Question

sample of 15 students and found a mean GPA of 3.01 with a standard deviation of 0.534 . a. For the given situation, identify the degrees of freedom and associated t-score for a 99% confidence interval. b. Using your answers from part (a), identify the margin of error for the given interval. c. Use your answer from part (c) to find the desired interval. Check your answer using the "TInterval" computation on your calculator.

Answer 1

a. The degrees of freedom for a sample of size 15 is 14. The associated t-score for a 99% confidence interval with 14 degrees of freedom is approximately 2.977. b. The margin of error for the given interval is 0.430. c. The desired interval is [2.58, 3.44].

Step-by-step explanation:

a. The degrees of freedom for a sample of size 15 can be calculated as (sample size - 1), which in this case is (15 - 1) = 14. For a 99% confidence interval, the associated t-score can be found using a t-table or a calculator. The t-score for a 99% confidence level with 14 degrees of freedom is approximately 2.977.

b. The margin of error can be calculated using the formula: margin of error = (t-score) * (standard deviation / √sample size). Substituting the values, the margin of error is 2.977 * (0.534 / √15) = 0.430.

c. To find the desired interval, we take the mean GPA (3.01) and add/subtract the margin of error calculated in part (b). The confidence interval is therefore [3.01 - 0.430, 3.01 + 0.430], which simplifies to [2.58, 3.44].