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 theif
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 mailing 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 Morrison Stewart. in the United States

Mar 31,2025

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 theif
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 mailing 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 Morrison Stewart. in the United States

Mar 31,2025

UpStudy AI · Accepted Answer

We start by noting the sample information: the number of donations is $$ x=4781 $$ out of a total of $$ n=100000 $$. The sample proportion is

$$
\hat{p}=\frac{4781}{100000}=0.04781.
$$

For a $$ 95\% $$ confidence interval for a proportion, we use the formula

$$
\hat{p}\pm z\sqrt{\frac{\hat{p}(1-\hat{p})}{n}},
$$

with $$ z=1.96 $$ for $$ 95\% $$ confidence.

1. First, we compute the standard error (SE):

$$
\text{SE}=\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}=\sqrt{\frac{0.04781(1-0.04781)}{100000}}.
$$

Since

$$
1-\hat{p}=0.95219,
$$

we have

$$
0.04781\times 0.95219\approx 0.04550.
$$

Thus,

$$
\text{SE}=\sqrt{\frac{0.04550}{100000}}=\sqrt{4.55\times10^{-7}}\approx 0.0006746.
$$

2. Next, find the margin of error (ME):

$$
\text{ME}=z\cdot \text{SE}=1.96 \times 0.0006746\approx 0.001321.
$$

3. The $$ 95\% $$ confidence interval for $$ p $$ is then

$$
\hat{p}\pm \text{ME}=0.04781\pm 0.001321.
$$

This gives us:

$$
\text{Lower bound}=0.04781-0.001321\approx 0.04649,
$$
$$
\text{Upper bound}=0.04781+0.001321\approx 0.04913.
$$

Converting to percentage form and rounding to two decimal places:

$$
\text{Lower bound} \approx 4.65\%, \quad \text{Upper bound} \approx 4.91\%.
$$

Thus, the $$ 95\% $$ confidence interval is from $$ 4.65\% $$ to $$ 4.91\% $$.

For part (b), a staff member suggests that the true rate is $$ 4.6\% $$. Since $$ 4.6\% $$ is below the lower limit of our confidence interval ($$ 4.65\% $$), it is not a plausible value for the true proportion.
The $$ 95\% $$ confidence interval for the true donation rate is from $$ 4.65\% $$ to $$ 4.91\% $$. Since $$ 4.6\% $$ is below the lower bound of this interval, the staff member's suggested rate of $$ 4.6\% $$ is not plausible.

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