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

a. 0.48

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

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{(\pi(1-\pi))/(n)}`

In which

z is the zscore that has a pvalue of
`1 - (\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 = (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 - (0.04)/(2) = 0.98`, so
`Z = 2.056`.

The lower limit of this interval is:

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

The upper limit of this interval is:

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

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