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

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

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 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.

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

To determine if the coating is beneficial, we can perform a hypothesis test. Let's perform the steps to find the t-value and p-value.

Step 1: Set up the hypotheses:
- Null hypothesis (H0): There is no difference between the corrosion on the coated and uncoated portions of the steel pipe. (d = 0)
- Alternative hypothesis (Ha): There is a difference between the corrosion on the coated and uncoated portions of the steel pipe. (d ≠ 0)

Step 2: Determine the test statistics:
The test statistic for a dependent sample t-test is given by:
t = (x̄ - 0) / (s / √n)

where x̄ is the mean difference, s is the standard deviation of the differences, and n is the sample size.

Given:
n = 48 (sample size)
σd = 238 (standard deviation of differences)

Step 3: Calculate the t-value:
First, we calculate the standard deviation of the mean difference (sd):
sd = σd / √n
= 238 / √48
≈ 34.156

Next, we calculate the mean difference (x̄):
x̄ = 0 (since we assume no difference under the null hypothesis)

Finally, we calculate the t-value:
t = (x̄ - 0) / sd
= 0 / 34.156
≈ 0

Step 4: Find the p-value:
To find the p-value, we need to reference the t-distribution with n - 1 degrees of freedom (48 - 1 = 47).

Using a t-distribution table or a statistical calculator, we find the p-value associated with a t-value of 0 and 47 degrees of freedom.
The p-value turns out to be 0.5000 (rounded to four decimal places).

Step 5: Make a decision:
- If the p-value is less than the significance level (α), we reject the null hypothesis.
- If the p-value is greater than the significance level (α), we fail to reject the null hypothesis.

Given α = 0.01 and the p-value of 0.5000, the p-value is greater than α.

Conclusion:
Since the p-value is greater than the significance level, we fail to reject the null hypothesis. Therefore, we do not have sufficient evidence to conclude that the coating is beneficial based on this random sample.