Question

Question 17 (1 point) Let's say that i take our class to be a SRS and gather data on everyone's weight. Taking that sample we see that our sample mean is 68 kilograms. Assume that we magically know that the true population standard deviation is 10 kilograms. Based on our data, gathered from 40 students, can you give me an estimate on the true population mean weight? Give me a 95% Confidence interval O (64.9,71.1) (66.4, 69.92) (62, 74.2) (58.1, 82.7)

Answer 1

17)

sample mean 'x̄=		68.000
sample size    n=		40
std deviation σ=		10.00
std error ='σx=σ/√n=10/√40=		1.581

for 95 % CI value of z=		1.960	from excel:normsinv((1+0.95)/2)
margin of error E=z*std error =		3.0990
lower bound=sample mean-E=		64.90
Upper bound=sample mean+E=		71.10
from above 95% confidence interval for population mean =(64.9,71.1)

18)

Ο Η :μ = 22 HA:μ > 22

22)

	1st group	2nd group
x=	47	62
n =	200	250
p̂=x/n=	0.2350	0.2480
estimated prop. diff =p̂1-p̂2    =		-0.0130
pooled prop p̂ =(x1+x2)/(n1+n2)=		0.2422
std error Se=√(p̂1*(1-p̂1)*(1/n1+1/n2) =		0.0406
test stat z=(p̂1-p̂2)/Se =		-0.320
P value   =	0.7490	(from excel:2*normsdist(-0.32)

since p value >0.10

The evidence suggests that there's no difference in the proportion of people with blue eyes between the 2 states.