The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free malling 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.
(a) Give a confidence interval for the true proportion of those from their entire maling list who may donate.
(b) A staff member thinks that the true rate is . Given the confidence interval you found, do you find that rate plausible?
(a) What is the confidence interval?
The confidence interval is from to .
(Round to two decimal places as needed.)

Ask by Harmon Lyons. in the United States

Mar 31,2025

Question

The Paralyzed Veterans of America is a philanthropic organization that relies on contributions. They send free malling 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.
(a) Give a confidence interval for the true proportion of those from their entire maling list who may donate.
(b) A staff member thinks that the true rate is . Given the confidence interval you found, do you find that rate plausible?
(a) What is the confidence interval?
The confidence interval is from to .
(Round to two decimal places as needed.)

Ask by Harmon Lyons. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We start by calculating the sample proportion:
$$
\hat{p} = \frac{4781}{100\,000} = 0.04781.
$$

Next, we use the formula for a $$95\%$$ confidence interval for a proportion:
$$
\hat{p} \pm z_{\alpha/2} \sqrt{\frac{\hat{p}(1-\hat{p})}{n}},
$$
where $$z_{0.025} = 1.96$$ (for a $$95\%$$ level) and $$n = 100\,000$$.

Now, compute the standard error:
$$
SE = \sqrt{\frac{0.04781(1-0.04781)}{100\,000}} = \sqrt{\frac{0.04781 \times 0.95219}{100\,000}}.
$$
Calculate the numerator:
$$
0.04781 \times 0.95219 \approx 0.04552.
$$
Thus,
$$
SE \approx \sqrt{\frac{0.04552}{100\,000}} = \sqrt{0.0000004552} \approx 0.0006747.
$$

Determine the margin of error:
$$
E = 1.96 \times SE \approx 1.96 \times 0.0006747 \approx 0.00132.
$$

Now, form the confidence interval:
$$
\hat{p} \pm E = 0.04781 \pm 0.00132.
$$
The lower limit is:
$$
0.04781 - 0.00132 \approx 0.04649,
$$
and the upper limit is:
$$
0.04781 + 0.00132 \approx 0.04913.
$$

When expressed as percentages and rounded to two decimal places, the $$95\%$$ confidence interval is from:
$$
4.65\% \text{ to } 4.91\%.
$$

(b) Since a staff member believes that the true rate is $$4.6\%$$, we compare this value with our confidence interval. The value $$4.6\%$$ is outside the interval $$4.65\%$$ to $$4.91\%$$, so based on this confidence interval, the rate of $$4.6\%$$ does not appear to be plausible.
(a) The $$95\%$$ confidence interval for the true donation rate is from $$4.65\%$$ to $$4.91\%$$. (b) The staff member's belief that the true rate is $$4.6\%$$ is not plausible based on the confidence interval.

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