Question

t-table questions

Answer 1

@JoannaBlackwelder

Answer 2

Use the appropriate student t-test to determine whether the two data sets are distinguishable at the 50%, 95%, and 99% confidence levels. Clearly state your conclusion.

Answer 3

I'll post more info about the two sets of data if requested.

Answer 4

Please do.

Answer 5

Its based on two different trials of paper weighing
Mine:
Average: 4.64419
Standard deviation: .07089
n=10
Given:
Average: 4.65215
Standard deviation: .0505
n=70

Answer 6

I also have spooled for them both.
Spooled=.0529

Answer 7

I know I need some kind of a table I guess. and I also need to know how to calculate tcalc. I have no idea what I'm doing. It's been awhile.

Answer 8

Change majors.

Answer 9

never

Answer 10

That's the spirit.

Answer 11

What was your sample size?

Answer 12

Oh, is that n?

Answer 13

yes =)

Answer 14

Ok, I am still figuring this out myself. But it looks like you need to look up a value in the t table for n-1 degrees of freedom and start with 50% confidence level

Answer 15

wouldn't I have to add both the n's? like n1+n2-2? maybe. I'm not sure but i thought I saw it somewhere.

Answer 16

This value is a. And can be plugged into the formula
\[c=\frac{ as _{2} }{ \sqrt{n}}\]

Answer 17

I'm not sure either....I never took stats. Just trying to figure it out.

Answer 18

I'm in analytical chemistry =( so yeah its kinda out of my realm of comfort.

Answer 19

I was a chemical engineering major. Way outta my comfort zone.

Answer 20

Oh man I feel you. I have a sheet with different equations from about a year ago, but I can't make sense of them.

Answer 21

Yeah, every little symbol has to be defined and then calculated. It takes a while to get the grasp of it!

Answer 22

The c is your margin of error. \[average \pm margin of error\] is confidence interval

Answer 23

yeah =\ its difficult to cram.
Umm I kinda need the degrees of freedom to continue though. I get the rest but I'm stuck there. I'm confused because I have two n's.

Answer 24

Yeah, still working on that one too.If @abb0t would care to be helpful instead of just sarcastic, that would be nice.

Answer 25

Haha if only

Answer 26

Haha! I think I may have found something. It is called a paired t test.

Answer 27

yeah? I have old notes where it says case 1 case 2 case 2b and case 3 so my notes don't go with anything online =( I hope you found something

Answer 28

This looks helpful. http://en.wikipedia.org/wiki/Student%27s_t-test#Equal_or_Unequal_sample_sizes.2C_unequal_variances

Answer 29

It has a formula for degrees of freedom.

Answer 30

It does actually.... I just need to figure out how to use it haha.

Answer 31

oh that formulas haha Let me try to work it out

Answer 32

Ok. I think that s^2 is stddev^2

Answer 33

yess I believe so

Answer 34

this is taking me awhile

Answer 35

No worries. I'm still working on what to do with t once you find it. :)

Answer 36

I think this becomes our "average"

Answer 37

I got 10.65

Answer 38

For t?

Answer 39

for confidence level its between 10 and 15 so I'm not sure what to use .700 or .691
That is for the degrees of freedom.

Answer 40

doesn't seem right

Answer 41

this is hopeless =/

Answer 42

I think we should use the t value given the formula on the page I linked. It gives t=0.342871

Answer 43

Hm, but then what about the 50%?

Answer 44

@SithsAndGiggles Are you good at stats?

Answer 45

yeah the t value doesn't correspond with anything I have

Answer 46

I calculated it with your stddev and averages.

Answer 47

@kropot72 Are you good with stats?

Answer 48

really? I'm behind somewhere what exactly did you get the t value from

Answer 49

Where it says "the t statistic to test whether the populations are different can be calculated as"

Answer 50

oh ok

Answer 51

But I think I am at a loss from here. I hope these guys can help you more than I have. Sorry!

Answer 52

its fine. I don't know what the heck I'm doing here either

Answer 53

Have you calculated the test statistic?

Answer 54

I got t=0.342871

Answer 55

To check if that agrees:
\[\large t=\frac{\bar{x}_1-\bar{x}_2}{{s_p}^2\sqrt{\dfrac{1}{n_1}+\dfrac{1}{n_2}}}\]
with
\[\large{s_p}^2=\sqrt{\frac{(n_1-1){s_1}^2+(n_2-1){s_2}^2}{n_1+n_2-2}}\]
I'm getting \(t\approx-0.444\).

http://www.wolframalpha.com/input/?i=%284.644-4.652%29%2F%28Sqrt%5B1%2F10%2B1%2F70%5DSqrt%5B%289*.07089%5E2%2B69*.0505%5E2%29%2F%2810%2B70-2%29%5D%29

Answer 56

My formula is a bit different, but I don't really know what I am doing. I am using the formula from this link.
http://en.wikipedia.org/wiki/Student%27s_t-test#Equal_or_Unequal_sample_sizes.2C_unequal_variances

Answer 57

The disparity could be from the fact that I'm assuming equal variances. You'll have to run an F-test/ANOVA table to determine if the variances are (statistically) equal or unequal.

Answer 58

Ok. I am using formulas that don't assume equal variances.

Answer 59

@Kkutie7 you can decide which t stat you think is best. @SithsAndGiggles Do you know what to do from here?

Answer 60

I used this one:
\[t=\frac{(X_1 ) ̅-(X_2 ) ̅}{√(\frac{S_1^2}{n_1} +\frac{S_2^2}n_2 )}\]

Answer 61

I would if I wasn't so tired :P

I can check back in the morning and try again. In the meantime, @Kkutie7 if it's not too much of a hassle, would you mind posting your individual data points? Maybe attach as a spreadsheet or something...

Answer 62

yeah sure.

Answer 63

I'm beat too. Happy studies and good night!

Answer 64

Thank you guys

Answer 65

Maybe repost in the statistics section if you need more help tonight.

Answer 66

First we use the F-test to determine if the variances are statistically different.

Our null hypothesis is that \({s_1}^2={s_2}^2\), and we use this F-statistic:
\[F=\frac{{s_1}^2}{{s_2}^2}=0.463\]
and we compare this to an appropriate table of values: http://mips.stanford.edu/courses/stats_data_analsys/lesson_5/ftable_05_2.gif

We use \(df_1=69\) and \(df_2=9\), and we find that the critical \(F\) is about 2.79. Since our calculated F statistic is less than 2.79, we conclude the variances are unequal (reject the null).

This means we will use the following t statistic:
\[t=\frac{\bar{x}_1-\bar{x}_2}{\sqrt{\dfrac{{s_1}^2}{n_1}+\dfrac{{s_2}^2}{n_2}}}=-0.101\]
with defrees of freedom (using the equation linked by @JoannaBlackwelder)
\[df\approx10\]
Use this to make a conclusion at the different significance levels.