Question

An obstetrician wanted to determine whether the mean number of births was the same for each of the five days of the week. She randomly selected eight days for each of the five weekdays and recorded the number of births on that day in the data table below. Refer to the ANOVA results from Excel.

Monday	Tuesday	Wednesday	Thursday	Friday
10023 10265 10283 10456 10691 11189 11198 11465	11045 11621 11753 11944 12509 12577 12927 13521	11346 11084 11593 11570 11216 12193 11875 11818	12023 11171 11903 11745 11233 12132 12433 12543	11749 11545 11627 12321 11624 12422 12192 12543

SUMMARY

Groups	Count	Sum	Average	Variance
Monday	8	85570	10696.25	279042.5
Tuesday	8	97897	12237.13	633537.8
Wednesday	8	92695	11586.88	136085.3
Thursday	8	95183	11897.88	252402.7
Friday	8	96023	12002.88	166186.1

ANOVA

Source Of Variation	SS	df	MS	F	P-Value	F Crit
Between Groups	11507633	4	2876908	9.803713	1.99E-05	2.641465
Within Groups	10270781	35	293450.9
Total	21778414	39

• a) Test the hypothesis that the mean number of births for each weekday is the same at the ? = 0.01 level of significance. Use the p-value method.

Please show thorough steps for hypothesis test I am having trouble understanding this.

Answer 1

H0: Null Hypothesis: ( the mean number of births was the same for each of the five days of the week) (Claim)

HA: Alternative Hypothesis: (At least one mean is different from other 4 means) ( the mean number of births was not the same for each of the five days of the week)

From the given data, the following Table is calculated:

	Monday	Tuesday	Wednesday	Thursday	Friday	Total
N	8	8	8	8	8	40
	85570	97897	92695	95183	96023	467368
Mean	85570/8=10696.25	97897/8=12237.125	92695/8=11586.875	95183/8=11897.875	96023/8=12002.875	467368/40=11684.2
Std. Dev.	528.2447	795.9509	368.8974	502.397	407.6593	747.2756

From the above Table, ANOVA Table is calculated as follows:

Source of Variation	Sum of Squares	Degrees of Fredom	Mean Square	F	P - Value	F Crit
Between groups	11507633.4	4	11507633.4/4 = 2876908.35	2876908.35/ 293450.8857 = 9.8037	0.00002	2.641465
Within groups	10270781	35	10270781/ 35 = 293450.8857
Total	21778414.4	39

F Ratio Value = 2876908.35/ 293450.8857 = 9.8037

Degtrees of Freedom for Numerator =4

Degrees of Freedom for denominaor = 35

By Technology, p - value = 0.00002

Since p - value = 0.00002 is less than = 0.01, the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the clam that the mean number of births was the same for each of the five days of the week.

= 0.01

Degtrees of Freedom for Numerator =4

Degrees of Freedom for denominaor = 35

By Technology, critical value of F = 2.641465.

Since calculated value of F = 9.8037 is greater than critical value of F = 2.641465., the difference is significant. Reject null hypothesis.

Conclusion:
The data do not support the clam that the mean number of births was the same for each of the five days of the week.