Math Problem Statement

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?

Solution

To determine the confidence level CC, we need to calculate the zz-score corresponding to the given confidence interval bounds. Here's the process:

Step 1: Identify the sample proportion and margin of error

The confidence interval is given as 0.5650.565 to 0.6950.695. From this:

  • The sample proportion (p^\hat{p}) is the midpoint of the interval: p^=0.565+0.6952=0.63\hat{p} = \frac{0.565 + 0.695}{2} = 0.63
  • The margin of error (MEME) is the distance from the sample proportion to either endpoint of the interval: ME=0.6950.63=0.065ME = 0.695 - 0.63 = 0.065

Step 2: Express the margin of error formula

The margin of error for a proportion is: ME=zp^(1p^)nME = z^* \cdot \sqrt{\frac{\hat{p}(1 - \hat{p})}{n}} Where:

  • zz^* is the critical value associated with the confidence level CC,
  • n=100n = 100 is the sample size,
  • p^=0.63\hat{p} = 0.63 is the sample proportion.

Step 3: Solve for zz^*

Substitute the known values into the margin of error formula: 0.065=z0.63(10.63)1000.065 = z^* \cdot \sqrt{\frac{0.63(1 - 0.63)}{100}} First, calculate the standard error: 0.63(10.63)100=0.630.37100=0.2331/100=0.0023310.0483\sqrt{\frac{0.63(1 - 0.63)}{100}} = \sqrt{\frac{0.63 \cdot 0.37}{100}} = \sqrt{0.2331 / 100} = \sqrt{0.002331} \approx 0.0483 Then solve for zz^*: z=0.0650.04831.346z^* = \frac{0.065}{0.0483} \approx 1.346

Step 4: Find the confidence level

The zz^*-value of approximately 1.3461.346 corresponds to a confidence level CC. Using standard normal distribution tables or software, we find that the area between 1.346-1.346 and 1.3461.346 is approximately 0.8240.824, or 82.4%.

Thus, the confidence level CC is 82.4%.


Would you like a detailed explanation of using the zz-table, or do you have further questions? 😊

Expanding Your Knowledge:

  1. How is the confidence interval affected by the sample size nn?
  2. What happens to the margin of error if the confidence level CC increases?
  3. How do you interpret a confidence interval in real-world terms?
  4. Can confidence intervals be used for means as well as proportions? If so, how?
  5. How do we determine the sample size needed for a specific margin of error?

Tip: Confidence intervals are a way to estimate population parameters using sample data—be sure to always check assumptions like random sampling and sufficient sample size!

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Math Problem Analysis

Mathematical Concepts

Confidence Intervals
Proportions
Standard Normal Distribution

Formulas

ME = z* * sqrt((p(1-p))/n)
Confidence Interval = p ± ME

Theorems

Properties of the Standard Normal Distribution
Calculation of z-scores

Suitable Grade Level

Grades 10-12