Question

In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home. Find a 96% confidence interval for the true proportion of voters who feel this way. a.Find a point estimate for p, the population proportion who thought US spends too little on fighting hunger at home.b.Find a 96% confidence interval for p.

Answer 1

Answer:

a. 0.48

b. The 96% confidence interval for p is (0.457, 0.503).

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 $1 - \frac{\alpha}{2}$.

In a poll of 2000 likely voters, 960 say that the US spends too little on fighting hunger at home.

This means that $n = 2000, \pi = \frac{960}{2000} = 0.48$.

This is the answer for question a.

96% confidence level

So $\alpha = 0.04$, z is the value of Z that has a pvalue of $1 - \frac{0.04}{2} = 0.98$, so $Z = 2.056$.

The lower limit of this interval is:

$\pi - z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 - 2.056\sqrt{\frac{0.48*0.52}{2000}} = 0.457$

The upper limit of this interval is:

$\pi + z\sqrt{\frac{\pi(1-\pi)}{n}} = 0.48 + 2.056\sqrt{\frac{0.48*0.52}{2000}} = 0.503$

The 96% confidence interval for p is (0.457, 0.503).