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;

$x^{bar}$ = $\frac{1}{n}$ ∑ $x_{i}$

where n is sample size and ∑ $x_{i}$ is summation of all the sample;

= $\frac{1}{13}$ ( 27 + 41 + 22 + 27 + 23 + 35 + 30 + 33 + 24 + 27 + 28 + 22 + 24 )

= $\frac{1}{13}$ ( 363

sample mean $x^{bar}$ = 27.9231

next we find the standard deviation

s = √( $\frac{1}{n-1}$ ∑( $x_{i}-x^{bar}$ )²

x ( $x_{i}-x^{bar}$ ) ( $x_{i}-x^{bar}$ )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 ∑( $x_{i}-x^{bar}$ )² = 378.9229

∴

s = √( $\frac{1}{13-1}$ ×378.9229 )

s = √31.5769

standard deviation s = 5.6193

now, the Test statistics

t = ( $x^{bar}$ - μ ) / $\frac{s}{\sqrt{n} }$

we substitute

t = ( 27.9231 - 25 ) / $\frac{5.6193}{\sqrt{13} }$

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

3 0

3 years ago

One solution has a formula C (n) H (2n) O (n) If this material weighs 288 grams, dissolves in weight 90 grams, the solution will

saw5 [17]

Explanation:

The given data is as follows.

Boiling point of water ( $T^{o}_{b}) = 100^{o}C$ = (100 + 273) K = 323 K,

Boiling point of solution ( $T_{b}) = 101.24^{o}C$ = (101.24 + 273) K = 374.24 K

Hence, change in temperature will be calculated as follows.

$\Delta T_{b} = (T_{b} - T^{o}_{b})$

= 374.24 K - 323 K

= 1.24 K

As molality is defined as the moles of solute present in kg of solvent.

Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

Let molar mass of the solute is x grams.

Therefore, Molality = $\frac{\text{weight of solute \times 1000}}{\text{molar mass of solute \times mass f solvent(g)}}$

m = $\frac{288 g \times 1000}{x g \times 90}$

= $\frac{3200}{x}$

As, $\Delta T_{b} = k_{b} \times molality$

$1.24 = 0.512 ^{o}C/m \times \frac{3200}{x}$

x = $\frac{0.512 ^{o}C/m \times 3200}{1.24}$

= 1321.29 g

This means that the molar mass of the given compound is 1321.29 g.

It is given that molecular formula is $C_{n}H_{2n}O_{n}$ .

As, its empirical formula is $CH_{2}O$ and mass is 30 g/mol. Hence, calculate the value of n as follows.

n = $\frac{\text{Molecular mass}}{\text{Empirical mass}}$

= $\frac{1321.29 g}{30 g/mol}$

= 44 mol

Thus, we can conclude that the formula of given material is $C_{44}H_{88}O_{44}$ .

4 0

3 years ago

Backpackers often use canisters of white gas to fuel a cooking stove’s burner. If one canister contains 1.45 L of white gas, and

rusak2 [61]

1 L ------- 1000 cm³
1.45 L ----- ???

1.45 * 1000 = 1450 cm³ ( volume )

Density = 0.710 g/cm³

mass = in Kg

m = D * V

m = 0.710 * 1450

m = 1029.5 g

1 Kg ------- 1000 g
kg -------- 1029.5 g

mass = 1029.5 / 1000

mass = 1.0295 Kg

hope this helps!

7 0

2 years ago

Choose the correct statement regarding the relative size of atoms:

Sonbull [250]

Answer:

The atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.

Explanation:

As we travel along a period in a periodic table then the atomic radii decreases

This is because as we travel along a period we have that the atomic number of the atoms increases which means the the number of proton's increased

But the electron's add to the same outer shell throughout the period , which means the effective nuclear charge increases which pulls the outer electrons toward's the nucleus and the size decreases.

Therefore the atoms on left side are larger than the atoms on the right side of the periodic table because those on the right have more proton's.

6 0

3 years ago