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slavikrds [6]
2 years ago
10

What is the 95% confidence interval of E(Y) at X = 4?

Mathematics
1 answer:
diamong [38]2 years ago
4 0

Based on the calculations, the 95% confidence interval of E(Y) is equal to 52.3% ± 3.26%.

<h3>How to construct a 95​% confidence interval?</h3>

Mathematically, a confidence interval of 95% is given by;

α = 1 - 0.95

α = 0.05.

α/2 = 0.05/2 = 0.025.

Next, we would determine the standard deviation of the mean:

Standard deviation of mean = 50/√800

Standard deviation of mean = 1.77.

From the z-table, the z-score of a 95% confidence interval is equal to 1.96.

Confidence interval (0.05, 50, 800) = 1.77 × 1.96

Confidence interval (0.05, 50, 800) = 3.46.

Mathematically, the confidence interval for a mean is given by:

Mean ± (t-critical × (standard deviation/√(sample size)))

Confidence interval = 52.3% ± 3.26%.

Read more on confidence interval here: brainly.com/question/25779324

#SPJ1

<u>Complete Question:</u>

A research team studied Y, the percentage of voters in favor of a candidate. The random variable Y had standard deviation o = 50%. A random sample of 800 voters was selected, and their average was y-bar(800) = 52.3%. What is the 95% confidence interval for E(Y)?

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\begin{gathered} \csc \theta=-\frac{6}{5} \\ \tan \theta>0 \\ \cos \frac{\theta}{2}=\text{?} \end{gathered}

Half Angle Formula

\cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}}\tan \theta>0\text{ and csc}\theta\text{ is negative in the third quadrant}\begin{gathered} \csc \theta=-\frac{6}{5}=\frac{r}{y} \\ x^2+y^2=r^2 \\ x=\pm\sqrt[\square]{r^2-y^2} \\ x=\pm\sqrt[\square]{6^2-(-5)^2} \\ x=\pm\sqrt[\square]{36-25} \\ x=\pm\sqrt[\square]{11} \\ \text{x is negative since the angle is on the 3rd quadrant} \end{gathered}\begin{gathered} \cos \theta=\frac{x}{r}=\frac{-\sqrt[\square]{11}}{6} \\ \cos \frac{\theta}{2}=\pm\sqrt[\square]{\frac{1+\cos\theta}{2}} \\ \cos \frac{\theta}{2}is\text{ also negative in the 3rd quadrant} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{1+\frac{-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{\frac{6-\sqrt[\square]{11}}{6}}{2}} \\ \cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}} \\  \\  \end{gathered}

Answer:

\cos \frac{\theta}{2}=-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}

Checking:

\begin{gathered} \frac{\theta}{2}=\cos ^{-1}(-\sqrt[\square]{\frac{6-\sqrt[\square]{11}}{12}}) \\ \frac{\theta}{2}=118.22^{\circ} \\ \theta=236.44^{\circ}\text{  (3rd quadrant)} \end{gathered}

Also,

\csc \theta=\frac{1}{\sin\theta}=\frac{1}{\sin (236.44)}=-\frac{6}{5}\text{ QED}

The answer is none of the choices

7 0
1 year ago
Find the function h(x) = f(x) + g(x) if f(x) = 4^x + 2 and g(x) = x - 4.
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4 0
3 years ago
How can you check that this answer is correct? 5,244 = 6= 874​
nignag [31]

I think you meant 5,244 ÷ 6 = 874.

Answer/Step-by-step explanation:

We can check if this 5,244 ÷ 6 = 874 is correct by doing it opposite.

Since it 5,244 divide 6 we can do 874 x 6.

  \left[8   7   4] \\

×    [6]

======

+ 5244

=======

5244

Hence, this answer is correct.

[RevyBreeze]

3 0
2 years ago
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