Question

The times of first sprinkler activation for a series of tests with fire prevention sprinkler systems using an aqueous film-forming were (in sec): 23, 27, 34, 26, 31, 36, 19, 28, 29, 21, 22, 39, 28. The system has been designed so that true average activation time is at most 25 sec under such conditions. Does the data strongly contradict the validity of this design specification? Test the relevant hypotheses at significance level .05 using the P-value approach.

A) No statistical significance B) Moderate statistical significance C) Very strong statistical significance D) Strong statistical significance

Answer 1

	Values ( X )	Σ ( Xi- X̅ )2
	23	24.2369
	27	0.8521
	34	36.9287
	26	3.6983
	31	9.4673
	36	65.2363
	19	79.6217
	28	0.0059
	29	1.1597
	21	47.9293
	22	35.0831
	39	122.6977
	28	0.0059
Total	363	426.9229

Mean X̅ = Σ Xi / n
X̅ = 363 / 13 = 27.9231
Sample Standard deviation S_X = √ ( (Xi - X̅ )² / n - 1 )
S_X = √ ( 426.9229 / 13 -1 ) = 5.9646

To Test :-
H0 :- µ <= 25
H1 :- µ > 25

Test Statistic :-
t = ( X̅ - µ ) / (S / √(n) )
t = ( 27.9231 - 25 ) / ( 5.9646 / √(13) )
t = 1.767

Decision based on P value
P - value = P ( t > 1.767 ) = 0.0513
Reject null hypothesis if P value < α = 0.05 level of significance
P - value = 0.0513 > 0.05 ,hence we fail to reject null hypothesis
Conclusion :- Fail to reject null hypothesis

No statistical significance

There is insufficient evidence to contradict the validity of this design specification at 5% level of significance.