Impact of College Roommates on Grades In this exercise, we investigate answers t

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Impact of College Roommates on Grades In this exercise, we investigate answers to the questions: How much of an efTect does your roommate have on your grades? Ir particular, does it mtter whether your roommate brings a videogame to college? A study was conducted involving 210 first-year students who were randomly assigned a roommate. Table 1 gives summary statistics on grade point average (GPA) for the first semester depending on whether the student andfor the roommate brought a videogame to campis brought Sample size Mean GPA St. Dev 0.590 0.689 0.699 0.639 ideogame ideogame BR 3.128 3,039 2.932 2.754 38 40 Table 1 Videagames and GPA Considering only students who do bring a videngame to campus, find a 95% confidence interval for the difference in mean GPA between students whose roommate does not bring a videogame and those whose roomate does bring a videngame Yes , where N represents the mean GPA for students whose roommate does not bring a videogame and ly represents the mean GPA for students whose roommate does bring a videogame. Raund your answers to three decimal places. 0.753 The 95% confidence interval is to Click If you would like to Show Work for this question: pen 5how Work

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=3.039
standard deviation , s.d1=0.689
number(n1)=44
y(mean)=2.754
standard deviation, s.d2 =0.639
number(n2)=40
I.
stanadard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((0.475/44)+(0.408/40))
= 0.145
II.
margin of error = t a/2 * (stanadard error)
where,
t a/2 = t -table value
level of significance, = 0.05
from standard normal table, two tailed and
value of |t | with min (n1-1, n2-1) i.e 39 d.f is 2.023
margin of error = 2.023 * 0.145
= 0.293
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (3.039-2.754) ± 0.293 ]
= [-0.008 , 0.578]
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DIRECT METHOD
given that,
mean(x)=3.039
standard deviation , s.d1=0.689
sample size, n1=44
y(mean)=2.754
standard deviation, s.d2 =0.639
sample size,n2 =40
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
where,
x1,x2 = mean of populations
sd1,sd2 = standard deviations
n1,n2 = size of both
a = 1 - (confidence Level/100)
ta/2 = t-table value
CI = confidence interval
CI = [( 3.039-2.754) ± t a/2 * sqrt((0.475/44)+(0.408/40)]
= [ (0.285) ± t a/2 * 0.145]
= [-0.008 , 0.578]
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interpretations:
1. we are 95% sure that the interval [-0.008 , 0.578] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 95% of these intervals will contains the true population proportion

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