help **Question:**Linh wants to advertise how many chocolate chips are in each banana chocolate muffin at her bakery. She randomly selects a sample of 13 muffins and finds that the number of chocolate chips per muffin in the sample has a mean of 14.5 and a sample standard deviation of 3.6. What is the 90% confidence interval for the number of chocolate chips per muffin? Assume the number of chocolate chips is normally distributed.**Detailed Explanation:**To calculate the 90% confidence interval for the number of chocolate chips per muffin, we will use the concept of confidence intervals in statistics. Given:- Sample size (n) = 13 muffins- Sample mean (x̄) = 14.5- Sample standard deviation (s) = 3.6Step-by-step process to determine the confidence interval:1. **Determining the critical value (t-value):** - For a 90% confidence level and a sample size of 13, the degrees of freedom (df) is n - 1, which is 13 - 1 = 12. - Using a t-distribution table or calculator, we find the t-value for 90% confidence level with 12 degrees of freedom. For a 90% confidence level, the t-value approximately is 1.782.2. **Calculating the standard error (SE):** - Standard error (SE) = s / √n = 3.6 / √13 ≈ 3.6 / 3.61 ≈ 0.997.3. **Determining the margin of error (ME):** - ME = t-value × SE ≈ 1.782 × 0.997 ≈ 1.78.4. **Calculating the confidence interval:** - Lower limit = x̄ - ME = 14.5 - 1.78 ≈ 12.72. - Upper limit = x̄ + ME = 14.5 + 1.78 ≈ 16.28.Therefore, the 90% confidence interval for the number of chocolate chips per muffin is approximately (12.72, 16.28). This means Linh can be 90% confident that the true mean number of chocolate chips per muffin in her bakery falls between 12.72 and 16.28.

Name: help **Question:**Linh wants to advertise how many chocolate chips are
Uploaded: 2024-03-05T11:52:21.000Z
Channel: Bartleby
Description: help **Question:**Linh wants to advertise how many chocolate chips are in each banana chocolate muffin at her bakery. She randomly selects a sample of 13 muffins and finds that the number of chocolate chips per muffin in the sample has a mean of 14.5