Question

An article in Fire Technology investigated two different foam-expanding agents that can be used in the nozzles of firefighting spray equipment. A random sample of five observations with an aqueous film-forming foam (AFFF) had a sample mean expansion of 4.7 and standard deviation of 0.6. A random sample of five observations with alcohol-type concentrates (ATC) had a sample mean expansion of 6.9 and a standard deviation of 0.8. You will assume σ 2 1 = σ 2 2 for this problem, so you’ll need to calculate a pooled estimate for σ 2 as S 2 p .

Answer 1

Answer:

Step-by-step explanation:

Hello!

The objective of this experiment is to test if two different foam-expanding agents have the same foam expansion capacity

Sample 1 (aqueous film forming foam)

n₁= 5

X[bar]₁= 4.7

S₁= 0.6

Sample 2 (alcohol-type concentrates )

n₂= 5

X[bar]₂= 6.8

S₂= 0.8

Both variables have a normal distribution and σ₁²= σ₂²= σ²= ?

The statistic to use to make the estimation and the hypothesis test is the t-statistic for independent samples.:

t= $\frac{(X[bar]_1 - X[bar]_2) - (mu_1 - mu_2)}{Sa*\sqrt{\frac{1}{n_1} + \frac{1}{n_2 } } }$

a) 95% CI

(X[bar]_1 - X[bar]_2) ± $t_{n_1 + n_2 - 2}$*$Sa* \sqrt{\frac{1}{n_1}+\frac{1}{n_2} }$

Sa²= $\frac{(n_1-1)S_1^2 + (n_2-1)S_2^2}{n_1 + n_2 - 2}$= $\frac{(5-1)0.36 + (5-1)0.64}{5 + 5 - 2}$= 0.5

Sa= 0.707ç

$t_{n_1 + n_2 -2: 1 - \alpha /2} = t_{8; 0.975} = 2.306$

(4.7-6.9) ± 2.306* $(0.707\sqrt{\frac{1}{5}+\frac{1}{5} })$

[-4.78; 0.38]

With a 95% confidence level you expect that the interval [-4.78; 0.38] will contain the population mean of the expansion capacity of both agents.

b.

The hypothesis is:

H₀: μ₁ - μ₂= 0

H₁: μ₁ - μ₂≠ 0

α: 0.05

The interval contains the cero, so the decision is to reject the null hypothesis.

Complete question

a. Find a 95% confidence interval on the difference in mean foam expansion of these two agents.

b. Based on the confidence interval, is there evidence to support the claim that there is no difference in mean foam expansion of these two agents?