Question

The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free mailing labels and greeting cards to potential donors on their list and ask for a voluntary contribution. To test a new campaign, they recently sent letters to a random sample of 100,000 potential donors and received 4781 donations. Give a 95% confidence interval for the true proportion of their entire mailing list who may donate.

Answer 1

Answer:

Confidence interval:  (0.04649,0.04913)

Step-by-step explanation:

We are given the following in the question:

Sample size, n =  100,000

Number of people who donated, x = 4781

$\hat{p} = \dfrac{x}{n} = \dfrac{4781}{100000} = 0.04781$

95% Confidence interval:

$\hat{p}\pm z_{stat}\sqrt{\dfrac{\hat{p}(1-\hat{p})}{n}}$

$z_{critical}\text{ at}~\alpha_{0.05} = 1.96$

Putting the values, we get:

$0.04781 \pm 1.96(\sqrt{\dfrac{0.04781 (1-0.04781 )}{100000}})\\\\=0.04781 \pm 0.00132\\\\=(0.04649,0.04913)$

is the required 95% confidence interval  for the true proportion of their entire mailing list who may donate.