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Anit [1.1K]
2 years ago
10

A sprinkler system is being installed in a newly renovated building on campus. The average activation time is supposed to be at

most 20 seconds. A series of 12 fire alarm/sprinkler system tests results in an average activation time of 21.5 seconds. Do these data indicate that the design specifications have not been met? The hypotheses to be tested are H0: m = 20 versus Ha: m > 20, where m = the true average activation time of the sprinkler system. Assume that activation times for this system are Normally distributed with s = 3 seconds.
(a) What is the value of the observed test statistic?
(b) What is the value of the P-value?
(c) Are the data statistically significant at the 5% significance level? Explain briefly.
(d) What does the decision you made mean with respect to the question "Do these data indicate that the design specifications have not been met?"
(e) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, what type of error would you have made?

Mathematics
1 answer:
melomori [17]2 years ago
4 0

Answer:

A) t = 1.73

B) p-value =  0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) The decision means that the design specifications are not met.

E) Type II error

Step-by-step explanation:

The hypotheses are:

H₀: μ = 20

H₁: μ > 20

A) Formula for the test statistic is;

t = (x' - μ)/(s/√n)

Now, we are given;

x' = 21.5

μ = 20

s = 3

n = 12

Thus;

t = (21.5 - 20)/(3/√12)

t = 1.73

B) we have our t-value as 1.73

Now, Degree of freedom(DF) = n - 1

So,DF = 12 - 1 = 11

Using significance level of α = 0.05, t-value = 1.73 and DF = 11, one tailed hypothesis, from online P-value calculator attached, we have;

p-value =  0.0558

C) Data is not statistically significant because the p-value of 0.0558 is more than the significance value of 0.05

D) We will not reject the null hypothesis. The decision means that the design specifications are not met.

E) If the true average activation time of the sprinkler system is, in fact, equal to 20 seconds, then the null hypothesis is false.

Since we did not reject the null hypothesis even though it is false, the error that was committed was therefore a type II error.

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