Question

Suppose we measure how much sleep students in first and second year at UVic get per night. A sample of 21 first year students slept an average of 7.9 hours per night with a standard deviation of 1.3 hours. A sample of 16 second year students slept an average of 6.8 hours per night with a standard deviation of 2.1 hours. Suppose we want to test if first year students sleep more than second year students. Assuming the number of hours slept is normally distributed for both first year students and second year students, what is the estimated standard error of the random variable that should be used for our test? Give your answer rounded to 4 decimal places. Answer:

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=7.9
standard deviation , s.d1=1.3
number(n1)=21
y(mean)=6.8
standard deviation, s.d2 =2.1
number(n2)=16
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
standard error = sqrt((1.69/21)+(4.41/16))
= 0.5967
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.05
from standard normal table,right tailed and
value of |t α| with min (n1-1, n2-1) i.e 15 d.f is 1.7531
margin of error = 1.753 * 0.5967
= 1.0461
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (7.9-6.8) ± 1.0461 ]
= [0.0539 , 2.1461]
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DIRECT METHOD
given that,
mean(x)=7.9
standard deviation , s.d1=1.3
sample size, n1=21
y(mean)=6.8
standard deviation, s.d2 =2.1
sample size,n2 =16
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 = [( 7.9-6.8) ± t a/2 * sqrt((1.69/21)+(4.41/16)]
= [ (1.1) ± t a/2 * 0.5967]
= [0.0539 , 2.1461]
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interpretations:
1. we are 95% sure that the interval [0.0539 , 2.1461] 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
Answer:
standard error = sqrt((1.69/21)+(4.41/16))
= 0.5967