Answer:
Step-by-step explanation:
Hello!
The variable of interest is
X: activation time of a sprinkler/fire alarm system.
The claim is that the new system has an average activation time of at most 20 seconds.
To test this claim, a sample of 12 test results of the fire alarm/sprinkler was taken, the resulting sample mean is X[bar]= 21.5 seconds. Assume that activation times for this system are Normally distributed with s = 3 seconds.
The hypotheses are:
H₀: μ = 20
H₁: μ > 20
Using α: 0.05
The statistic for this test is a one sample t-test:
[tex]t_{H_0}= \frac{X[bar]-Mu}{\frac{S}{\sqrt{n} } } = \frac{21.5-20}{\frac{3}{\sqrt{12} } } = 1.73[/tex]
The p-value for this test is p-value: 0.0558
The p-value is greater than the level of significance so the decision is to not reject the null hypothesis.
If the average activation time is, in fact, 20 seconds then the null hypothesis is false.
The situation is that the null hypothesis was not rejected given that it is false. The error that was committed was a type II error.
I hope it helps!
