Question

In a hospital the doctor wish to decide to hire staff. He knows that he needs to have more ward-boys every weekend on other days, but he is trying to find if the need them constant across the rest of the week. He collects data for the number of counters needed day for two months. Use α= 5%. Here are the data: (Hint: ANNOVA TEST)

Mondays:        296,    243,    318,    276,    297,    329,    332,    285,    331
Tuesdays:       263,    299,    321,    305,    294,    263,    248,    318,    275
Wednesdays: 308,    312,    330,    287,    263,    313,    275,    293     295
Thursdays:      274,    299,    261,    247,    298,    296,    276,    282    

Answers should be in Word Version Format

Answer 1

	Monday	Tuesday	Wednesday	Thursday	Total
Sum	2707	2586	2676	2233	10202
Count	9	9	9	8	35
Mean, sum/n	300.7778	287.3333	297.3333	279.125
Sum of square, Ʃ(xᵢ-x̅)²	7219.556	5450	3450	2440.875

Null and Alternative Hypothesis:

Ho: µ1 = µ2 = µ3 = µ4

H1: At least one mean is different

Number of treatment, k = 4

Total sample Size, N = 35

df(between) = k-1 = 3

df(within) = k(n-1) = 31

df(total) = N-1 = 34

SS(between) = (Sum1)²/n1 + (Sum2)²/n2 + (Sum3)²/n3 + (Sum4)²/n4 - (Grand Sum)²/ N = 2462.312

SS(within) = SS1 + SS2 + SS3 + SS4 = 18560.43

SS(total) = SS(between) + SS(within) = 21022.74

MS(between) = SS(between)/df(between) = 820.7708

MS(within) = SS(within)/df(within) = 598.7236

F = MS(between)/MS(within) = 1.3709

p-value = F.DIST.RT(1.3709, 3, 31) = 0.2699

Critical value Fc = F.INV.RT(0.05, 3, 31) = 2.9113

Decision:

Do not reject the null hypothesis

ANOVA
Source of Variation	SS	df	MS	F	P-value
Between Groups	2462.3123	3	820.7708	1.3709	0.2699
Within Groups	18560.4306	31	598.7236
Total	21022.7429	34