Question

2.Construct the confidence interval for μ 1 − μ 2 for the level of confidence and the data from independent samples given. 90% confidence: n 1 = 28, x - 1 = 212, s 1 = 6 n 2 = 23, x - 2 = 198, s 2 = 5 99% confidence: n 1 = 14, x - 1 = 68, s 1 = 8 n 2 = 20, x - 2 = 43, s 2 = 3

Answer 1

Q1.
given that,
mean(x)=212
standard deviation , s.d1=6
sample size, n1=28
y(mean)=198
standard deviation, s.d2 =5
sample size,n2 =23
few text books follow a different method in calculating df.
df = ( sd1 ^2 / n1 + sd2 ^2 /n2 )^2 / (s1^4 / n1^2 ( n1-1)) + (s2^4 / n2^2 ( n2-1))
df = (( (6^2/28)+(5^2/23) ))^2 / (( 6^4 / (28^2 ( 28 - 1 )) ) + (5^4/(23^2(23-1))))
df = 48.98 ~ 49,
the value of t α at 0.1 los with 49 df is +1.6766 and -1.6766          
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 = [( 212-198) ± t a/2 * sqrt((36/28)+(25/23)]
= [ (14) ± t a/2 * 1.54]
= [ (14) ± 1.6766 * 1.54]
= [11.418 , 16.58]
we are 90% sure that the interval [11.355 , 16.645] contains the true population mean
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Q2. n 1 = 14, x - 1 = 68, s 1 = 8 n 2 = 20, x - 2 = 43, s 2 = 3
given that,
mean(x)=68
standard deviation , s.d1=8
sample size, n1=14
y(mean)=43
standard deviation, s.d2 =3
sample size,n2 =20
few text books follow a different method in calculating df.
df = ( sd1 ^2 / n1 + sd2 ^2 /n2 )^2 / (s1^4 / n1^2 ( n1-1)) + (s2^4 / n2^2 ( n2-1))
df = (( (8^2/14)+(3^2/20) ))^2 / (( 8^4 / (14^2 ( 14 - 1 )) ) + (3^4/(20^2(20-1))))
df = 15.5820 ~16,
the value of t α at 0.01 los with 16 df is +2.9208 and -2.9208  
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 = [( 68-43) ± t a/2 * sqrt((64/14)+(9/20)]
= [ (25) ± t a/2 * 2.241]
= [ (25) ± 2.9208 * 2.241]
= [18.454 , 31.545]