Sampling Distribution of the Sample Mean. Example a Let X denote the lifetime of a battery Suppose the distribution of battery battery lifetimes has 

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Presentation transcript:

Sampling Distribution of the Sample Mean

Example a Let X denote the lifetime of a battery Suppose the distribution of battery battery lifetimes has  = 400 –mean lifetime,  = 400 hours  40 –standard deviation of lifetimes,  40 hours

n A Sample of Size n is Taken nCalculate the mean of these n batteries Then another sample of size n batteries is taken n –The mean of this second sample of n batteries is calculated This is done again, and again, and again, and ….

Distribution of aX (Life of a battery) is normal, and σ is known

Central Limit Theorem 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. n = 30 is usually large enough to use this approximation.

Example When Distribution of X is Normal a X = the life of a 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

Observation of X X  = 40 μ = 400 x = 350

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

AN OBSERVATION OF Suppose 4 batteries are selected and their lives are: 420, 450, 380, 390 –Their average = ( )/4 = 410 This is an observation of a random variable for the Sample Mean

DISTRIBUTION OF WHEN X IS NORMAL

_X_X_X_X Sampling Distribution and an Observed Value

Observation of X When X is Not Normal  = 400  = 40 x = 350 f(x) X

AN OBSERVATION OF Suppose 4 batteries are selected and their lives are: (410, 450, 360, 360) –Their average = ( )/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

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_X_X Sampling Distribution and an Observed Value

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  = X 0 Z =.4207

P(X >408) X Not Normal  = 400  = 40 x = 408 X ?

_ P(X >408) X Normal, n = 16 _X_X_X_X Z =.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_X_X_X 400 _ P(X >408) X Normal, n = Z =.0228

_X_X_X_X 400 _ P(X >408) X Not Normal, n = Z =.0228

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

Review Given a distribution X, with  and  known -- for samples of size n: If X is normal and  is known If X is not normal