STATS POP DISTRIBUTION QUESTION

Question

When the population distribution is ??(368, 152) and the sample size is n=25, find a symmetric interval around µ that will include 95% of the sample means.

 

PLEASE SHOW HELP WITH STEPS

Answer 1

So first things first, lets clear up what those parentheses at the beginning of the question mean.  The first number 368 is the mean of the sample set, the second number 152 is the standard deviation of the sample set.  To do anything of interest here we're going to need to find a t-value distribution that includes 95% of the data.  This can only be accomplished by referencing the student's t-table (which you can find online or in your statistics book).  But, to read this table, you need to understand what they mean by "degrees of freedom".  Degrees of freedom (Df) is calculated by taking the size of your sample and subtracting one.

 

So, Df = n-1 = 25-1 = 24.

 

Once you know that Df = 24, you find where on the table Df = 24 meets t_.975.  But wait, I thought we were looking for 95% of the data?  Unfortunately, most of the tables are one-tailed by default, but this question is asking us to find an interval centered on the mean, so it wants a two-tailed answer.  Since t_.975 means that .025 of the data is beyond that t-value, it also means that .025 is beyond the other tail of the data.  Adding up to a total of .05.

 

On the table I found online, this intersection gives us t = 2.065

 

But what does that mean,

 

Well, "t" is a measurement of how many standard deviations we are away from the mean, so we are 2.065 standard deviations in both directions from the mean for our upper and lower limits.

 

Mathematically, that is 2.065*152 in both directions.

 

Or, 313.88 in both directions from 368.  

 

368 - 313.88 = 54.12

 

368 + 313.88 = 681.88

 

Therefore, 95% of our data lies in the interval between 54.12 and 681.88.

 

Hope that was helpful, that's a pretty complicated answer to type out.