Please Help! If someone could just show me how to do a problem like this!

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 = 37, Ód = 231, Ód2 = 6193

(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 = 6193- (231)^2/37

SSd = 4750.811

Sd = sqrt(SSd/n-1)

Sd = sqrt(4750.811/36))

Sd = 11.488

Mean = 231/37 = 6.243

t = 6.243/11.488/sqrt((37))
t = 3.31
p-value = .0021

To solve this problem, we need to perform a dependent t-test since we have paired or dependent samples. The paired or dependent t-test allows us to compare the means of two related samples, in this case, the corrosion on the coated and uncoated portions of the steel pipe.

To find the t-value and p-value, we will use the following formulas:

t = d̄ / (Ód / √n)
p-value = P(T > t) + P(T < -t)

Where:
d̄ = mean difference of corrosion (d) between coated and uncoated portions
Ód = standard deviation of the difference (d) between coated and uncoated portions
n = sample size

Now let's calculate the values:

(a) Finding the t-value:

First, we need to find the mean difference (d̄):
d̄ = 0 (since we are comparing whether the coating is beneficial)

Next, we calculate the standard error of the mean difference (SEM):
SEM = Ód / √n
SEM = 231 / √37 = 37.4947 (rounded to four decimal places)

Now we can calculate the t-value:
t = d̄ / SEM
t = 0 / 37.4947 = 0 (rounded to two decimal places)

(ii) Finding the p-value:

To find the p-value, we need to use statistical software or a t-distribution table. However, since we already have the t-value as 0, the p-value will be 0.5000 (rounded to four decimal places).

The p-value is the probability of observing a t-value as extreme as the one calculated (or more extreme) under the null hypothesis, which suggests that the coating has no effect.

Therefore, the p-value of 0.5000 (rounded to four decimal places) means that we do not have sufficient evidence to conclude that the coating is beneficial, as it is not significantly different from zero.