A newspaper reporter asked an SRS of 100 residents in a large city for their opinion about the mayor’s job performance. Using the results from the sample, the C% confidence interval for the proportion of all residents in the city who approve of the mayor’s job performance is 0.565 to 0.695. What is the value of C?
Do your best to show me an equation that you used to determine the value of C.
The Correct Answer and Explanation is:
To determine the value of C, which represents the confidence level of the given interval, we need to understand how confidence intervals are constructed and how we can work backward from a confidence interval to find C.
Given:
- Sample size n=100n = 100
- Confidence interval for population proportion pp: [0.565, 0.695]
- Sample proportion p^=0.565+0.6952=0.63\hat{p} = \frac{0.565 + 0.695}{2} = 0.63
- Margin of error ME=0.695−0.63=0.065ME = 0.695 – 0.63 = 0.065
Step 1: Use the formula for Margin of Error
The formula for the margin of error for a population proportion is: ME=z∗⋅p^(1−p^)nME = z^* \cdot \sqrt{\frac{\hat{p}(1 – \hat{p})}{n}}
Where:
- z∗z^* is the critical value (z-score) corresponding to the confidence level CC,
- p^=0.63\hat{p} = 0.63 (sample proportion),
- n=100n = 100,
- ME=0.065ME = 0.065
Step 2: Plug in known values and solve for z∗z^*
0.065=z∗⋅0.63(1−0.63)1000.065 = z^* \cdot \sqrt{\frac{0.63(1 – 0.63)}{100}} 0.065=z∗⋅0.2331100=z∗⋅0.0023310.065 = z^* \cdot \sqrt{\frac{0.2331}{100}} = z^* \cdot \sqrt{0.002331} 0.065=z∗⋅0.048280.065 = z^* \cdot 0.04828 z∗=0.0650.04828≈1.346z^* = \frac{0.065}{0.04828} \approx 1.346
Step 3: Find the confidence level corresponding to z∗=1.346z^* = 1.346
A z-score of 1.346 corresponds to a cumulative probability of approximately 0.910 (91.0%) on one side. Since confidence intervals are two-tailed: C=2⋅0.910−1=0.82 or 82%C = 2 \cdot 0.910 – 1 = 0.82 \text{ or } 82\%
✅ Correct Answer: 82% Confidence Level
Explanation (300+ words)
A confidence interval is used in statistics to estimate a range within which a population parameter (like a proportion or mean) is likely to fall. The confidence level (C) indicates how certain we are that the true parameter lies within this interval. In this problem, a simple random sample (SRS) of 100 residents was surveyed, and the confidence interval for the proportion who approve of the mayor’s performance was calculated to be between 0.565 and 0.695.
To find the confidence level C, we first determine the sample proportion (p^\hat{p}) by averaging the endpoints of the interval: p^=0.565+0.6952=0.63\hat{p} = \frac{0.565 + 0.695}{2} = 0.63
Next, we calculate the margin of error (ME), which is the distance from the sample proportion to either endpoint of the interval: ME=0.695−0.63=0.065ME = 0.695 – 0.63 = 0.065
The formula for the margin of error for a proportion is: ME=z∗⋅p^(1−p^)nME = z^* \cdot \sqrt{\frac{\hat{p}(1 – \hat{p})}{n}}
Using the values of p^\hat{p} and n=100n = 100, we plug in and solve for the critical value z∗z^*, which turns out to be approximately 1.346. A z-score of 1.346 corresponds to a cumulative probability of about 91.0%. Because this is a two-tailed interval, the total confidence level is: C=2⋅0.91−1=0.82C = 2 \cdot 0.91 – 1 = 0.82
Thus, the confidence level for this interval is 82%, meaning that if we repeated this sampling method many times, about 82% of the intervals we compute would contain the true proportion of city residents who approve of the mayor’s performance.