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.

Question

n = 48, Ód = 238, Ód2 = 6202

(a) Find t. (Give your answer correct to two decimal places.)


(ii) Find the p-value. (Give your answer correct to four decimal places.)

Answer 1

SSd = 6202 - (238)^2/48

SSd = 5201.917

Sd = sqrt(SSd/n-1)

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

Answer 2

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.

Answer 3

This is mean = 238/48 = 4.985

I am looking for t.

Answer 4

Typo
238/48 = 4.958

Answer 5

But how do you get the p-value, 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 p-value 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.

Answer 6

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/n-1)

Sd = 10.337

Find p-value you use your ti-83 plus

Go to stat
Tests
2-T-Test
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