Question

	Race
	White	All others	Black
Neonatal Mortalities	12164	6564	5920
	75.10%	12.600%	12.30%

You are interested in determining if neonatal mortality is equally likely across races in the United States. You know that at the time of analysis, the United States was composed of 75.1% white individuals, 12.3% black individuals, and 12.6% of other races and you can presume that births match these proportions. If neonatal mortality is not equally likely across all races, then determine which categories differ from other categories. Number of neonatal mortalities in 1999 by race are provided in the table above

*I will have to solve this using Excel. Specifically what should I do? What functions could I use? For the question, I have to

1. A null and alternative hypothesis stated, as appropriate and for each hypothesis tested. may involve several hypothesis tests.

2. Choose the most appropriate test. Explain how you have met the assumptions of the test or why the test is robust to violations of the assumptions.

3. State explicitly what test(s) you are using.

4. If you fail to reject the null hypothesis, calculate the power of the test.

Answer 1

1)

Null hypothesis : p1 = p2 = p3

Alternate hypothesis : all proportions are not equal

2)

We can test this by chi-square test of independence

For getting actual frequency

formulas

All expected frequencies are greater than 5, hence assumptions are satisfied

3)

Now we can conduct hypothesis

You can just copy paste below

formulas

		c1	c2	c3	sum
	r1	9135.164	827.064	728.16	=(C2+D2+E2)
	r2	3028.836	5736.936	5191.84	=(C3+D3+E3)
	sum	=(C2+C3)	=(D2+D3)	=(E2+E3)	=(F2+F3)
	Eij	1	2	3
expected	1	=(F2*C5)/F5	=(F2*D5)/F5	=(F2*E5)/F5
	2	=(F3*C5)/F5	=(F3*D5)/F5	=(F3*E5)/F5
	Oi	=C2	=D2	=E2	=C3	=D3	=E3
	Ei	=(C5*F2)/F5	=(D5*F2)/F5	=(E5*F2)/F5	=(C5*F3)/F5	=(D5*F3)/F5	=(E5*F3)/F5
								sum
	(Oi-Ei)^2/Ei	=(C10-C11)^2/C11	=(D10-D11)^2/D11	=(E10-E11)^2/E11	=(F10-F11)^2/F11	=(G10-G11)^2/G11	=(H10-H11)^2/H11	=SUM(C13:H13)
							alpha	0.05
							critical value	=CHISQ.INV(1-I15,2)
							p-value	=CHISQ.DIST.RT(I13,2)

4)

p-value = 0.000 < alpha (0.05)

hence we reject the null hypothesis