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3 years ago

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The confidence interval is reported as follows: Lower 95% CI < p < Upper 95% CI For this question you will be calculating

the confidence interval but only reporting the Lower 95% CI using the Agresti-Coull Method. ( (you'll need to know the whole 95% CI for a future question) Do NOT use calculations from previous questions, this is a new scenario. In the Invisible Gorilla Experiment, 23 students were watching the video, only 11 noticed the gorilla. Calculate the 95% CI using the Agresti-Coull method, but only report the Lower 95% CI. Report your answer to 3 decimals.

1 answer:

monitta3 years ago

7 0

Answer:

lower 95% Cl is 0.292

Step-by-step explanation:

p^ = 0.4783

n = 23

for 95% Cl,

z = 1.9600

Inputting all the available data into P as attached below

LCL = 0.292

UCL = 0.670

therefore, lower 95% Cl = 0.292

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Step-by-step explanation:

We have been given that Luke bought 256 ounces of strawberries. He divided them evenly between 8 different pies. We are asked to find the pounds of strawberries that Luke put on each pie.

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3 years ago

Determine whether the sequences converge.

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$a_n=\sqrt{\dfrac{(2n-1)!}{(2n+1)!}}$

Notice that

$\dfrac{(2n-1)!}{(2n+1)!}=\dfrac{(2n-1)!}{(2n+1)(2n)(2n-1)!}=\dfrac1{2n(2n+1)}$

So as $n\to\infty$ you have $a_n\to0$ . Clearly $a_n$ must converge.

The second sequence requires a bit more work.

$\begin{cases}a_1=\sqrt2\\a_n=\sqrt{2a_{n-1}}&\text{for }n\ge2\end{cases}$

The monotone convergence theorem will help here; if we can show that the sequence is monotonic and bounded, then $a_n$ will converge.

Monotonicity is often easier to establish IMO. You can do so by induction. When $n=2$ , you have

$a_2=\sqrt{2a_1}=\sqrt{2\sqrt2}=2^{3/4}>2^{1/2}=a_1$

Assume $a_k\ge a_{k-1}$ , i.e. that $a_k=\sqrt{2a_{k-1}}\ge a_{k-1}$ . Then for $n=k+1$ , you have

$a_{k+1}=\sqrt{2a_k}=\sqrt{2\sqrt{2a_{k-1}}\ge\sqrt{2a_{k-1}}=a_k$

which suggests that for all $n$ , you have $a_n\ge a_{n-1}$ , so the sequence is increasing monotonically.

Next, based on the fact that both $a_1=\sqrt2=2^{1/2}$ and $a_2=2^{3/4}$ , a reasonable guess for an upper bound may be 2. Let's convince ourselves that this is the case first by example, then by proof.

We have

$a_3=\sqrt{2\times2^{3/4}}=\sqrt{2^{7/4}}=2^{7/8}$
$a_4=\sqrt{2\times2^{7/8}}=\sqrt{2^{15/8}}=2^{15/16}$

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Clearly, $a_1=2^{1/2}$ . Let's assume this is the case for $n=k$ , i.e. that $a_k$ . Now for $n=k+1$ , we have

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and so by induction, it follows that $a_n$ for all $n\ge1$ .

Therefore the second sequence must also converge (to 2).

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