What is the sampling distribution of the mean percent x_bar that the sample spends on housing if null is true?

The Census bureau repots that households spend an average of 31% of their total spendings on housing. A homebuilders association in Cleveland interviews an SRS of 40 households in Cleveland metropolitan area to learn what percent of their spending goes toward housing. suppose we know that spending on housing in cleveland follows a Normal distribution with standard deviantion sigma=9.6%








What is the sampling distribution of the mean percent x_bar that the sample spends on housing if null hypothesis is true?What is the sampling distribution of the mean percent x_bar that the sample spends on housing if null is true?
For any normal random variable X with mean 渭 and standard deviation 蟽 , X ~ Normal( 渭 , 蟽 ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( 渭 , 蟽虏 ). Most software denotes the normal with just the standard deviation.)





You can translate into standard normal units by:


Z = ( X - 渭 ) / 蟽





Where Z ~ Normal( 渭 = 0, 蟽 = 1). You can then use the standard normal cdf tables to get probabilities.





If you are looking at the mean of a sample, then remember that for any sample with a large enough sample size the mean will be normally distributed. This is called the Central Limit Theorem.





If a sample of size is is drawn from a population with mean 渭 and standard deviation 蟽 then the sample average xBar is normally distributed





with mean 渭 and standard deviation 蟽 /鈭?n)





An applet for finding the values


http://www-stat.stanford.edu/~naras/jsm/鈥?/a>





calculator


http://stattrek.com/Tables/normal.aspx





how to read the tables


http://rlbroderson.tripod.com/statistics鈥?/a>





In this question we have


Xbar ~ Normal( 渭 = 0.31 , 蟽虏 = 0.009216 / 40 )


Xbar ~ Normal( 渭 = 0.31 , 蟽虏 = 0.0002304 )


Xbar ~ Normal( 渭 = 0.31 , 蟽 = 0.096 / sqrt( 40 ) )


Xbar ~ Normal( 渭 = 0.31 , 蟽 = 0.01517893 )