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4 years ago

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A state biologist is investigating whether the proportion of frogs in a certain area that are bullfrogs has increased in the pas

t ten years. The proportion ten years ago was estimated to be 0.20. From a recent random sample of 150 frogs in the area, 36 are bullfrogs. The biologist will conduct a test of H0:p. What is the test statistic for the appropriate test?

1 answer:

kumpel [21]4 years ago

4 0

Here's the answer. Hope it helps.

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In a biscuit tin, there are 10 chocolate and 4 shortbread biscuits. What proportion are, a) Chocolate ?

amid [387]

Answer:

Proportion of chocolate in biscuit tin = 5:7

Step-by-step explanation:

Given:

Number of chocolate in biscuit tin = 10

Number of shortbread biscuits in biscuit tin = 4

Find:

Proportion of chocolate in biscuit tin

Computation:

Proportion of chocolate in biscuit tin = Number of chocolate in biscuit tin / Total product in biscuit tin

Proportion of chocolate in biscuit tin = 10 / [10 + 4]

Proportion of chocolate in biscuit tin = 10 / 14

Proportion of chocolate in biscuit tin = 5:7

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Helppppp me thank you

Basile [38]

Answer:

y = 5/6x + 4

Step-by-step explanation:

y = mx + b

b = 4 because when x = 0, y = 4

m = (y2 -y1) / (x2 - x1)

where two points on the equation are (x1, y1) & (x2, y2)

Let's pick points (6,9) & (-6,-1)

m = (-1 - 9) / (-6 - 6)

m = -10 / -12 = 5/6

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Which is the best meaning of the word nice?

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We would like to use the power series method to find the general solution to the differential equation d 2y dx2 − 4x dy dx + 12y

Feliz [49]

$y=\displaystyle\sum_{n\ge0}a_nx^n$

$\dfrac{\mathrm dy}{\mathrm dx}=\displaystyle\sum_{n\ge1}na_nx^{n-1}\implies4x\dfrac{\mathrm dy}{\mathrm dx}=4\sum_{n\ge1}na_nx^n=4\sum_{n\ge0}na_nx^n$

$\dfrac{\mathrm d^2y}{\mathrm dx^2}=\displaystyle\sum_{n\ge2}n(n-1)a_nx^{n-2}=\sum_{n\ge0}(n+2)(n+1)a_{n+2}x^n$

Substituting into the ODE

$\dfrac{\mathrm d^2y}{\mathrm dx^2}-4x\dfrac{\mathrm dy}{\mathrm dx}+12y=0$

gives

$\displaystyle\sum_{n\ge0}\bigg((n+2)(n+1)a_{n+2}-4na_n+12a_n\bigg)x^n=0$

so that the coefficients of the series are given according to

$\begin{cases}a_0=y(0)\\a_1=y'(0)\\a_{n+2}=\dfrac{4(n-3)a_n}{(n+2)(n+1)}&\text{for }n\ge0\end{cases}$

We can shift the index in the recursive part of this definition to get

$a_n=\dfrac{4(n-5)a_{n-2}}{n(n-1)}$

for $n\ge2$ . There's dependency between coefficients that are 2 indices apart, so we can consider 2 cases:

If $n=2k$ , where $k\ge0$ is an integer, then

$k=0\implies n=0\implies a_0=a_0$

but since $y(0)=0$ , we have $a_0=0$ and $a_{2k}=0$ for all $k\ge0$ .

If $n=2k+1$ , then

$k=0\implies n=1\implies a_1=a_1$

$k=1\implies n=3\implies a_3=\dfrac{4(-2)a_1}{3\cdot2}=-\dfrac43a_1$

$k=2\implies n=5\implies a_5=0$

and so $a_{2k+1}=0$ for all $k\ge2$ . If $y'(0)=1$ , we then have $a_1=1$ and $a_3=-\dfrac43$ .

So the ODE has solution

$y(x)=\displaystyle\sum_{k\ge0}(a_{2k}x^{2k}+a_{2k+1}x^{2k+1})\implies\boxed{y(x)=x-\dfrac43x^3}$

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3 years ago