math
posted by tracy10 on .
The corrosive effects of various soils on coated and uncoated steel pipe was tested by using a dependent sampling plan. The data collected are summarized below, where d is the amount of corrosion on the coated portion subtracted from the amount of corrosion on the uncoated portion. Does this random sample provide sufficient reason to conclude that the coating is beneficial? Use á = 0.01 and assume normality.
n = 48, Ód = 238, Ód2 = 6202
(a) Find t. (Give your answer correct to two decimal places.)
(ii) Find the pvalue. (Give your answer correct to four decimal places.)

SSd = 6202  (238)^2/48
SSd = 5201.917
Sd = sqrt(SSd/n1)
Sd = sqrt(5021.917/47) =
Sd = 10.337
t = 4.958/10.337/sqrt(48)
t = 3.323
Two tail = .0017
right tail = 8.6885e^4
Left tail = .9991 
Have a question, where did you get the 4.958 and what did you look at for it. Thanks this has really helped so far, just unsure on that.

This is mean = 238/48 = 4.985
I am looking for t. 
Typo
238/48 = 4.958 
But how do you get the pvalue, I have worked it out by hand and also put into calculator and it comes up wrong. 0.0017 was the last answer I got. and 0.9991 before that. I have several questions like this asking for pvalue so if you could work one out and I could use as an example I would appreciate it. Thank you for all your help. I got the t as 3.32 and that was the right answer on that one.

First of all, you have to calculate your mean, standard deviation
sum of d is 238
n is 48
Mean = 238/48 = 4.958
Variance = 6202 238^2/48 = 5201.92
standard derivation sqrt((variance/n1)
Sd = 10.337
Find pvalue you use your ti83 plus
Go to stat
Tests
2TTest
Enter mean = 4.958
Go to down to Sx then
Standard deviation or sx = 10.337
n =48
Select Not equal
Go to down and select calculate enter
It will you info