# AP Statistics Curriculum 2007 Bayesian Normal

(Difference between revisions)
 Revision as of 01:56, 26 May 2009 (view source)JayZzz (Talk | contribs) (New page: ''Normal Example:'' It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, wi...)← Older edit Current revision as of 21:09, 28 June 2010 (view source)Jenny (Talk | contribs) (→Motivational example) (11 intermediate revisions not shown) Line 1: Line 1: - ''Normal Example:'' + ==[[EBook | Probability and Statistics Ebook]] - Normal Example== - It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, with a standard deviation of 10 mph.  Knowing that his car is capable of much higher speeds, the owner took the car to the shop. After a checkup, the speedometer was replaced with a new one, which gave a new estimate of 220 mph with a standard deviation of 4 mph. The errors are assumed to be normally distributed. + ===Motivational example=== - We can say that the owner S’s prior beliefs about the top speed of his car were represented by: + It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, with a standard deviation of 10 mph.  Knowing that his car is capable of much higher speeds, the owner take the car to the shop. After a checkup, the speedometer is replaced with a better one, which gives a new estimate of 220 mph with a standard deviation of 4 mph. The errors are assumed to be normally distributed. -
'''µ ~ N($\mu_0$, $\phi_0$) = µ ~ N(185,$10^2$)'''
+ We can say that the owner '''S’s''' prior beliefs about the top speed of his car are represented by: + + : $\mu \sim N \mu_0$, $\phi_0$, i.e., $\mu \sim N(185,10^2) We could then say that the measurements using the new speedometer result in a measurement of: We could then say that the measurements using the new speedometer result in a measurement of: - ''' x ~ N([itex]\mu$, $\phi$) = x ~ N(µ,$4^2)''' + :x \sim N(\mu, \phi)$, i.e., $x \sim N(\mu, 4^2)$ - We note that the observation '''x''' turned out to be 210, and we see that S’s posterior beliefs about '''µ''' should be represented by: + We note that the observation '''x''' turned out to be 210, and we see that '''S’s''' posterior beliefs about '''µ''' should be represented by: -
'''µ | x ~ N($\mu_1$, $\phi_1 )''' + :\mu | x \sim N(\mu_1, \phi_1)$ where (rounded) where (rounded) -
'''$\phi_1$ = $(10^{-2} + 4^{-2})^{-1}$ = 14 = $4^2 ''' + :[itex]\phi_1 = (10^{-2} + 4^{-2})^{-1} = 14 = 4^2$ -
'''$\mu_1$ = $14(185/10^2 + 220/4^2) = 218$'''
+ :$\mu_1 = 14(185/10^2 + 220/4^2) = 218$. Therefore, the posterior for the top speed is: Therefore, the posterior for the top speed is: -
'''$\mu$ | x ~ N($218,4^2$)'''
+ :$\mu | x \sim N(218,4^2)$. Meaning 218 +/- 4 mph. - Meaning 218 +/- 4 mph. + If the new speedometer measurements were considered by another person '''S’''' who had no knowledge of the readings from the first speedometer, but still had a vague idea (from knowledge of the stock speedometer) that the top speed was about 200 +/- 30 mph. Then: - If the new speedometer measurements were considered by another person S’ who had no knowledge of the readings from the first speedometer, but still had a vague idea (from knowledge of the stock speedometer) that the top speed was about 200 +/- 30 mph, + :$\mu \sim N(200,30^2)$. - Then: + - + -
'''$\mu$ ~ N($200,30^2$)'''
+ Then '''S’''' would have a posterior variance: Then '''S’''' would have a posterior variance: -
'''$\phi_1 = (30^{-2} + 4^{-2})^{-1} = 16 = 4^2$'''
+ : $\phi_1 = (30^{-2} + 4^{-2})^{-1} = 16 = 4^2$. '''S’''' would have a posterior mean of: '''S’''' would have a posterior mean of: + + :$\mu_1 = 16(200/30^2 + 220/4^2) = 224$. + + Therefore, the distribution of '''S’''' would be: + + :$\mu | x \sim N(224,4^2)$. Meaning 224 +/- 4 mph. + + This calculation has been carried out assuming that the prior information we have is rather vague, and therefore the posterior is almost entirely determined by the data. + + The situation is summarized as follows: +
+ {| class="wikitable" style="text-align:center; width:25%" border="1" + |- + ! Speed || Prior Distribution || Likelihood from Data || Posterior Distribution + |- + | S || $N(185 , 10^2)$  || || $N(218 , 4^2)$ + |- + |- + | || || $N(220 , 4^2)$ || + |- + | S’ || $N(200,30^2)$|| || $N(224,4^2)$ + |} +
+ + ==See also== + * [[EBook#Chapter_III:_Probability |Probability Chapter]] + + ==References== + +
+ * SOCR Home page: http://www.socr.ucla.edu + + {{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=AP_Statistics_Curriculum_2007_Bayesian_Normal}}

## Probability and Statistics Ebook - Normal Example

### Motivational example

It is known that the speedometer that comes with a certain new sports car is not very accurate, which results in an estimate of the top speed of the car of 185 mph, with a standard deviation of 10 mph. Knowing that his car is capable of much higher speeds, the owner take the car to the shop. After a checkup, the speedometer is replaced with a better one, which gives a new estimate of 220 mph with a standard deviation of 4 mph. The errors are assumed to be normally distributed.

We can say that the owner S’s prior beliefs about the top speed of his car are represented by:

$\mu \sim N \mu_0$, φ0, i.e., $\mu \sim N(185,10^2)$

We could then say that the measurements using the new speedometer result in a measurement of:

$x \sim N(\mu, \phi)$, i.e., $x \sim N(\mu, 4^2)$

We note that the observation x turned out to be 210, and we see that S’s posterior beliefs about µ should be represented by:

$\mu | x \sim N(\mu_1, \phi_1)$

where (rounded)

φ1 = (10 − 2 + 4 − 2) − 1 = 14 = 42
μ1 = 14(185 / 102 + 220 / 42) = 218.

Therefore, the posterior for the top speed is:

$\mu | x \sim N(218,4^2)$. Meaning 218 +/- 4 mph.

If the new speedometer measurements were considered by another person S’ who had no knowledge of the readings from the first speedometer, but still had a vague idea (from knowledge of the stock speedometer) that the top speed was about 200 +/- 30 mph. Then:

$\mu \sim N(200,30^2)$.

Then S’ would have a posterior variance:

φ1 = (30 − 2 + 4 − 2) − 1 = 16 = 42.

S’ would have a posterior mean of:

μ1 = 16(200 / 302 + 220 / 42) = 224.

Therefore, the distribution of S’ would be:

$\mu | x \sim N(224,4^2)$. Meaning 224 +/- 4 mph.

This calculation has been carried out assuming that the prior information we have is rather vague, and therefore the posterior is almost entirely determined by the data.

The situation is summarized as follows:

Speed Prior Distribution Likelihood from Data Posterior Distribution
S N(185,102) N(218,42)
N(220,42)
S’ N(200,302) N(224,42)