# t critical value

• Oct 31st 2009, 07:46 PM
Pengu
t critical value

A bookstore recorded the amount spent by a subset of its members for its online gift-shop. The average spent this year was $10.90, with a standard deviation of$2.10 for 510 members observed. The average spent last year was $10.30 with standard deviation$3.14 for 520 members observed. Test whether the average amount spent by members last year compared to this year is not significantly different.

So i think that we use t-test, and i can find the test statistic, but i dont know how to calculate the t critical value as im unsure of what the degrees of freedom is ( i think either 9 or 519 or 509?). Can someone please help me?
• Oct 31st 2009, 08:07 PM
matheagle
with both n's so large, you should just use the normal distribution.
The question I was going to ask you, can we assume equal variances in th etwo populations?
But again, with large n's, I would use the central limit theorem.
• Oct 31st 2009, 08:11 PM
matheagle
$\displaystyle {(\bar X_1-\bar X_2)-0\over \sqrt{{s_1^2\over n_1}+{s_2^2\over n_2}}}\approx N(0,1)$
• Oct 31st 2009, 08:50 PM
Pengu
well i got:
$\displaystyle \frac{0.6}{\sqrt{\frac{(2.1)^2}{510} + \frac{(3.14)^2}{520}}} = 3.61106$

i dont really know how to use CLT sorry. But after i found that value, then i thought that i was supposed to compare it to the critical value and if it is greater than the critical then i reject the null hypothesis...
but if there is another way to find out if it is significantly different using the CLT can you please show me?

oh and for the variances, they're not equal?
• Oct 31st 2009, 08:55 PM
Pengu
Ohhh am i supposed to do

P(Z<3.61106) and look this up on the table and compare to $\displaystyle \alpha(0.5)$ ?
• Oct 31st 2009, 09:16 PM
matheagle
this is a 2 sided test

p-value = $\displaystyle 2P(Z>3.611)$

which can be found at Free Two Tailed Area Under the Standard Normal Curve Calculator
3.611 gives a p-value of...0.0003051