Questions: A philanthropic organisation sent free mailing labels and greeting cards to a random sample of 100,000 potential donors on their mailing list and received 4608 donations. (a) Give a 95% 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 4.7%. Given the confidence interval you found, do you find that rate plausible?

Solution Steps

The sample proportion of donations is calculated as follows:

\[ \hat{p} = \frac{x}{n} = \frac{4608}{100000} = 0.04608 \]

To find the \(95\%\) confidence interval for the true proportion, we use the formula:

\[ \hat{p} \pm z \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} \]

Where:

• \(z\) is the z-score corresponding to the \(95\%\) confidence level, which is approximately \(1.96\).
• \(n\) is the sample size, \(100000\).

Calculating the margin of error:

\[ \text{Margin of Error} = 1.96 \cdot \sqrt{\frac{0.04608(1 - 0.04608)}{100000}} \approx 0.002 \]

Thus, the confidence interval is:

\[ 0.04608 \pm 0.002 = (0.04408, 0.04808) \]

Expressing this in percentage terms:

\[ (0.04408 \times 100\%, 0.04808 \times 100\%) = (4.408\%, 4.808\%) \]

The staff member's proposed rate is \(4.7\%\). We check if this rate falls within the calculated confidence interval:

\[ 4.408\% \leq 4.7\% \leq 4.808\% \]

Since \(4.7\%\) lies within the interval, it is considered plausible.

Final Answer