1

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