There are 300 students enrolled in Business Statistic. Historically, exam scores are normally distributed with a standard deviation of 30.9. Your instructor randomly selected a sample of 30 examinations and finds a mean of 74.2. Determine a 90% confidence interval for the mean score for all students taking the course.
50 to 60
To determine a confidence interval for the mean score of all students taking the course, you can use the formula:
Confidence Interval = Sample Mean ± Margin of Error
To calculate the Margin of Error, we need to find the standard error, which can be calculated using the formula:
Standard Error = Standard Deviation / √Sample Size
Given:
Sample Mean (x̄) = 74.2
Standard Deviation (σ) = 30.9
Sample Size (n) = 30
Confidence Level (CL) = 90% (which corresponds to α = 0.10)
1. Calculate the standard error:
Standard Error = 30.9 / √30 ≈ 5.64 (rounded to two decimal places)
2. Compute the critical value, also known as the Z-Score, corresponding to the desired confidence level. For a 90% confidence interval, the Z-Score is found from the standard normal distribution table (or using a statistical calculator). The Z-Score for a 90% confidence level is approximately 1.645.
3. Calculate the margin of error:
Margin of Error = Z-Score * Standard Error
= 1.645 * 5.64
≈ 9.28 (rounded to two decimal places)
4. Calculate the lower and upper bounds of the confidence interval:
Lower Bound = Sample Mean - Margin of Error
= 74.2 - 9.28
≈ 64.92 (rounded to two decimal places)
Upper Bound = Sample Mean + Margin of Error
= 74.2 + 9.28
≈ 83.48 (rounded to two decimal places)
Therefore, the 90% confidence interval for the mean score of all students taking the course is approximately (64.92, 83.48). This means we are 90% confident that the true mean score lies between these two values.