A certain test preparation course is designed to help students improve their scores on the GRE exam. A mock exam is given at the beginning and end of the course to determine the effectiveness of the course. The following measurements are the net change in 7 students' scores on the exam after completing the course: \[ 24,6,21,8,16,10,18 \] Using these data, construct a \( 80 \% \) confidence interval for the average net change in a student's score after completing the course. Assume the population is approximately normal. Step 2 of 4 : Calculate the sample standard deviation for the given sample data. Round your answer to one decimal place.

To find the sample standard deviation, we first need to calculate the mean of the scores and then compute the standard deviation using the following formula: 1. **Calculate the mean (\( \bar{x} \)):** \[ \bar{x} = \frac{(24 + 6 + 21 + 8 + 16 + 10 + 18)}{7} = \frac{103}{7} \approx 14.7 \] 2. **Calculate the squared deviations from the mean:** \[ \text{Deviations} = (24 - 14.7)^2, (6 - 14.7)^2, (21 - 14.7)^2, (8 - 14.7)^2, (16 - 14.7)^2, (10 - 14.7)^2, (18 - 14.7)^2 \] \[ = (9.3)^2, (-8.7)^2, (6.3)^2, (-6.7)^2, (1.3)^2, (-4.7)^2, (3.3)^2 \] \[ = 86.49, 75.69, 39.69, 44.89, 1.69, 22.09, 10.89 \] 3. **Sum of squared deviations:** \[ \text{Sum} = 86.49 + 75.69 + 39.69 + 44.89 + 1.69 + 22.09 + 10.89 = 281.42 \] 4. **Calculate the sample variance \( s^2 \):** \[ s^2 = \frac{\text{Sum of squared deviations}}{n - 1} = \frac{281.42}{7 - 1} = \frac{281.42}{6} \approx 46.9 \] 5. **Calculate the sample standard deviation \( s \):** \[ s = \sqrt{46.9} \approx 6.8 \] Hence, the sample standard deviation rounded to one decimal place is **6.8**.