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Fofino [41]
3 years ago
5

You may need to use the appropriate technology to answer this question. Consider the following hypothesis test. H0: μd ≤ 0 Ha: μ

d > 0 (a) The following data are from matched samples taken from two populations. Compute the difference value for each element. (Use Population 1 − Population 2.) Element Population Difference 1 2 1 21 19 2 Correct: Your answer is correct. 2 28 25 3 Correct: Your answer is correct. 3 18 15 3 Correct: Your answer is correct. 4 20 19 1 Correct: Your answer is correct. 5 26 25 1 Correct: Your answer is correct. (b) Compute d. 2 Correct: Your answer is correct. (c) Compute the standard deviation sd. 0.89 Incorrect: Your answer is incorrect. (d) Conduct a hypothesis test using α = 0.05. Calculate the test statistic. (Round your answer to three decimal places.) 5 Incorrect: Your answer is incorrect. Calculate the p-value. (Round your answer to four decimal places.) p-value = 0.9999 Incorrect: Your answer is incorrect. What is your conclusion? Do not reject H0. There is insufficient evidence to conclude that μd > 0. Do not Reject H0. There is sufficient evidence to conclude that μd > 0. Reject H0. There is sufficient evidence to conclude that μd > 0. Reject H0. There is insufficient evidence to conclude that μd > 0. Correct: Your answer is correct.
Mathematics
1 answer:
Lena [83]3 years ago
5 0

Answer

hello ,

it looks your question on this site is not clear please upload for me a word document with full details of your question to my g mail which is [email protected] for clarity and accuracy.

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Find the general solution of the differential equation and check the result by differentiation. (Use C for the constant of integ
atroni [7]

Answer: y=Ce^(^3^t^{^9}^)

Step-by-step explanation:

Beginning with the first differential equation:

\frac{dy}{dt} =27t^8y

This differential equation is denoted as a separable differential equation due to us having the ability to separate the variables. Divide both sides by 'y' to get:

\frac{1}{y} \frac{dy}{dt} =27t^8

Multiply both sides by 'dt' to get:

\frac{1}{y}dy =27t^8dt

Integrate both sides. Both sides will produce an integration constant, but I will merge them together into a single integration constant on the right side:

\int\limits {\frac{1}{y} } \, dy=\int\limits {27t^8} \, dt

ln(y)=27(\frac{1}{9} t^9)+C

ln(y)=3t^9+C

We want to cancel the natural log in order to isolate our function 'y'. We can do this by using 'e' since it is the inverse of the natural log:

e^l^n^(^y^)=e^(^3^t^{^9} ^+^C^)

y=e^(^3^t^{^9} ^+^C^)

We can take out the 'C' of the exponential using a rule of exponents. Addition in an exponent can be broken up into a product of their bases:

y=e^(^3^t^{^9}^)e^C

The term e^C is just another constant, so with impunity, I can absorb everything into a single constant:

y=Ce^(^3^t^{^9}^)

To check the answer by differentiation, you require the chain rule. Differentiating an exponential gives back the exponential, but you must multiply by the derivative of the inside. We get:

\frac{d}{dx} (y)=\frac{d}{dx}(Ce^(^3^t^{^9}^))

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*\frac{d}{dx}(3t^9)

\frac{dy}{dx} =(Ce^(^3^t^{^9}^))*27t^8

Now check if the derivative equals the right side of the original differential equation:

(Ce^(^3^t^{^9}^))*27t^8=27t^8*y(t)

Ce^(^3^t^{^9}^)*27t^8=27t^8*Ce^(^3^t^{^9}^)

QED

I unfortunately do not have enough room for your second question. It is the exact same type of differential equation as the one solved above. The only difference is the fractional exponent, which would make the problem slightly more involved. If you ask your second question again on a different problem, I'd be glad to help you solve it.

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