Question

This deals with t- test.
d. For each method, is the Aluminum content different at 95% confidence level for the two Lots? Which technique offers the superior precision?
I've done majority of the work for this and will provide certain answers if requested. I don't know how to use the t-table to answer this.

Answer 1

https://controls.engin.umich.edu/wiki/images/0/0a/DOF.jpg
this is the t-table I'm supposed to use I guess

Answer 2

what are parts a,b,c, , i need more information

Answer 3

it's just the answers for mean, standard deviation, spooled, and t-calc.

Answer 4

I wanna know how to use the table and the other things I've listed to answer the question.

Answer 5

lololol okay was so lost. Just saw the link, give me a sec to think :)

Answer 6

alright

Answer 7

@kirbykirby
@zepdrix

Answer 8

I'm sorry.... I'm trying to figure out what you're asking, but you're using terms that I don't recognize. Your question, if it is a simple math problem, is worded in a very bizarre way.... :( I'm really sorry

Answer 9

It's analytical chemistry but can be counted at statistics which is math oriented.

Answer 10

I think you should post the entire question... there is not enough being said from part d) only to know what the context of the problem is.

Answer 11

You say "for each method" for example.. we don't even know what these methods are

Answer 12

alright.
4. An analyst was using two different methods (I and II) to monitor the content of Al in two different lots of an Aluminum nitride power (AIN) that his company was producing. Method I involved the use of a gravimetric method to determine the % Al in the AIN. The % Al results from Method II were obtained using an atomic emission spectroscopic method called inductively coupled plasma (ICP).
Method I:

Lot 1: 65.27% Lot 2: 65.98%
65.22% 65.80%
65.44% 65.67%
65.50% 65.75%

Method II:
Lot 1: 65.5% Lot 2: 66.0%
65.1% 65.5%
65.4% 66.5%
65.8% 65.7%

Answer 13

a. Calculate the mean and standard deviation for each lot in each of the two methods.
b. Calculate the pooled standard deviations for two lots for each method (I and II).
c. Calculate the student’s t values (talc) for each method (I and II).

Answer 14

calculate the student's what ?

Answer 15

hey you can upload the whole question, use a screenshot or

Answer 16

you dont have to type it, just take a picture with a camera

Answer 17

Do you know how to set up a 95% confidence interval? You just need to apply the formula. I'm guessing you are using the pooled standard deviation as well for this.

If we just doing a hypothesis test that the 2 means are different, then you can use the confidence interval:
\((\bar{x_1}-\bar{x_2})\pm t_{\text{crit}}*(S_p\sqrt{1/n_1+1/n_2}) \) , where S_p your pooled std. deviation and \(t_{crit}\) is the value you look for in your t-table. So look at the 95% column, and then the degrees of freedom corresponds to the sample size total - 2, so \(4+4-2\)

Answer 18

there did you get 7.1 for your critical t value?

Answer 19

I used the t calc i calculated... I'm guess that was wrong it should have been the first number I wrote from the t-table?

Answer 20

t=2.447

Answer 21

shoot you are right so this then?
\[(65.63-65.80)\pm 2.447*(.00145\sqrt{1/4+1/4})\]
\[-.17\pm.00355\sqrt{.5}\]
\[-.17\pm.00251\]

Answer 22

I believe so, assuming the other numbers are correct from a)...c)

Answer 23

And the better method, I believe, should give a smaller confidence interval

Answer 24

well the one offering more precision

Answer 25

ok but how do i see if they are different?

Answer 26

Sorry to interject, but are we sure we're comparing means and not proportions? The data seems to indicate the latter, but I don't know for sure.

Answer 27

Ohh I think you might be right on that one actually. Hm my brain is really tired tonight after that probability problem xD

Answer 28

I don't know =/ its been awhile since I've done work like this

Answer 29

answer to a:
Method I
Mean Standard Deviation
Lot 1 65.63 .001158
Lot 2 65.80 .001138

Method II
Mean Standard Deviation
Lot 1 65.45 0.002887
Lot 2 65.93 0.004349

Answer 30

answer to b:
Method I:
\[pooled=√(((4-1) 〖.001158〗^2+(4-1) 〖.001138〗^2)/(4+4-2))\]
\[=√((3(〖.001158〗^2 )+3(〖.001138〗^2))/6)\]\[=√((3(.00000134)+3(.00000130))/6)\]\[=√.00000132=.00145\]
Method II:

\[pooled=√(((4-1) 〖.002887〗^2+(4-1) 〖.004349〗^2)/(4+4-2))\]
\[=√((3(〖.002887〗^2 )+3(〖.004349〗^2))/6)\]\[=√((3(.00000833)+3(.0000190))/6)\]\[=√.0000137=.00370\]

Answer 31

and c:
Method I:
\[t_calc=|65.80-65.63|/√(.001158/4+.001138/4)\]
\[=|.17|/√(.0002895+.0002845)\]
\[=.17/√.000574=.17/.023958=7.1\]


Method II:
\[t_calc=|65.93-65.45|/√(0.002887/4+0.004349/4)\]
\[=|.48|/√(.000722+.001087)\]
\[=.48/√.001809=.48/.042532=11.3\]

Answer 32

more information...

Answer 33

whats up with the question marks?

Answer 34

noticed it tends to stay like that when I first input it but afterawhile it goes back to normal on from what I see. No question marks on my screen. @perl are you any good at chemistry?

Answer 35

yes a bit

Answer 36

whats the q.

Answer 37

@perl http://openstudy.com/study#/updates/54111f13e4b0670f10120589