Question

An article in Fire Technology describes the investigation of 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 of 4.340 and a standard deviation of 0.508. A random sample of five observations with alcohol-type concentrates (ATC) had a sample mean of 7.091 and a standard deviation of 0.430. Assume that both populations are well represented by normal distributions with the same standard deviations. (a) Is there evidence to support the claim that there is no difference in mean foam expansion of these two agents? Use a fixed-level test with α=0.10. (b) Calculate the P-value for this test. (c) Construct a 90% CI for the difference in mean foam expansion. Explain how this interval confirms your finding in part (a).

Answer 1

Answer:

a) There is no evidence to support the claim that there is no difference in mean foam expansion of these two agents.

b) P=0

c)  90% CI

$-3.2406\leq \mu_1-\mu_2 \leq -2.2614$

This CI tells us that there is a 90% confidence that the real value of the difference between the means is between this two values. We see that both are negative values, so the value 0 is left out of the interval.

That means we can be almost sure both means don't have the same value, confirming the results of the previous test.

Step-by-step explanation:

The null and alternative hypothesis are:

$H_0: \mu_1-\mu_2=0\\\\H_a: \mu_1-\mu_2\neq 0$

The significance level is 0.10 and it will be a two-tailed test.

The difference of the sample means is:

$M_d=M_1-M_2=4.340-7.091=-2.751$

As the sample size is equal for both samples, the estimated standard error of the difference between means is calculated as:

$s_M_d=\sqrt{\frac{s_1^2+s_2^2}{N}}= \sqrt{\frac{0.508^2+0.430^2}{5}}=\sqrt{\frac{0.442964}{5} }= \sqrt{0.0886}=0.2976$

Then, the statistic z is:

$z=\frac{M_d-(\mu_1-\mu_2)}{s_M_d}=\frac{-2.751-0}{0.2976}=-9.24\\\\P(|z|>9.24)=0$

The P-value (P=0) is much lower than the significance level, so the null hypothesis is rejected. The means are different.

For a 90% confidence interval for the difference of the means, we use a z=1.645.

Then the confidence interval is defined as:

$M_d-z*s_M_d\leq \mu_1-\mu_2 \leq M_d-z*s_M_d\\\\-2.751-1.645*0.2976\leq \mu_1-\mu_2 \leq -2.751+1.645*0.2976\\\\-3.2406\leq \mu_1-\mu_2 \leq -2.2614$

This CI tells us that there is a 90% confidence that the real value of the difference between the means is between this two values. We see that both are negative values, so the value 0 is left out of the interval.

That means we can be almost sure both means don't have the same value.