The problem is to perform an ANOVA test to determine if there are significant differences in the average ratings of three types of candy: hard candy, chewable candy, and chocolate. The average ratings are given as $\bar{X}_1 = 3.13$, $\bar{X}_2 = 4.20$, and $\bar{X}_3 = 4.40$, respectively. Each type of candy was rated by 30 people. We are given the sum of squares between groups ($SS_B = 27.82$) and the sum of squares within groups ($SS_W = 1091.30$). We need to calculate the F-statistic and compare it to a critical value (which we cannot do since the significance level, $\alpha$, is not given) to determine if there are significant differences in the average candy ratings.

1. Problem Description

The problem is to perform an ANOVA test to determine if there are significant differences in the average ratings of three types of candy: hard candy, chewable candy, and chocolate. The average ratings are given as $\bar{X}_1 = 3.13$, $\bar{X}_2 = 4.20$, and $\bar{X}_3 = 4.40$, respectively. Each type of candy was rated by 30 people. We are given the sum of squares between groups ($SS_B = 27.82$) and the sum of squares within groups ($SS_W = 1091.30$). We need to calculate the F-statistic and compare it to a critical value (which we cannot do since the significance level, $\alpha$, is not given) to determine if there are significant differences in the average candy ratings.

2. Solution Steps

First, we need to calculate the degrees of freedom between groups ($df_B$) and within groups ($df_W$).

$df_B = k - 1$, where $k$ is the number of groups. In this case, $k = 3$ (hard candy, chewable candy, chocolate).

$df_B = 3 - 1 = 2$

$df_W = N - k$, where $N$ is the total number of observations. In this case, each candy type was rated by 30 people, so $N = 30 \times 3 = 90$.

$df_W = 90 - 3 = 87$

Next, we calculate the Mean Square Between ($MS_B$) and Mean Square Within ($MS_W$).

$MS_B = \frac{SS_B}{df_B} = \frac{27.82}{2} = 13.91$

$MS_W = \frac{SS_W}{df_W} = \frac{1091.30}{87} = 12.54367816$

Now, we calculate the F-statistic.

$F = \frac{MS_B}{MS_W} = \frac{13.91}{12.54367816} = 1.1089$

3. Final Answer

The F-statistic is 1.

1

0

8

9. Since we don't have a significance level to compare with a critical value, we can only report the test statistic. We can conclude there is no evidence of a significant difference in the average ratings of the three candy types. However, this conclusion is limited by the lack of a defined significance level to determine what a significant difference is.