# T-score vs. Z-score

• Dec 6th 2009, 11:33 PM
hello_math_
T-score vs. Z-score
Random variables (RVs) ‘X’ and ‘Y’ represent annual incomes of well established unmarried young professional men and women. These two RVs are normally distributed and are statistically independent. The mean values of ‘X’ and ‘Y’ are $100K and$80K with standard deviations of $20K and$15K respectively. Here ‘K’ represents units of thousands.

a.A random sample of 25 professional men is taken to find their average income. What is the probability that this mean income of the sample is less than $90K. For professional men, the population standard deviation given is considered reliable. If this event does occur, what conclusion can you draw? b.A random sample of 20 professional women is taken to find their average income. However, the population standard deviation is considered unreliable and is estimated for this sample to be$22K. What is the probability that this mean income of the sample is more than $91.3932K. If this event does occur, what conclusion can you draw? -------------------------------------------------------------------- In the answers, a) uses the Z-score and b) uses the t-score. I'm confused because I thought the rule of thumb was for samples under 30 that the t-score should be used. Can someone further explain this to me please? • Dec 6th 2009, 11:57 PM matheagle People get this confused all the time. You use Z, when you have $\sigma$ If the underlying population is normal, i.e., $X_1,...,X_n i.i.d\sim N(\mu ,\sigma^2)$, then ${\bar X-\mu\over \sigma/\sqrt{n}}\sim N(0,1)$ BUT in real life, how would we know the st deviation when we usually don't know the $\mu$. So if we estimate $\sigma$ with $S$ and .... the underlying population is normal, i.e., $X_1,...,X_n i.i.d\sim N(\mu ,\sigma^2)$, then ${\bar X-\mu\over s/\sqrt{n}}\sim t_{n-1}$ --------------------------------------------------------------------- NOW for the part of having a large sample. This n larger than 30 I assume is from the Berry-Esseen Theorem, which tells you how close you are to a normal distribution. http://en.wikipedia.org/wiki/Berry%E...Esseen_theorem If you have i.i.d random variables, then both ${\bar X-\mu\over \sigma/\sqrt{n}}$ and ${\bar X-\mu\over s/\sqrt{n}}$ approach a standard normal random variable as n goes to INFINITY (CLT). Now, we never sample an infinite number of items, but the larger the better. Note that a t density approaches a normal as n goes to infinity, so it doesn't make much difference which table you use for very large n. • Dec 7th 2009, 12:08 AM hello_math_ but for b) we know the standard deviation to be$22k but it still uses the t-score in the answer.

Ps. thanks for answering so fast!
• Dec 7th 2009, 01:05 AM
mr fantastic
Quote:

Originally Posted by hello_math_
but for b) we know the standard deviation to be $22k but it still uses the t-score in the answer. Ps. thanks for answering so fast! *Ahem* ..... Quote: However, the population standard deviation is considered unreliable and is estimated for this sample to be$22K.
How was it estimated I wonder ....?