How do you find the p-value, on a problem like below.
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.)
Correct: Your answer is correct.3.323 .
(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
P-value = 0.0017
t = 3.32
To find the p-value, we need to perform a t-test using the given information. The formula for the t-statistic is:
t = (mean difference - hypothesized mean difference) / (standard deviation / sqrt(n))
In this case, the hypothesized mean difference is 0 (since we want to test if the coating is beneficial), the mean difference is unknown, standard deviation is given as standard deviation of d (Ód) = 238, and n = 48.
Using the given information, we can calculate the t-statistic:
t = (mean difference - 0) / (238 / sqrt(48))
Next, we need to find the degrees of freedom (df). Since this is a dependent sampling plan, the degrees of freedom is equal to the sample size minus 1:
df = n - 1 = 48 - 1 = 47
Now, we can use the t-distribution table or a calculator to find the p-value. However, since the given table is not provided, we will use a t-distribution calculator.
Using the calculator with df = 47 and t = 3.323 (which we calculated earlier), we find that the p-value is approximately 0.0012.
Therefore, the p-value is 0.0012.
To find the p-value in this problem, you need to follow these steps:
Step 1: Define the null and alternative hypotheses.
In this case, the null hypothesis (H0) is that the coating is not beneficial, and the alternative hypothesis (Ha) is that the coating is beneficial.
Step 2: Calculate the test statistic (t).
The test statistic (t) is calculated using the formula:
t = (x̄ - μ) / (σ / √n)
where x̄ is the mean difference (in this case, the mean difference between the amount of corrosion on the uncoated portion and the coated portion, d), μ is the population mean difference (assumed to be 0 under the null hypothesis), σ is the standard deviation of the difference scores, and n is the sample size.
In this problem, the mean difference (x̄) is not provided directly but can be calculated using the formula:
x̄ = Ód / n
The test statistic (t) can be calculated as:
t = (x̄ - μ) / (σ / √n) = (x̄ - 0) / (σ / √n) = x̄ / (σ / √n)
Using the provided values:
x̄ = Ód / n = 238 / 48 = 4.9583 (rounded to four decimal places)
σ is not directly given, but we can estimate it using the sample standard deviation (s) based on the formula:
s = √(Òd2 - (Òd)^2 / n) = √(6202 - (238)^2 / 48) ≈ 10.4262 (rounded to four decimal places)
n = 48
Substituting the values into the formula for t:
t = 4.9583 / (10.4262 / √48) ≈ 3.323
Therefore, the calculated value of t is 3.323.
Step 3: Calculate the p-value.
The p-value represents the probability of observing a test statistic as extreme as the one calculated (or more extreme) under the assumption that the null hypothesis is true.
To find the p-value in this case, you can use a t-distribution table or a statistical software. Given that the degrees of freedom (df) for this problem are n - 1 = 48 - 1 = 47, you need to determine the area in the tail(s) of the t-distribution that corresponds to the calculated value of t.
The p-value is the probability of observing a t-value as extreme as the calculated t, multiplied by 2 (because it is a two-tailed test) since we are looking for extreme values in both tails of the t-distribution.
Using a t-distribution table (or a statistical software), you can find that the p-value for t = 3.323 with 47 degrees of freedom (df) is approximately 0.0012.
Therefore, the p-value for this problem is approximately 0.0012 (rounded to four decimal places).