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ivann1987 [24]
2 years ago
7

The p-value of a statistical test depends on all of the following, except:

Chemistry
1 answer:
vaieri [72.5K]2 years ago
4 0

Answer:

The correct option is e.

Explanation:

p-value is the probability value for a given statistical model, the probability that, when the null hypothesis is true.

For two two samples the formula of test statistics is

t=\frac{(\overline{x}_1-\overline{x}_2)-(\mu_1-\mu_2)}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}

where,

\overline{x} is sample mean

\mu is population mean.

s is standard deviation.

n is sample size.

Variance is the square of standard deviation.

It means variance, mean, numbers of samples is used in calculation of p-value.

Degree of freedom define the shape of the t-distribution that your t-test uses to calculate the p-value.

d.f.=n_1+n_2-2

p-value of a statistical test depends on all of the following, except median.

Therefore the correct option is e.

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Assuming that all the energy given off in the reaction goes to heating up only the air in the house, determine the mass of metha
nirvana33 [79]

Answer:

0.92 kg

Explanation:

The volume occupied by the air is:

35.0m\times 35.0m \times 3.2m \times \frac{10^{3}L }{1m^{3} } =3.9 \times 10^{6} L

The moles of air are:

3.9 \times 10^{6} L \times \frac{1.00mol}{22.4L} =1.7 \times 10^{5}mol

The heat required to heat the air by 10.0 °C (or 10.0 K) is:

1.7 \times 10^{5}mol \times \frac{30J}{K.mol} \times 10.0 K = 5.1 \times 10^{7}J

Methane's heat of combustion is 55.5 MJ/kg. The mass of methane required to heat the air is:

5.1 \times 10^{7}J \times \frac{1kgCH_{4}}{55.5 \times 10^{6} J } =0.92kgCH_{4}

3 0
2 years ago
C5H12 + O2 = CO2 + H2O
Lady_Fox [76]

Answer: 241.6 grams of CO2

Explanation: you take 84.3 grams C5H12 and divide it by 72.15 grams of C5H12(which is the molar mass) you take that answer and calculate the mols of CO2 by multiplying the 1.168 you got before and multiply it by 5. You take the answer you get from that and multiply it by the molar mass of CO2 and get the theoretical yield and then you just plug it in. 94= (x/257.02)x100 and solve to find x which is the actual yield.

3 0
3 years ago
(( PLEASE HELP QUICK ))Which expression can be used to calculate the ratio of the rate of effusion of gas A to the rate of effus
borishaifa [10]

Answer:

The square root of the molar mass of B ÷ the square root of the molar mass of A  

Explanation:

Graham’s Law applies to the effusion of gases:

The rate of effusion (r) of a gas is inversely proportional to the square root of its molar mass (M).

r \propto \dfrac{1}{\sqrt{M}}

If you have two gases A and B, the ratio of their rates of effusion is

\dfrac{r_{\text{A}}}{r_{\text{B}}} = \sqrt{\dfrac{M_{\text{B}}}{M_{\text{A}}}}

3 0
3 years ago
The volume of a fixed amount of gas is doubled, and the absolute temperature is doubled. According to the ideal gas law, how has
neonofarm [45]

Answer:

Option A. It has stayed the same.

Explanation:

To answer the question given above, we assumed:

Initial volume (V₁) = V

Initial temperature (T₁) = T

Initial pressure (P₁) = P

From the question given above, the following data were:

Final volume (V₂) = 2V

Final temperature (T₂) = 2T

Final pressure (P₂) =?

The final pressure of the gas can be obtained as follow:

P₁V₁/T₁ = P₂V₂/T₂

PV/T = P₂ × 2V / 2T

Cross multiply

P₂ × 2V × T = PV × 2T

Divide both side by 2V × T

P₂ = PV × 2T / 2V × T

P₂ = P

Thus, the final pressure is the same as the initial pressure.

Option A gives the correct answer to the question.

3 0
3 years ago
Adding an oxygen atom to a carbon atom in an organic molecule causes the carbon atom to become more?
Marat540 [252]

Answer:

diluted  or stronger

Explanation:this is caused by the addition of oxygen and it is normally overbearing of ends up changing the whole molecule

3 0
3 years ago
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