sampling distribution of the sample mean
DESCRIPTION
Sampling Distribution of the Sample Mean. Example. Let X denote the lifetime of a battery Suppose the distribution of battery battery lifetimes has mean lifetime, = 400 hours standard deviation of lifetimes, s = 40 hours. A Sample of Size n is Taken. - PowerPoint PPT PresentationTRANSCRIPT
Example
Let X denote the lifetime of aa battery
Suppose the distribution of battery battery lifetimes has
– mean lifetime, = 400 = 400 hours– standard deviation of lifetimes, 4040
hours
A Sample of Size nn is Taken• Calculate the mean of these nn batteries
• Then another sample of size n batteries is taken– The mean of this second sample of nn batteries is
calculated
• This is done again, and again, and again, and ….
.σX
μX
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:if normal is batteries) n of life (average X•X (Life of aa battery) is normal, and•σ is known
Central Limit Theorem
Even if X does not have a normal distribution,Even if X does not have a normal distribution,
will be approximately normal if n is large.will be approximately normal if n is large.X
n = 30 is usually large enough to use this approximation.
Example When Distribution of X is Normal
X = the life of aa battery
• Assume battery life is:– Distributed normal – Mean battery life = 400 hours– Standard deviation of battery life = 40 hours
• We choose a battery at random– The battery lasts 350 hours
• This is an observation of X
The Random Variable
• Suppose random samples of size n = 4 batteries are selected independently and their sample means calculated
• These are observations of the random variable
X
X
AN OBSERVATION OF
• Suppose 4 batteries are selected and their lives are: 420, 450, 380, 390– Their average = (420 + 450 + 380 + 390)/4 = 410
• This is an observation of a random variable for the Sample Mean
X
X
AN OBSERVATION OF
• Suppose 4 batteries are selected and their lives are: (410, 450, 360, 360)– Their average = (410 + 450 + 360 + 360)/4 = 395
• This is an observation of a random variable for the Sample Mean
• But the sample size is small so we do not know the distribution of -- we can’t plot it
X
X
X
Using A Larger Sample Size
• Suppose 100 batteries are selected and their lives are: (420, 450, 380, 350,…., 415)– Their average = (420 + … + 415)/100 = 408
• This is an observation of the random variable (where n = 100)
• Because this is a large sample, the distribution of is approximately normal
X
X
EXAMPLES
• Answer the following questions assuming:– Battery life is distributed normal– Battery life distribution is not normal (or
unknown)
• What is the probability:– a random battery will last longer than 408 hours?– the average of life of 16 batteries be longer than
408 hours?– the average life of 100 batteries will be longer than
408 hours?
_ P(X >408) X Normal, n = 16
10
16
40 σX
__XX400400
knownσ normal, is X
because normal is X
4084080 Z0 Z.80.80
.7881.7881
1 - .7881 1 - .7881 = .2119= .2119
_ P(X >408) X Not Normal, n = 16
• Can’t do• X is Not Normal and n is small
– So we do not know the distribution of the Sample Mean
X
4100
40 σX
__XX400400
_ P(X >408) X Normal, n = 100
knownσ normal, is X
because normal is X
0 Z0 Z2.002.00408408
.9772.9772
1 - .9772 = .0228 .0228
4100
40 σX
__XX400400
_ P(X >408) X Not Normal, n = 100
0 Z0 Z2.002.00408408
.9772.9772
1 - .9772 = .0228 .0228
Theorem)Limit (Central large isn
because normalely approximat is X
Using Excel• As long as it can be assumed that the distribution of
the sample mean is normal, NORMDIST and NORMINV can be used to give probabilities except:– Instead of using , put in /n
• Let Excel do the arithmetic
– Example: Find the probability the average of 100 batteries exceeds 408 hours
nσ/