Question

A researcher with the Department of Education followed a cohort of students who graduated from high school in a certain year, monitoring the progress the students made toward completing a bachelor's degree. One aspect of his research was to determine whether students who first attended community college took longer to attain a bachelor's degree than those who immediately attended and remained at a 4-year institution. The data in the table attached below summarize the results of his study. Complete parts a) through e) below.

(a) Determine the test statistic (t = ?). Round to two decimal places as needed.

(b) Construct a 90% confidence interval for μ_{community college} − μ_{no transfer} to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) • ​100% confidence interval about μ₁ − μ₂ can be found using the following​ formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

(c) Construct a 95​% confidence interval for μ_{community college} − μ_{no transfer} to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) • ​100% confidence interval about μ₁ − μ₂ can be found using the following​ formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

(d) Construct a 98​% confidence interval for μ_{community college} − μ_{no transfer} to approximate the mean additional time it takes to complete a​ bachelor's degree if you begin in community college. The bounds of a ​(1 − α​) • ​100% confidence interval about μ₁ − μ₂ can be found using the following​ formulas, where t_α/2 is the critical value. Round to six decimal places as needed.

(e) Do the results of parts​ (b), (c) and​ (d) imply that community college causes you to take extra time to earn a​ bachelor's degree?

Answer 1

hinh that W = 243, ng =1138 1 = 5.33, ), = 4.32 Si=1. 133 S, -10ls a) t-lest Ti-T 2 = 13.48 2 5.33-4.32 (1.133)2 + (1.018) (273 + n2 1138 b) t212 ,df +0.40,1409 = 1.646 907. of C-I = (-2) + Wie na =(5-33-4-32) + 1.646 (1.13)2 273 + (1.01) 113 8. =1.01 + 0.12325729 = (0.88674, 1.13325) talledf = to.05, 1409 = 1.962 95-1. of C.I. - (R-T) +42127 + ni 52V (1.01) = (5.33-4-32) + 1.9621(1-133) 273 -1.01 + 0.1469 2029 = (0.86307, 1. 15692)

d) tx12, df =to.o2 02,1409 = 2.330 98-1. OR CI: = 61,-1, IL-112 S² + nz - (5.33 -4.32) + 2.3301 (1.133)24 (1.015)2 273 1138 = 1.01 t 0.1744772 - (0-83552, 1-18447)