Solving for a Confidence Interval: Algebra 2 points possible (graded) In the problems on this page, we will continue building the confidence interval of asymptotical level 95% by solving for p as in the video. Recall that R1,…,Rn∼iidBer(p) for some unknown parameter p , and we estimate p using the estimator p^=R¯¯¯¯n=1n∑i=1nRi.
As in the method using a conservative bound, our starting point is the result of the central limit theorem:
In this second method, we solve for values of P that satisfy the inequality volves penat che non esito para polcomp R -P
To do this, we manipulate - ulate | " Vp(1-) 5 < 90/2 into an inequality involving a quadratic function App + Bp+C where A > 0, B, C la/2 into an inequality in depend on 13, 4a/2, and R. Which of the following is the correct inequality?
(We will use find the values of A, B, and C in the next problem.)
1. Ap^2 + Bp + C<0 where A >0.
2. Ap^2 + Bp+C>Owhere A >0.
Let P1 and P2 with 0 a. (P P2)
b. P
c. o
d. P2
e. o

Question

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Respuesta :

shallomisaiah19 shallomisaiah19 · Answer 1 · 2020-06-10T18:29:53+02:00

Answer:

Step-by-step explanation:

1) The given inequality is

[tex]|\sqrt{n} \frac{(\bar R_n-p)}{\sqrt{p(1-p)} } |<q_{\alpha /2}| \\\\ \to(\frac{(\sqrt{n} \bar R_n-p)}{\sqrt{p(1-p)} })<q^2_{\alpha /2}[/tex]

[tex]\to n( \bar R _n - p)^2<p(1-p)q^2_{\alpha /2}[/tex]

[tex]\to n\bar R +np^2-2nR_np<q^2_{\alpha /2 p- q^2_{\alpha /2}p^2[/tex]

Arranging the terms with p² and p, we get

[tex]p^2(n+q^2_{\alpha /2)-p(2n \bar R _n+q^2_{\alpha / 2})+n \bar R ^2 _n <0[/tex]

Hence, the inequality is of the form

Ap² + Bp + c < 0

2. A quadratic equation of the form

Ap² + Bp + c < 0 with A > 0 looks like

Check the attached image

The region where the values are negative lies between p₁ and p₂ ...

The p₁ < p < p₂