Atoms like sodium and lithium often lose electrons, which makes them
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This question is incomplete, the complete question is;
The times of first sprinkler activation (in seconds) for a series of tests of fire-prevention sprinkler systems that use aqueous film-forming foam is as follows; 27 41 22 27 23 35 30 33 24 27 28 22 24
( see " use of AFFF in sprinkler systems," Fire technology, 1975: 5)
The system has been designed so that the true average activation time is supposed to be at most 25 seconds.
Does the data indicate the design specifications have not been met?
Test the relevant hypothesis at significance level 0.05 using the P-value approach
Answer:
since p-value (0.042299) is lesser than the level of significance ( 0.05)
We Reject Null Hypothesis
Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds
Explanation:
Given the data in the question;
lets consider Null and Alternative hypothesis;
Null hypothesis H₀ : There is sufficient evidence that the true average activation time is at most 25 seconds
Alternative hypothesis H₁ : There is no sufficient evidence that the true average activation time is at most 25 seconds
i.e
Null hypothesis H₀ : μ ≤ 25
Alternative hypothesis H₁ : μ > 25
level of significance σ = 0.05
first we determine the sample mean;
=
∑
where n is sample size and ∑ is summation of all the sample;
= ( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )
= ( 363
sample mean = 27.9231
next we find the standard deviation
s = √( ∑(
)²
x () (
)²
27 -0.9231 0.8521
41 13.0769 171.0053
22 -5.9231 35.0831
27 -0.9231 0.8521
23 -4.9231 24.2369
35 7.0769 50.0825
30 2.0769 4.3135
33 5.0769 25.7749
24 -3.9231 15.3907
27 -0.9231 0.8521
28 0.0769 0.0059
22 -5.9231 35.0831
24 -3.9231 15.3907
sum 378.9229
so ∑()² = 378.9229
∴
s = √( ×378.9229 )
s = √31.5769
standard deviation s = 5.6193
now, the Test statistics
t = ( - μ ) /
we substitute
t = ( 27.9231 - 25 ) /
t = 2.9231 / 1.5585
t = 1.88
now degree of freedom df = n - 1 = 13 - 1 = 12
next we calculate p-value
p-value = 0.042299 ( using Execl's ( = TDIST(1.88,12,1)))
Here x=1.88, df=12, one tail
now we compare the p-value with the level of significance
since p-value (0.042299) is lesser than the level of significance ( 0.05)
We Reject Null Hypothesis
Hence, There is no sufficient evidence that the true average activation time is at most 25 seconds