Statistics: Is Y normally distributed? Explain.?

According to recent General Social Surveys, in the United States the distribution of Y the number of sex partners you have had in the past 12 months has a mean of about 1.0 and a standard deviation of about 1.0.





b) For a random sample of 100 adults, find the sampling distribution of Ybar.


c) Based on the information in part (b), report an interval within which the sample mean would almost surely fall.

Statistics: Is Y normally distributed? Explain.?
For any normal random variable X with mean μ and standard deviation σ , X ~ Normal( μ , σ ), (note that in most textbooks and literature the notation is with the variance, i.e., X ~ Normal( μ , σ² ). Most software denotes the normal with just the standard deviation.)





You can translate into standard normal units by:


Z = ( X - μ ) / σ





Where Z ~ Normal( μ = 0, σ = 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 μ and standard deviation σ then the sample average xBar is normally distributed





with mean μ and standard deviation σ /√(n)





An applet for finding the values


http://www-stat.stanford.edu/~naras/jsm/...





calculator


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





how to read the tables


http://rlbroderson.tripod.com/statistics...





In this question we have


Xbar ~ Normal( μ = 1 , σ² = 1 / 100 )


Xbar ~ Normal( μ = 1 , σ² = 0.01 )


Xbar ~ Normal( μ = 1 , σ = 1 / sqrt( 100 ) )


Xbar ~ Normal( μ = 1 , σ = 0.1 )








Confidence intervals are used to find a region in which we are 100 * ( 1 - α )% confident the true value of the parameter is in the interval.





For large sample confidence intervals about the mean you have:





xBar ± z * sx / sqrt(n)





where xBar is the sample mean


z is the zscore for having α% of the data in the tails, i.e., P( |Z| %26gt; z) = α


sx is the sample standard deviation


n is the sample size





The sample mean xbar = 1


The sample standard deviation sx = 1


The sample size n = 100





The z score for a 0.99 confidence interval is the z score such that 0.005 is in each tail.


z = 2.575829


The confidence interval is:





( xbar - z * sx / sqrt( n ) , xbar + z * sx / sqrt( n ) )


( 0.742417 , 1.257583 )





this interval tells us that we are 99% confident the true value of the number of sex partners is contained in the interval.