o€¦ · a confidence interval is a range of numbers believed to include an unknown population...
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STAT 400 UIUC
7.1 Confidence Intervals for Means 7.4 Sample Size Planning
Stepanov Dalpiaz Nguyen
Example 1: Suppose the lifetime of a particular brand of light bulbs is normally distributed with standard deviation of s = 75 hours and unknown mean. a) What is the probability that in a random sample of n = 49 bulbs, the average lifetime is within 21 hours of the overall average lifetime? b) Suppose the sample average lifetime of n = 49 bulbs is = 843 hours. Construct a 95% confidence interval for the overall average lifetime for light bulbs of this brand. A confidence interval is a range of numbers believed to include an unknown population parameter. Associated with the interval is a measure of the confidence we have that the interval does indeed contain the parameter of interest. A (1 - a) 100% confidence interval
for the population mean µ
when s is known and sampling is done from a normal
population, or with a large sample, is
X
x
÷÷ø
öççè
æ×a+×a-ssnznz , 2
2
XX
ypop . standard
dev.
→ to pop. mean ,sample mean
=
=← R .V.- X , . . . . . Xn
'
'id N ( µ , 6=75)My n N ( µ , Ern )
I ( pi -al CI i put 21 )= I (KYLIE f z s (M%M_)
= I f - 1.96 f Z S 1.96) = 0.950
£an realization of the R.
V.
( value)
( a, b) such that 95% confident that pi is in Ca, b)
( I - 21 , I t 21 ) = 1822,864€
959. ↳ t I (a b)is either O f or I
a constant
o.oSt
.dev . of I
2
±
± e
estimate (point estimate)
margin of error
e
e =
Example 1: (continued)
b) Suppose the sample average lifetime of n = 49 bulbs is = 843 hours. Construct a 95% confidence interval for the overall average lifetime for light bulbs of this brand. c) Construct a 90% confidence interval for the overall average lifetime for light bulbs. d) Construct a 92% confidence interval for the overall average lifetime for light bulbs.
X nz
s×a
2 X
nz
s×a
2
x
µanestimate for µ
Z- score . stdev. ofI
0
006=75n = 49 Normal pop .
A 95% confidence interval for µ is :
( I - ¥q% ,I t Zaz %) = ( 843 - 1.96 . ,
843+1.96 -7¥)2=0.
I (zs*)2-
5=0.025 =(822,8⑦I off Zo
.025=1.91
A 90% confidence interval for pe is :
( I -Jaz Fn ,I +ZazG) = 1825375860.625€
u
i- 0.05I = =0.05=0.952-0.05 = 1.645←
A 92% CI for pi :
( I - Zfz¥ ' It Zaz%) = 1824.2586175€2¥ 0¥ = 0.04
2-0.04
= 1.75
3
Minimum required sample size in estimating the population mean µ to within e
with ( 1 – a ) 100 % confidence is
.
Always round n up.
Example 2: How many test runs of an automobile are required for determining its average miles-per-gallon rating on the highway to within 0.5 miles per gallon with 95% confidence, if a guess is that the variance of the population of miles per gallon is about 6.25? Example 1: (continued) e) What is the minimum sample size required if we wish to estimate the overall average lifetime for light bulbs to within 10 hours with 90% confidence?
2
εσz 2α
núúû
ù
êêë
é ×=
I
margin
of error
( MOE)
E ⇐ 0.5 mpg 959. confi ⇒ 2=0.05 ⇒ Z±= 1.962
62=6.25 ⇒ 6=2.5
n=fZ%g 12=[1.96%252] ! 96.04⇒ n --⑤
E = 10 6=75
90% CL ⇒ 2=0. I ⇒ Zzz = 1.645
n = [ ZEe) ! [ (1-645,1175)-1 ! 152.2139
⇒ n=l5⑦