Question

Measuring Treadwear. Refer to Exercise 10.158 and find a 95% confidence interval for the mean difference in measurement by the weight and groove methods.

Answer 1

 

TRADITIONAL METHOD
Assumed values,
mean(x)=251
standard deviation , s.d1=54
number(n1)=140
y(mean)=228
standard deviation, s.d2 =52
number(n2)=80
I.
standard error = sqrt(s.d1^2/n1)+(s.d2^2/n2)
where,
sd1, sd2 = standard deviation of both
n1, n2 = sample size
stanadard error = sqrt((2916/140)+(2704/80))
= 7.391
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, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 79 d.f is 1.99
margin of error = 1.99 * 7.391
= 14.708
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (251-228) ± 14.708 ]
= [8.292 , 37.708]

DIRECT METHOD
given that,
mean(x)=251
standard deviation , s.d1=54
sample size, n1=140
y(mean)=228
standard deviation, s.d2 =52
sample size,n2 =80
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 = [( 251-228) ± t a/2 * sqrt((2916/140)+(2704/80)]
= [ (23) ± t a/2 * 7.391]
= [8.292 , 37.708]

interpretations:
1. we are 95% sure that the interval [8.292 , 37.708] 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
conclusion:
95% confidence interval for the mean difference in measurement by the weight and groove methods
95% sure that the interval [8.292 , 37.708]