Statistics Math Help

Question

Students in a college class were asked to time how long it takes them in minutes to get ready in the morning. Here were the responses: 14, 28, 5, 12, 56, 32, 2, 17Construct a 90% confidence interval for the time it takes college students to get ready in the morning. (Hint: Use t-distribution)

Answer 1

Write Sample Mean as (14 + 28 + 5 +12 + 56 + 32 + 2 + 17) ÷ 8 equal to 166/8 or 20.75.

Write Sample Standard Deviation as {[(14 – 20.75)² + (28 – 20.75)² + (5 – 20.75)² + (12 – 20.75)² +

(56 – 20.75)² + (32 – 20.75)² + (2 – 20.75)² + (17 – 20.75)²] ÷ (8 − 1)}^0.5 which simplifies to 17.55603274.

To find or construct the 90% Confidence Interval , compute the Standard Error Of The Mean as 17.55603274 ÷ √8 equal to 6.206994901.

With 0.05 of the Total Area under "Student's" T Distribution Curve in each "tail", at (8 −1) or 7 Degrees Of Freedom (df), the Critical Values Of t are tabulated as ±1.895.

The limits of the 90% Confidence Interval For The Mean are then given by [20.75 − 1.895(6.206994901),20.75 + 1.895(6.206994901)] or [8.987744663,32.51225534].