E = 5.415 / 0.4057 ≈ 13.35

Step 6: Calculate p - values

Using F - distribution with (df1, df2):
- For Factor A: F(1, 144), p - value ≈ 0.724 (using F - table or calculator)
- For Factor B: F(2, 144), p - value ≈ 0.0000 (very small)
- For Interaction: F(2, 144), p - value ≈ 0.0000 (very small)

Now, filling the ANOVA table:

| Source of Variation | Sum of Squares | Degrees of Freedom | Mean Square | F | p - value |
|---------------------|----------------|--------------------|-------------|---|-----------|
| Factor A (Time Pressure) | 0.05 (approx) | 1 | 0.05 | 0.12 | 0.724 |
| Factor B (Knowledge) | 58.21 (approx) | 2 | 29.10 | 71.73 | 0.0000 |
| Interaction (A×B) | 10.83 (approx) | 2 | 5.42 | 13.35 | 0.0000 |
| Error | 58.41 (approx) | 144 | 0.41 | - | - |
| Total | 127.50 | 149 | - | - | - |

For the p - value of Factor A: approximately 0.724 (in 4 decimals: 0.7240)

Answer:

The p - value for Factor A is approximately \(\boxed{0.7240}\) (this is an approximate value based on the calculations above; more precise calculation may be needed with software or calculator).