Answer:
a. 0.498 < p < 0.642
b. We are 96% sure that the error of estimator ^p = 0.57 will not exceed 0.07001
Explanation:
Given;
Sample, n = 200 voters
Let x represent those that support annexation suit.
x = 114
First, we'll calculate the probability of supporting annexation suit.
Let p represent the probability of supporting annexation suit.
p = x/n
p = 114/200
p = 0.57
From elementary probability;
p + q = 1 where q represent probability of failure.
In this case, q represent probability of not supporting 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 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 substituton, 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
b. Here, we'll make use of the following theorem;
Using ^p as an estimate
We are 100%(1 - α) confident that the error will not exceed z(α/2) √(pq/n)
From (a), we have.
z(α/2) = 2.055, p = 0.57, q = 0.43, n = 200
By substituton, z(α/2) √(pq/n) becomes
2.055 * √(0.57 * 0.43/200)
= 2.055 * 0.071939677073920
= 0.070014284256857 ---- Approximate
= 0.07001
We are 96% sure that the error of estimator ^p = 0.57 will not exceed 0.07001