Question

Big fish: A sample of

108

one-year-old spotted flounder had a mean length of

120.25

millimeters with a sample standard deviation of

19.97

millimeters, and a sample of

122

two-year-old spotted flounder had a mean length of

136.08

millimeters with a sample standard deviation of

28.52

millimeters. Construct a

90

%

confidence interval for the mean length difference between two-year-old flounder and one-year-old flounder. Let

μ1

denote the mean length of two-year-old flounder and round the answers to at least two decimal places.

A 90 % confidence interval for the mean length difference, in millimeters, between two-year-old flounder and one-year-old flounder is <<−μ1μ2 .

Answer 1

since sample size is larger than 40 in both samples; we can use large sample z interval:

		Pop 1	Pop 2
sample mean x =		136.08	120.25
std deviation σ=		28.520	19.970
sample size n=		122	108
std error σx1-x2=√(σ21/n1+σ22/n2)    =			3.219

Point estimate of difference '=x1-x2 =			15.830
for 90 % CI value of z=			1.645	from excel:normsinv((1+0.9)/2)
margin of error E=z*std error =			5.295
lower bound=(x1-x2)-E =			10.54
Upper bound=(x1-x2)+E =			21.12
from above 90% confidence interval for population mean =(10.54 <µ1-µ2< 21.12)

(please try (10.49 ,21.17) if z distribution to be used)