Question

Find the sample size needed to estimate the percentage of adults who can wiggle their ears. Use a margin of error of 4 percentage points and use a confidence level of 90%.

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Assume that p and q are unknown.
n= ☐
(Round up to the nearest integer.)

Answer 1

To estimate the percentage of adults who can wiggle their ears with a 4% margin of error and 90% confidence, a sample size of approximately 68 is needed, rounded up to the nearest integer.

To find the sample size (n) needed to estimate the percentage of adults who can wiggle their ears with a margin of error of 4 percentage points and a confidence level of 90%, you can use the formula:

`\[ n = \frac{{Z^2 \cdot p \cdot q}}{{E^2}} \]`

where:

- Z is the Z-score corresponding to the desired confidence level (for 90%, it's approximately 1.645).

- p is the estimated proportion (use 0.5 for maximum variability, as it gives the maximum sample size needed).

- q is 1 - p.

- E is the margin of error (expressed as a decimal).

Substitute the values into the formula:

`\[ n = \frac{{1.645^2 \cdot 0.5 \cdot 0.5}}{{0.04^2}} \]`

Calculate this expression to find the sample size needed. After calculating, round up to the nearest integer.

`\[ n \approx \frac{{1.645^2 \cdot 0.25}}{{0.0016}} \]\[ n \approx \frac{{0.429 \cdot 0.25}}{{0.0016}} \]\[ n \approx \frac{{0.10725}}{{0.0016}} \]\[ n \approx 67.03125 \]`

Round up to the nearest integer:

`\[ n \approx 68 \]`

Therefore, the sample size needed is approximately 68.