A study utilized the random roommate assignment process of a small college to in
ID: 3223659 • Letter: A
Question
A study utilized the random roommate assignment process of a small college to investigate the interracial mix of friends among students in college. As part of this study, the researchers looked at 238 white students who were randomly assigned a roommate in their first year and recorded the proportion of their friends (not including the first-year roommate) who were black. The following table summarizes the results, broken down by roommate race, for the middle of the first and third years of college. Middle of First Year: Middle of Third Year: For each year use conservative df to construct a 95% confidence interval (plusminus 0.001) for the difference in means mu_1 - mu_2. For the first year: a 95% confidence interval is from to For the third year: a 95% confidence interval is from toExplanation / Answer
a.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=0.0802
Standard deviation( sd1 )=0.136
Sample Size(n1)=41
Mean(x2)=0.0611
Standard deviation( sd2 )=0.112
Sample Size(n2)=197
CI = [ ( 0.0802-0.0611) ±t a/2 * Sqrt( 0.018496/41+0.012544/197)]
= [ (0.0191) ± t a/2 * Sqrt( 0.0005) ]
= [ (0.0191) ± 2.021 * Sqrt( 0.0005) ]
= [-0.0268 , 0.065]
b.
CI = x1 - x2 ± t a/2 * Sqrt ( sd1 ^2 / n1 + sd2 ^2 /n2 )
Where,
x1 = Mean of Sample 1, x2 = Mean of sample2
sd1 = SD of Sample 1, sd2 = SD of sample2
a = 1 - (Confidence Level/100)
ta/2 = t-table value
CI = Confidence Interval
Mean(x1)=0.1499
Standard deviation( sd1 )=0.2411
Sample Size(n1)=41
Mean(x2)=0.06
Standard deviation( sd2 )=0.156
Sample Size(n2)=197
CI = [ ( 0.1499-0.06) ±t a/2 * Sqrt( 0.05812921/41+0.024336/197)]
= [ (0.0899) ± t a/2 * Sqrt( 0.0015) ]
= [ (0.0899) ± 2.021 * Sqrt( 0.0015) ]
= [0.0106 , 0.1692]
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