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puteri [66]
4 years ago
8

A company's old antacid formula provided relief for 70% of the people who used it. The company tests a new formula to see if it

is better and gets a P-value of 0.27. Is it reasonable to conclude that the new formula and the old one are equally effective?
A)
Yes, because there is only a 27% chance of seeing the observed effectivness just from natural sampling. There is evidence the new formula is more effective therefore we can conclude equal effectiveness.

B)
No, we can only say there is 27% chance of seeing the observed effectiveness from natural sampling variation. There is no evidence the new formula is more effective but we cannot conclude equal effectiveness.

C)
Yes, because there is only a 83% chance of seeing the observed effectivness just from natural sampling. There is evidence the new formula is more effective therefore we can conclude equal effectivness.

D)
No, we can only say there is a 83% chance of seeing the observed effectiveness just from natural sampling variation. There is no evidence the new formula is more effective but we cannot conclude equal effectiveness.
Mathematics
1 answer:
klemol [59]4 years ago
6 0

Answer:

B)

No, we can only say there is 27% chance of seeing the observed effectiveness from natural sampling variation. There is no evidence the new formula is more effective but we cannot conclude equal effectiveness.

Step-by-step explanation:

Hello!

The company compared the old antiacid formula against the new one. The claim is that the new formula is more effective.

The hypotheses are

H₀: μ₁ ≤ μ₂

H₁: μ₁ > μ₂

Where the subfix 1 represents the new formula and the subfix 2 represents the old formula.

The statistical analysis threw a p-value of 0.27.

Remember if the p-value ≥ α, n the decision is to not reject the null hypothesis.

If p-value < α, the decision is to reject the null hypothesis.

Let's say α: 0.1 ⇒ you'd decide to not reject the null hypothesis.

Then there would not be enough evidence to say the new formula is better than the old one (μ₁ > μ₂) instead you'd conclude that the new formula is at most as effective as the old one (μ₁ ≤ μ₂). To know if it is equally effective as the old one or less effective a new test should be made.

In simple words, the p-value is the probability of obtaining the value of the statistic under the null hypothesis. In this case, there is a 27% of possibility of observing the effectiveness of the new antiacid formula from a sampling error than because the new antiacid formula is, in fact, effective.

I hope it helps!

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igor_vitrenko [27]

Answer:

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7

So, the solution of the equation is

y = Ae^{m_{1} t} + Be^{m_{2} t}

where m₁ = 3 and m₂ = -7.

So,

y = Ae^{3t} + Be^{-7t}

Also,

y' = 3Ae^{3t} - 7e^{-7t}

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1      (1)

y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}

Substituting A into (1) above, we have

\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1      \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}

Substituting B into A, we have

A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}

Substituting A and B into y, we have

y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

So the solution to the differential equation is

y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}

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Given:

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The side length cannot be negative. So, the only value of x is 5\sqrt{2}.

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