a random sample of 200 voters in a town is selected, and 114 are found to support annexation suit. find the 96% confidence interval for the fraction of the voting population favoring the suit.

Respuesta :

The 96% confidence interval for the fraction of the voting population favoring the suit is 0.498< p< 0.642.

Here we have to find the confidence interval.

Given

Sample, n = 200 voters

Let x represent those that support the annexation suit.

x = 114

First, we have to calculate the probability of supporting the annexation suit.

Let p represent the probability of supporting the annexation suit.

p = x/n

p = 114/200

p = 0.57

From elementary probability;

p + q = 1 where q represents the probability of failure.

In this case, q represent the probability of not supporting the annexation suit

Substitute 0.57 for p

0.57 + q = 1

q = 1 - 0.57

q = 0.43

To find the 96% confidence interval for the fraction of the voting population favoring the suit;

The confidence interval is bounded by the following;

^p - z(α/2) √(pq/n) < p < ^p + z(α/2) √(pq/n)

At this point, we have values for p,q, and n.

Next is to solve z(α/2)

First, we'll find the value of α/2 using

C.I = 100%(1 - α) where C.I = 96%

96% = 100%(1 - α)

1 - α = 96%

1 - α = 0.96

α = 1 - 0.96

α = 0.04

So,

α/2 = 0.04/2

α/2 = 0.02

So, z(α/2) = z(0.02)

Using a normal probability table

z0.02 = 2.055 --- This is the closest value which leaves an area of 0.02 to the right and 0.98 to the left

Recalling our formula to solve 96% interval;

^p - z(α/2) √(pq/n) < p < ^p + z(α/2) √(pq/n)

By substitution, we have

0.57 - 2.055 * √(0.57*0.43/200) < p < 0.57 + 2.055 * √(0.57*0.43/200)

0.57 - 0.071939677073920 < p < 0.57 + 0.071939677073920

0.498060322926079 < p < 0.641939677073920 ---- Approximate

0.498 < p < 0.642

Therefore the confidence interval is 0.498<p<0.642.

To know more about the confidence interval refer to the link given below:

https://brainly.com/question/17212516

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