Question

There are three health clinics in Bangkok, Thailand, area. The following data show the number of outpatient surgeries performed on Monday, Tuesday, Wednesday, Thursday, and Friday at each hospital last week. At the 0.01 significance level, can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

	Number of Surgeries Performed
Day	St. Luke's	St. Vincent	Mercy
Monday	45	31	28
Tuesday	26	35	19
Wednesday	21	35	24
Thursday	23	31	37
Friday	25	26	23

1. Set up the null hypothesis and the alternate hypothesis.

For Treatment:

Null hypothesis

• H₀: µ_{St. Luke's} = µ_{St. Vincent} = µ_Mercy

• H₀: µ_{St. Luke's} ≠ µ_{St. Vincent} ≠ µ_Mercy

2. Alternate hypothesis

• H₁: Not all means are equal.

• H₁: All means are equal.

3. For blocks:

Null hypothesis

1. A. H₀: µ_Mon = µ_Tue = µ_Wed = µ_Thu = µ_Fri

2. B. H₀: µ_Mon ≠ µ_Tue ≠ µ_Wed ≠ µ_Thu ≠ µ_Fri

• Choice A

• Choice B

4. Alternate hypothesis

• H₁: All means are equal.

• H₁: Not all means are equal.

5. State the decision rule for 0.01 significance level. (Round your answers to 2 decimal places.)

For Treatment:

For blocks:

6. Complete the ANOVA table. (Round your SS, MS and F to 2 decimal places.)

7. What is your decision regarding the null hypothesis?

The decision for the F value (Treatment) at 0.01 significance is:

• Reject H₀

• Do not reject H₀

8. The decision for the F value (Block) at 0.01 significance is:

• Reject H₀

• Do not reject H₀

9. Can we conclude there is a difference in the mean number of surgeries performed by hospital or by day of the week?

Answer 1

For Treatment:

Null hypothesis

H₀: µ_{St. Luke's} = µ_{St. Vincent} = µ_Mercy

Alternate hypothesis

H₁: Not all means are equal.

--

For blocks:

Null hypothesis

A. H₀: µ_Mon = µ_Tue = µ_Wed = µ_Thu = µ_Fri

Alternate hypothesis

H₁: Not all means are equal.

--

Decision rule for 0.01 significance level:

For Treatment: Critical value , Fc = F.INV(0.01, 2, 8) = 8.65

For blocks: Critical value , Fc = F.INV(0.01, 4, 8) = 7.01

--

ANOVA table:

Anova: Two-Factor Without Replication
SUMMARY	Count	Sum	Average	Variance
Monday	3	104	34.66667	82.33333
Tuesday	3	80	26.66667	64.33333
Wednesday	3	80	26.66667	54.33333
Thursday	3	91	30.33333	49.33333
Friday	3	74	24.66667	2.333333
St. Luke's	5	140	28	94
St. Vincent	5	158	31.6	13.8
Mercy	5	131	26.2	46.7
ANOVA
Source of Variation	SS	df	MS	F	P-value
Blocks	188.27	4	47.07	0.88	0.518695
Treatment	75.6	2	37.8	0.70	0.522979
Error	429.73	8	53.72
Total	693.6	14

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The decision for the F value (Treatment) at 0.01 significance is: Do not reject H₀

--

The decision for the F value (Block) at 0.01 significance is: Do not reject H₀