- Identify the margin of error formula: \(ME = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\).
- Substitute the given values to solve for the critical value \(z^*\): \(z^* = \frac{0.03}{\sqrt{\frac{0.62(1-0.62)}{800}}} \approx 1.748\).
- Find the area under the normal distribution curve for \(z^* \approx 1.748\), which corresponds to a confidence level of approximately \(92\%\).
- The final confidence level is \(\boxed{92\%}\).
Explanation
- Analyze the problem and identify the formula
To determine the confidence level used in the poll, we start with the given data: the sample size is \(n = 800\), the sample proportion is \(\hat{p} = 0.62\), and the margin of error (ME) is \(0.03\). The margin of error for a sample proportion is calculated using the formula:
\[ME = z^* \sqrt{\frac{\hat{p}(1-\hat{p})}{n}}\]
where \(z^*\) is the critical value corresponding to the confidence level \(C\). Our objective is to find \(z^*\), and then convert that value into the corresponding confidence level.
- Calculate the critical value \(z^*\)
First, we isolate the critical value \(z^*\) by rearranging the margin of error formula:
\[z^* = \frac{ME}{\sqrt{\frac{\hat{p}(1-\hat{p})}{n}}}\]
Substituting the known values into this equation, we get:
\[z^* = \frac{0.03}{\sqrt{\frac{0.62 \times (1 - 0.62)}{800}}}\]
Calculating the denominator:
\[\frac{0.62 \times 0.38}{800} = \frac{0.2356}{800} = 0.0002945\]
\[\sqrt{0.0002945} \approx 0.017161\]
Now, calculate \(z^*\):
\[z^* \approx \frac{0.03}{0.017161} \approx 1.748\]
This value, \(1.748\), represents the number of standard deviations from the mean in a standard normal distribution.
- Determine the confidence level
The confidence level \(C\) is the area under the standard normal curve between \(-z^*\) and \(z^*\). Mathematically, this is expressed as:
\[C = P(-1.748 < Z < 1.748)\]
Using the standard normal distribution, we calculate the cumulative probability:
\[P(Z < 1.748) \approx 0.95978\]
\[P(Z < -1.748) \approx 0.04022\]
\[C = 0.95978 - 0.04022 = 0.91956\]
Converting this decimal to a percentage, we find a confidence level of approximately \(91.96\%\), which rounds to \(92\%\).
Examples
In real life, this statistical method is used by researchers and pollsters to quantify the reliability of their findings. When a news agency reports an approval rating, knowing the margin of error and sample size allows us to calculate how "confident" they are that the true population proportion falls within that reported range. For instance, a 92% confidence level means that if we conducted this poll 100 times, 92 of those polls would likely yield a result within 3% of the true population approval rating.