Question

A poll was conducted by the marketing department of a video game company to determine the popularity of a new game that was targeted to be launched in three months. Telephone interviews with 1000 young adults were conducted which revealed that 50% said they would purchase the new game. The margin of error was ±4 percentage points. What is the needed sample size to obtain a 90% confidence interval estimate of the percentage of the targeted young adults who will purchase the new game by allowing the same level of margin of error?

Answer 1

To determine the needed sample size to obtain a 90% confidence interval estimate with the same level of margin of error (±4 percentage points), we can use the formula for the sample size for estimating proportions:

\[
n = \frac{Z^2 \cdot p \cdot (1 - p)}{E^2}
\]

Where:
- \( n \) is the sample size,
- \( Z \) is the z-score corresponding to the desired confidence level,
- \( p \) is the estimated proportion (from the poll),
- \( E \) is the margin of error.

**Step 1: Identify the values:**
- For a 90% confidence level, the z-score \( Z \) is approximately \( 1.645 \) (this can be found from standard normal distribution tables).
- The estimated proportion \( p \) is 0.50 (50%).
- The margin of error \( E \) is 0.04 (4 percentage points).

**Step 2: Plug the values into the formula:**

\[
n = \frac{(1.645)^2 \cdot 0.5 \cdot (1 - 0.5)}{(0.04)^2}
\]

Calculating the components step by step:

\[
Z^2 = (1.645)^2 \approx 2.706
\]

\[
p \cdot (1 - p) = 0.5 \cdot (1 - 0.5) = 0.5 \cdot 0.5 = 0.25
\]

Now substituting those values back into our formula:

\[
n = \frac{2.706 \cdot 0.25}{0.0016}
\]

Calculating the numerator:

\[
2.706 \cdot 0.25 = 0.6765
\]

Now calculate:

\[
n = \frac{0.6765}{0.0016} \approx 422.8125
\]

Since the sample size must be a whole number, we round up to ensure the sample is sufficiently large:

\[
n \approx 423
\]

**Conclusion:**
The needed sample size to obtain a 90% confidence interval estimate of the percentage of the targeted young adults who will purchase the new game with a margin of error of ±4 percentage points is **423**.