Question

Race	Weight	Systolic BP		1. Construct confidence intervals for the following situations. You will be prompted to indicate the: critical values, margin of error, confidence limits, and an interpretation.    (18 points)
Hispanic	117	137
Caucasian	107	127		a. Determine the proportion (percentage) of Asians for your point estimate. Then construct a 90% confidence interval for the true proportion of all Asian patients.
Caucasian	101	119
Hispanic	123	143		Sample Proportion
Hispanic	121	141		Critical Value
Caucasian	157	174		Margin of Error
Hispanic	168	188		Lower Limit
Asian	160	180		Upper Limit
African American	156	176		Interpretation
Caucasian	149	175
Caucasian	140	160
Asian	113	133		b. Determine the average weight of all 100 patients for your point estimate. Then construct a 95% confidence interval for the true mean weight of all patients.
Caucasian	133	153
Caucasian	126	152		Sample Mean
Hispanic	151	171		Standard Deviation
Caucasian	148	168		Critical Value
Asian	163	183		Margin of Error
Asian	128	140		Lower Limit
Caucasian	125	145		Upper Limit
Hispanic	162	182		Interpretation
Caucasian	133	153
African American	133	150

Answer 1

a)

x=   4   COUNTIF(A3:A24,"Asian")
n=   22   COUNTA(A3:A24)

Sample Proportion	x/n =	=4/22	0.181818
Critical Value	z(a/2)	=NORMSINV(1-0.1/2)	1.644854
Margin of Error	z(a/2)*sqrt(p*(1-p)/n) =	=1.645*SQRT(0.1818*(1-0.1818)/22) =	0.135264
Lower Limit	p - z(a/2)*sqrt(p*(1-p)/n)	=0.1818-1.645*SQRT(0.1818*(1-0.1818)/22) =	0.046536
Upper Limit	p + z(a/2)*sqrt(p*(1-p)/n)	=0.1818+1.645*SQRT(0.1818*(1-0.1818)/22) =	0.317064
Interpretation	I am 90% confident that estimated proportion of all Asian patients lie in the interval (0.046, 0.317)

b)

Sample Mean	137	AVERAGE(B3:B24)
Standard Deviation	19.56674	STDEV(B3:B24)
Critical Value	1.959964	NORMSINV(1-0.05/2) = z(A/2)
Margin of Error	7.758547	1.96*18.56674/SQRT(22) = z(A/2)*sd/sqrt(n)
Lower Limit	129.2415	137-7.758547 = mean-ME
Upper Limit	144.7585	137+7.758547 = mean + ME
Interpretation	I am 95% confident that estimated true mean weight of all patients lie in the interval (129.2415, 144.7585)