using excel A pharmaceutical company has developed a drug that is expected to reduce hunger. To test the drug, three samplesof rats are selected with n = 10 in each sample. The first sample receives the drug every day. The second sample isgiven the drug once a week, and the third sample receives no drug at all (the control group). The dependentvariables is the amount of food eaten by each rat over a 1-month period. These data are analyzed by an ANOVA, andthe results are reported in the following summary table. Fill in all missing values in the table. (Hint: Start with the dfcolumn.)S.S.d.f.M.S.Between3.58Within3.56TOTALUse the =FDIST() function in Excel to locate the p-value for this ANOVA:p-value =Report p-value accurate to at least 6 decimal places.If you use a significance level of a = .05, what would you conclude about these treatments?OThese data do not provide evidence of a difference between the treatmentsOThere is a significant difference between treatments

Name: using excel A pharmaceutical company has developed a drug that is expe
Uploaded: 2024-03-20T07:07:26.000Z
Channel: Bartleby
Description: using excel A pharmaceutical company has developed a drug that is expected to reduce hunger. To test the drug, three samplesof rats are selected with n = 10 in each sample. The first sample receives the drug every day. The second sample isgiven the d

Total observations = 10×3 = 30 d.f. for Between Treatments = k - 1 d.f. for Between Treatments = 3 - 1 d.f. for Between Treatments = 2 d.f. for Within Treatments = n - k d.f. for Within Treatments = 30-3 d.f. for Within Treatments = 27 Total degrees of freeddom = 30-1 Total degrees of freeddom = 29 F = M.S. BetweenM.S. Within 3.58 = M.S. Between3.56 M.S. Between = 12.7448M.S. Between = S.S. Betweend.f. Between 12.7448 = S.S. Between2 S.S. Between = 25.4896 M.S. Within = S.S. Withind.f. Within S.S. Within = 3.56 × 27 S.S. Within = 96.12 S.S. Total = S.S. Between + S.S. Within S.S. Total = 25.4896 + 96.12 S.S. Total = 121.6096 For P-value using Excel : =FDISTx, deg_freedom1, deg_freedom2 =FDIST3.58, 2, 27 P-value = 0.041775 Significance level, α = 0.05 Here, P-value = 0.041775 < significance level = 0.05. reject the null hypothesis. B. There is a significant difference between treatments.