PLEASE help ! :) How do I construct a 98% confidence interval with two independant simple samples with normally distributed populations with sigma1=sigma2 (o1=o2)?? How do I figure out the t score when it is 0.01 significance level and degrees of freedom=394??
I do not understand (ノಠ益ಠ)ノ彡┻━┻ [FLIP TABLE RAGE!]
shh it's ok. no need to spam... @Dbzfan836
can't help though sorry. i don't know statistics...
oh, ok ┬──┬ ノ( ゜-゜ノ) [put back the table]
wait, what!?!? ┻━┻ ︵ヽ(`Д´)ノ︵ ┻━┻ [DOUBLE TABLE FLIP RAGE!]
you could get suspended for typing like that...
never mind... ┬──┬ ノ( ゜-゜ノ) [put back the table]
sorry
Its pretty much the same as you would with just a single sample.
Do you own a graphing calculator?
no
Im really just having a hard time understanding what and where to find the t value to use in the formulas
I ask if you have a graphing calculator because, if it is a TI, then it could probably calculate it for you. Your sample size is much greater than 30, why not just use a z-interval?
Because i need t for the formula i am using
http://mips.stanford.edu/courses/stats_data_analsys/principles/t_table.html http://allianthawk.org/victionary/showdef.php?word=217 Look at the critical values of the second page, and then look at the t scores corresponding to the popular significance levels. See how when df becomes close to 30, the t table values converge to the z table values? This is the Law of Large Numbers in play.
I dont use z scores
I need to use a t score for this formula
What is the formula? Is it (x1 − x2) ± t* × √(s1 ⁄ n1 + s2/ n2) ?
Basically, what I am saying, is that when df > 30, there is no significant difference between using t-scores and z-scores.
When you have a left tailed test, and find your t score, Do you change it to become negative?
Your t-score will be negative, since it is below the mean difference.
okay thanks!
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