I so a production supervisor for the LTD company. His name is Melvin and he was just to compare production levels at the companies for plants. The weekly levels tons are collected over seven week. Production tons for the four locations are the following location a is 51.3 location B is 38.3 location C is 47.1 and location is 40.2. I have to use Alpha 0.05 number for Table For between sample treatment is 761.4 and within sample error is 1514.7 so I need to figure out the degree of freedom forever MSE and then I have to determine if there’s a difference between the production levels at the facilities because the calculated value is 4.02 so it’s true

To find the degrees of freedom (df) for the between-sample treatment (df_between) and within-sample error (df_within), we use the formula:

df_{between} = k - 1
df_{within} = N - k

where:
- k is the number of treatment groups (locations)
- N is the total number of observations (weeks)

Given k = 4 treatment groups (locations) and N = 7 weeks, we can calculate the degrees of freedom:

df_{between} = 4 - 1 = 3
df_{within} = 7 - 4 = 3

Next, we need to determine the mean square (MSE) for both between-sample treatment and within-sample error:

MSE_{between} = \frac{SS_{between}}{df_{between}} = \frac{761.4}{3} = 253.8
MSE_{within} = \frac{SS_{within}}{df_{within}} = \frac{1514.7}{3} = 504.9

To calculate the F-statistic, we use the formula:

F = \frac{MSE_{between}}{MSE_{within}} = \frac{253.8}{504.9} \approx 0.5026

Given that the calculated F-statistic is 4.02, we need to compare it with the critical F-value from the F-table with 3 and 3 degrees of freedom for between and within samples, respectively, at a significance level of 0.05.

Since the calculated F-statistic 4.02 is greater than the critical F-value (0.5026 > F_critical), we reject the null hypothesis. This means that there is a significant difference between the production levels at the facilities.

\boxed{Answer: \text{Significant difference in production levels at the facilities.}}