# Type 3 ANOVA in R (with Examples)

Type 3 ANOVA in R can be calculated using `Anova()` function from CAR. The following shows the steps of doing it in R.

## Step 1: Prepare data in R

The data has two categorical IVs (cities and stores) and one DV (sales).

```x_1 = rep(c('City1','City2'),each=5)
x_2 = rep(c('store1','store2'), 5)
sales=c(10,20,20,50,30,10,5,4,12,4)

df <- data.frame (cities  = x_1,
stores = x_2,
sales=sales)
print(df)```

Output:

```   cities stores sales
1   City1 store1    10
2   City1 store2    20
3   City1 store1    20
4   City1 store2    50
5   City1 store1    30
6   City2 store2    10
7   City2 store1     5
8   City2 store2     4
9   City2 store1    12
10  City2 store2     4```

## Step 2: Use Anova() for Type 3 ANOVA in R

The following is the R code using ANOVA() to calculate Type 3 ANOVA in R. Note that, it states type=3, which suggests it is a Type 3 ANOVA.

```result5<-car::Anova(lm(sales ~ cities*stores, data = df),type=3)
print(result5)```

Output:

```Anova Table (Type III tests)

Response: sales
Sum Sq Df F value  Pr(>F)
(Intercept)   1200.00  1 10.3078 0.01835 *
cities         158.70  1  1.3632 0.28728
stores         270.00  1  2.3193 0.17861
cities:stores  183.75  1  1.5784 0.25569
Residuals      698.50  6
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1```

## Step 3: Interpretation the output

The p-value for the interaction is 0.256, which is greater than 0.05. Thus, the interactin effect of cities*stores is not significant.

Further, we can see that 158.70+270.00+183.75+698.50 = 1310.95. Thus, it is not equal to SST, which is 1878.5 (see here regarding how to calculate SST). Thus, it is consistent with what I discussed in another post, namely SSA | B, AB + SSB | A, AB + SSAB | A, B + SSE ≠ SST for Type 3 ANOVA.

## Step 4 (optional): Type 2 vs. Type 3 ANOVA in R

If you test the same data for Type 2 ANOVA, you will see the following output.

```Anova Table (Type II tests)

Response: sales
Sum Sq Df F value  Pr(>F)
cities        984.15  1  8.4537 0.02707 *
stores         93.75  1  0.8053 0.40408
cities:stores 183.75  1  1.5784 0.25569
Residuals     698.50  6
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1```

If you use You might have question why Type 2 ANOVA dose not have the output for the intercept here, where there is intercept in Type 3 ANOVA in R.

This is due to the fact that Type III ANOVA is equal to the linear regression model below. However, this is not the case for Type I or Type II ANOVA.

`sales = b0 + b1 cities + b2 stores +b3 cities*stores`

To verify that, we can test the same data using linear regression. That is, the following is the linear model we are going to test.

```# Linear regression for the same data
estimated_coefficients <- lm(sales ~ cities*stores, data = df)

summary(estimated_coefficients)```

Output:

```> summary(estimated_coefficients)

Call:
lm(formula = sales ~ cities * stores, data = df)

Residuals:
Min      1Q  Median      3Q     Max
-15.000  -3.125  -1.000   3.875  15.000

Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept)                20.000      6.229   3.211   0.0184 *
citiesCity2               -11.500      9.850  -1.168   0.2873
storesstore2               15.000      9.850   1.523   0.1786
citiesCity2:storesstore2  -17.500     13.929  -1.256   0.2557
---
Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 10.79 on 6 degrees of freedom
Multiple R-squared:  0.6282,	Adjusted R-squared:  0.4422
F-statistic: 3.379 on 3 and 6 DF,  p-value: 0.09539```

We can see that all the p-values in linear regression are consistent with Type 3 ANOVA, but not Type 2 ANOVA.