Question

The thickness of a plastic film (in mils) on a substrate material is thought to be influenced by the temperature at which the coating is applied. A completely randomized experiment is carried out. Eleven substrates are coated at 125F, resulting in a sample mean coating thickness of and a sample standard deviation of s1 5.08. Another 13 substrates are coated at 150F, for which and s2 20.15 are observed. It was originally suspected that raising the process temperature would reduce mean coating thickness. Do the data support this claim

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective is to test if rising the process temperature reduces the thickness of the plastic film that coats a substrate material To do so, two samples of substrates are coated at different temperatures:

Sample 1

X₁: Thickness of the plastic film after the substrate is coated at 125F

n₁=11

X[bar]₁= 101.28

S₁= 5.08

Sample 2

X₂: Thickness of the plastic film after the substrate is coated at 150F

n₂= 13

X[bar]₂= 101.70

S₂= 20.15

Does the data support this claim? Use the P-value approach and assume that the two population standard deviations are not equal.

Now if the higher the heat, the thinner the thickness of the plastic coating, then the average thickness of the coating done at 150F should be less than the average thickness of the coating done at 125F, symbolically: μ₂ < μ₁

Then the hypotheses are:

H₀: μ₂ ≥ μ₁

H₁: μ₂ < μ₁

α:0.05 (there is no α level stated so I've chosen the most common one)

Assuming that both variables have a normal distribution since the population standard deviations are not equal, the statistic to use is the Welch's t-test:

$t= \frac{(X[bar]_2-X[bar]_1)-(Mu_1-Mu_2)}{\sqrt{\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} } } ~~t_w$

$t_{H_0}= \frac{(101.7-101.28)-0}{\sqrt{\frac{(20.15)^2}{13} +\frac{(5.08)^2}{11} } } = 0.072$

This test is one-tailed to the left, meaning that you'll reject the null hypothesis at small values of t. The p-value is also one-tailed and has the same direction as the test. To calculate it you have to first calculate the degrees of freedom of the Welch's t:

$Df_w= \frac{(\frac{S^2_1}{n_1} +\frac{S^2_2}{n_2} )^2}{\frac{(\frac{S^2_1}{n_1})^2 }{n_1-1}+\frac{(\frac{S^2_2}{n_2} )^2}{n_2-1} }$

$Df_w= \frac{(\frac{5.08^2}{11} +\frac{20.15^2}{13} )^2}{\frac{(\frac{5.08^2}{11})^2 }{10} +\frac{(\frac{20.15^2}{13} )^2}{12} } = 13.78$

The distribution is a Student's t with 13 degrees of freedom, then you can calculate the p-value as:

P(t₁₃≤0.072)= 0.4718

Using the p-value approach, the decision rule is:

If the p-value ≤ α, the decision is to reject the null hypothesis.

If the p-value > α, the decision is to not reject the null hypothesis.

The p-value is greater than the significance level, so the decision is to nor reject the null hypothesis.

Using a significance level of 5%, there is no significant evidence to support the claim that the average thickness pf the plastic coat processed at 150F is less than the average thickness pf the plastic coat processed at 125F.

I hope you have a SUPER day!