Question

​If, based on a sample size of 950​, a political candidate finds that 563 people would vote for him in a​ two-person race, what is the 99​% confidence interval for his expected proportion of the​ vote? Would he be confident of winning based on this​ poll?

Answer 1

Answer:

The 99​% confidence interval for his expected proportion of the​ vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.

Step-by-step explanation:

In a sample with a number n of people surveyed with a probability of a success of $\pi$, and a confidence level of $1-\alpha$, we have the following confidence interval of proportions.

$\pi \pm z\sqrt{\frac{\pi(1-\pi)}{n}}$

In which

z is the zscore that has a pvalue of $1 - \frac{\alpha}{2}$.

For this problem, we have that:

$n = 950, p = \frac{563}{950} = 0.5926$

99% confidence level

So $\alpha = 0.01$, z is the value of Z that has a pvalue of $1 - \frac{0.01}{2} = 0.995$, so $Z = 2.575$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 - 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.5516$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.5926 + 2.575\sqrt{\frac{0.5926*0.4074}{950}} = 0.6336$

The 99​% confidence interval for his expected proportion of the​ vote is (0.5516, 0.6336). Since the interval for the proportion is above 50%, he would be confident of winning based on the pool.