The problem asks us to perform an ANOVA test to determine if there are statistically significant differences in the average prices of products across three different stores. We are given the average prices for each store: $\bar{x}_1 = \$3.67$, $\bar{x}_2 = \$2.80$, and $\bar{x}_3 = \$5.33$. We are also given the total sum of squares ($SST = 540.80$) and the sum of squares within groups ($SSW = 491.06$). We need to complete the ANOVA table and perform a 4-step hypothesis test. There are 15 products sampled at each store.

1. Problem Description

The problem asks us to perform an ANOVA test to determine if there are statistically significant differences in the average prices of products across three different stores. We are given the average prices for each store: \bar{x}_1 = \3.67$,$\bar{x}_2 = \$2.80$, and \bar{x}_3 = \5.33$. We are also given the total sum of squares ($SST = 540.80$) and the sum of squares within groups ($SSW = 491.06$). We need to complete the ANOVA table and perform a 4-step hypothesis test. There are 15 products sampled at each store.

2. Solution Steps

Step 1: State the null and alternative hypotheses.

$H_0: \mu_1 = \mu_2 = \mu_3$ (The average prices are the same across all stores)

$H_1$: At least one $\mu_i$ is different (The average prices are not all the same)

Step 2: Determine the critical value.

First, we need to calculate the degrees of freedom.

df_between = k - 1, where k is the number of groups (stores)

$df_{between} = 3 - 1 = 2$

df_within = N - k, where N is the total number of observations and k is the number of groups

$N = 15 \times 3 = 45$

$df_{within} = 45 - 3 = 42$

We need to choose a significance level $\alpha$. Let's assume $\alpha = 0.05$.

We look up the F-critical value in an F-distribution table with $df_{between} = 2$ and $df_{within} = 42$ and $\alpha = 0.05$.

$F_{critical} \approx 3.22$ (Using a table or calculator)

Step 3: Calculate the test statistic (F-statistic).

First, we need to calculate the sum of squares between groups (SSB).

$SSB = SST - SSW$

$SSB = 540.80 - 491.06 = 49.74$

Next, we calculate the mean square between groups (MSB) and the mean square within groups (MSW).

$MSB = \frac{SSB}{df_{between}} = \frac{49.74}{2} = 24.87$

$MSW = \frac{SSW}{df_{within}} = \frac{491.06}{42} = 11.6919$

Now we can calculate the F-statistic:

$F = \frac{MSB}{MSW} = \frac{24.87}{11.6919} \approx 2.127$

Step 4: Make a decision.

Since our calculated F-statistic ($2.127$) is less than the critical F-value ($3.22$), we fail to reject the null hypothesis.

ANOVA Table:

| Source of Variation | Sum of Squares (SS) | Degrees of Freedom (df) | Mean Square (MS) | F |

|---|---|---|---|---|

| Between Groups (Stores) | 49.74 | 2 | 24.87 | 2.127 |

| Within Groups (Error) | 491.06 | 42 | 11.6919 | |

| Total | 540.80 | 44 | | |

3. Final Answer

Since the calculated F-statistic (2.127) is less than the critical F-value (3.22), we fail to reject the null hypothesis. Therefore, there is not enough evidence to conclude that there are statistically significant differences in the average prices of products across the three stores at a significance level of 0.

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