Please so formula ### Constructing a 95% Confidence Interval for Population Mean**Scenario:**You are given a task to construct a 95% confidence interval for the population mean, denoted as $ \mu $. Assume the population follows a normal distribution. A sample of 20 college students yields the following statistics:- Mean annual earnings: $3120- Standard deviation: $677**Options:**You are provided with four potential confidence intervals. Select the correct one:- A. ($1324, $1567)- B. ($2657, $2991)- C. ($2803, $3437)- D. ($2135, $2567)**Steps to Solve:**1. **Identify Given Information:** - Sample size (n): 20 - Sample mean ($ \bar{x} $): $3120 - Sample standard deviation (s): $6772. **Determine the Confidence Level:** Since the confidence level is 95%, the corresponding z-value (critical value) for a two-tailed test is approximately 1.96 (from z-tables).3. **Calculate the Margin of Error (ME):** The formula for the margin of error in this context is given by: \[ ME = z \times \left(\frac{s}{\sqrt{n}}\right) \]4. **Substitute Given Values into Formula:** \[ ME = 1.96 \times \left(\frac{677}{\sqrt{20}}\right) \]5. **Compute Margin of Error:** \[ ME = 1.96 \times \left(\frac{677}{4.47}\right) \approx 1.96 \times 151.45 \approx 296.84 \]6. **Determine the Confidence Interval:** The confidence interval can be found using: \[ \left( \bar{x} - ME, \bar{x} + ME \right) \]7. **Substitute Values:** \[ \left( 3120 - 296.84, 3120 + 296.84 \right) \approx \left( 2823.16, 3416.84 \right) \]Therefore, the correct answer from the options given is:- C. ($2803

Answered: Image /qna-images/answer/7306fb8f-20e0-44ba-a67d-3dda8d36c56f.jpg

Construct a 95% confidence interval for the population mean, μ. Assume the population has a normal distribution. A sample of 20 college students had mean annual earnings of $3120 with a standard deviation of $677. contents OA. ($1324,$1567) B. ($2657,$2891) C. ($2803,$3437) D. ($2135,$2567) e Resourc r Success or Success media Library F estly cloudy Ask my instructor H 4 Clear all Check answer