Question

1. A high-school administrator who is concerned about the amount of sleep the students in his district are getting selects a random sample of 14 seniors in his district and asks them how many hours of sleep they get on a typical school night. He then uses school records to determine the most recent grade-point average (GPA) for each student. His data and a computer regression output are given below. (remember to do ALL parts).

Sleep (hrs) 9 8.5 9 7 7.5 6 7 8 5.5 6 8.5 6.5 8 8

GPA 3.8 3.3 3.5 3.6 3.4 3.3 3.2 3.2 3.2 3.4 3.6 3.1 3.4 3.7

(a) Do these data provide convincing evidence of a linear relationship between the hours of sleep students typically get and their academic performance, as measured by their GPA? Carry out a significance test at the α = 0.05 level. (10 points)

(b) Construct and interpret at 95% confidence interval for the slope of the regression of GPA on hours of sleep for seniors in this school district. (5 points)

(c) Can we conclude from these data that students’ GPA will improve if they get more sleep? Explain. (

Answer 1

Procedure:
data -> data analytics -> regression
Y input: GPA
X input: Sleep

Output:

a)
Null hypothesis, Ho: there is no significant linear relationship between the hours of sleep students typically get and their academic performance, as measured by their GPA
Alternative hypothesis, h1: there is a linear relationship between the hours of sleep students typically get and their academic performance, as measured by their GPA
With F(1, 12) = 5.4789, p<5%, i reject Ho at 5% level of signficance and conclude that there is a linear relationship between the hours of sleep students typically get and their academic performance, as measured by their GPA

b)
95% confidence interval for the slope:
b1 +- t(a/2,n-1)*SE_b1
lower limit = 0.1018 - 2.1788*0.0435 = 0.0070222
upper limit = 0.1018 + 2.1788*0.0435 = 0.1965778

c)
Since the confidence interval has all values greater than 0, i can say with 95% confidence that with 1 hr increase in sleep, the GPA is estimated to increase by (0.0070222, 0.1965778)