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): 24, 19, 17, 21, 24, 28, 22, 27, 15, 23, 26, 20, 21. 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. O A. Moderate statistical significance O B. No statistical significance O C. Strong statistical significance O D. Very strong statistical significance Reset Selection

Answer 1

The t-statistic is computed as follows: V X – uo s/vn 21.3077 – 25 == -3.354 3.9692/V13 (4) Decision about the null hypothesis Since it is observed that t = -3.354 < tc = -2.681, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0029, and since p = 0.0029 < 0.01, it is concluded that the null hypothesis is rejected. 2-0.Oy (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 25, at the 0.01 significance level.

.(3) Test Statistics The t-statistic is computed as follows: t= X – Mo s//m 21.3077 – 25 3.9692/13 = -3.354 (4) Decision about the null hypothesis Since it is observed that t = -3.354 < to = -1.782, it is then concluded that the null hypothesis is rejected. Using the P-value approach: The p-value is p=0.0029, and since p= 0.0029 < 0.05, it is concluded that the null hypothesis is rejected. a=o.os (5) Conclusion It is concluded that the null hypothesis Ho is rejected. Therefore, there is enough evidence to claim that the population mean u is less than 25, at the 0.05 significance a level
Very stron statistical significance:

p value less then 0.05 and 0.01 also..

i attached image check...

for 0.05 and 0.01