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damaskus [11]
3 years ago
13

Suppose a researcher is testing the hypothesis Upper H 0​: pequals0.4 versus Upper H 1​: pless than0.4 and she finds the​ P-valu

e to be 0.33. Explain what this means. Would she reject the null​ hypothesis? Why? Choose the correct explanation below. A. If the​ P-value for a particular test statistic is 0.33​, she expects results no more extreme than the test statistic in about 33 of 100 samples if the null hypothesis is true. B. If the​ P-value for a particular test statistic is 0.33​, she expects results no more extreme than the test statistic in exactly 33 of 100 samples if the null hypothesis is true. C. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in about 33 of 100 samples if the null hypothesis is true. D. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true. Choose the correct conclusion below. A. Since this event is​ unusual, she will reject the null hypothesis. B. Since this event is not​ unusual, she will reject the null hypothesis. C. Since this event is​ unusual, she will not reject the null hypothesis. D. Since this event is not​ unusual, she will not reject the null hypothesis.
Mathematics
1 answer:
andre [41]3 years ago
3 0

Answer:

D. If the​ P-value for a particular test statistic is 0.33​, she expects results at least as extreme as the test statistic in exactly 33 of 100 samples if the null hypothesis is true.

D. Since this event is not​ unusual, she will not reject the null hypothesis.

Step-by-step explanation:

Hello!

You have the following hypothesis:

H₀: ρ = 0.4

H₁: ρ < 0.4

Calculated p-value: 0.33

Remember: The p-value is defined as the probability corresponding to the calculated statistic if possible under the null hypothesis (i.e. the probability of obtaining a value as extreme as the value of the statistic under the null hypothesis).

In this case, you have a 33% chance of getting a value as extreme as the statistic value if the null hypothesis is true. In other words, you would expect results as extreme as the calculated statistic in 33 about 100 samples if the null hypothesis is true.

You didn't exactly specify a level of significance for the test, so, I'll use the most common one to make a decision: α: 0.05

Remember:

If p-value ≤ α, then you reject the null hypothesis.

If p-value > α, then you do not reject the null hypothesis.

Since 0.33 > 0.05 then I'll support the null hypothesis.

I hope it helps!

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Let A(t) denote the amount of salt (in ounces, oz) in the tank at time t (in minutes, min).

Salt flows in at a rate of

\dfrac{dA}{dt}_{\rm in} = \left(17 (1 + 15 \sin(t)) \dfrac{\rm oz}{\rm gal}\right) \left(8\dfrac{\rm gal}{\rm min}\right) = 136 (1 + 15 \sin(t)) \dfrac{\rm oz}{\min}

and flows out at a rate of

\dfrac{dA}{dt}_{\rm out} = \left(\dfrac{A(t) \, \mathrm{oz}}{180 \,\mathrm{gal} + \left(8\frac{\rm gal}{\rm min} - 8\frac{\rm gal}{\rm min}\right) (t \, \mathrm{min})}\right) \left(8 \dfrac{\rm gal}{\rm min}\right) = \dfrac{A(t)}{180} \dfrac{\rm oz}{\rm min}

so that the net rate of change in the amount of salt in the tank is given by the linear differential equation

\dfrac{dA}{dt} = \dfrac{dA}{dt}_{\rm in} - \dfrac{dA}{dt}_{\rm out} \iff \dfrac{dA}{dt} + \dfrac{A(t)}{180} = 136 (1 + 15 \sin(t))

Multiply both sides by the integrating factor, e^{t/180}, and rewrite the left side as the derivative of a product.

e^{t/180} \dfrac{dA}{dt} + e^{t/180} \dfrac{A(t)}{180} = 136 e^{t/180} (1 + 15 \sin(t))

\dfrac d{dt}\left[e^{t/180} A(t)\right] = 136 e^{t/180} (1 + 15 \sin(t))

Integrate both sides with respect to t (integrate the right side by parts):

\displaystyle \int \frac d{dt}\left[e^{t/180} A(t)\right] \, dt = 136 \int e^{t/180} (1 + 15 \sin(t)) \, dt

\displaystyle e^{t/180} A(t) = \left(24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t)\right) e^{t/180} + C

Solve for A(t) :

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) + C e^{-t/180}

The tank starts with A(0) = 15 oz of salt; use this to solve for the constant C.

\displaystyle 15 = 24,480 - \frac{66,096,000}{32,401} + C \implies C = -\dfrac{726,594,465}{32,401}

So,

\displaystyle A(t) = 24,480 - \frac{66,096,000}{32,401} \cos(t) + \frac{367,200}{32,401} \sin(t) - \frac{726,594,465}{32,401} e^{-t/180}

Recall the angle-sum identity for cosine:

R \cos(x-\theta) = R \cos(\theta) \cos(x) + R \sin(\theta) \sin(x)

so that we can condense the trigonometric terms in A(t). Solve for R and θ :

R \cos(\theta) = -\dfrac{66,096,000}{32,401}

R \sin(\theta) = \dfrac{367,200}{32,401}

Recall the Pythagorean identity and definition of tangent,

\cos^2(x) + \sin^2(x) = 1

\tan(x) = \dfrac{\sin(x)}{\cos(x)}

Then

R^2 \cos^2(\theta) + R^2 \sin^2(\theta) = R^2 = \dfrac{134,835,840,000}{32,401} \implies R = \dfrac{367,200}{\sqrt{32,401}}

and

\dfrac{R \sin(\theta)}{R \cos(\theta)} = \tan(\theta) = -\dfrac{367,200}{66,096,000} = -\dfrac1{180} \\\\ \implies \theta = -\tan^{-1}\left(\dfrac1{180}\right) = -\cot^{-1}(180)

so we can rewrite A(t) as

\displaystyle A(t) = 24,480 + \frac{367,200}{\sqrt{32,401}} \cos\left(t + \cot^{-1}(180)\right) - \frac{726,594,465}{32,401} e^{-t/180}

As t goes to infinity, the exponential term will converge to zero. Meanwhile the cosine term will oscillate between -1 and 1, so that A(t) will oscillate about the constant level of 24,480 oz between the extreme values of

24,480 - \dfrac{267,200}{\sqrt{32,401}} \approx 22,995.6 \,\mathrm{oz}

and

24,480 + \dfrac{267,200}{\sqrt{32,401}} \approx 25,964.4 \,\mathrm{oz}

which is to say, with amplitude

2 \times \dfrac{267,200}{\sqrt{32,401}} \approx \mathbf{2,968.84 \,oz}

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