Chapter 10, Section 10.3, Problem 23c Incorrect. To compare two programs for tra

Question

Chapter 10, Section 10.3, Problem 23c Incorrect. To compare two programs for training industrial workers to perform a skilled job, 20 workers are included in an experiment. Of these, 10 are selected at random and trained by method 1; the remaining 10 workers are trained by method 2. After completion of training, all the workers are subjected to a time-and-motion test that records the speed of performance of a skilled job. The following data are obtained Time (minutes) Method 114 21 10 23 15 22 11 19 17 24 Method 2 22 32 13 20 23 17 28 27 24 28 Construct a 95% confidence interval for the population mean difference in job times between the two methods. 4.12 7.48 The 95% confidence interval in this case is ( ) [Round your answers to 2 decimal places.]

Explanation / Answer

TRADITIONAL METHOD
given that,
mean(x)=17.6
standard deviation , s.d1=4.9933
number(n1)=10
y(mean)=23.4
standard deviation, s.d2 =5.6999
number(n2)=10
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((24.933/10)+(32.489/10))
= 2.396
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 9 d.f is 2.262
margin of error = 2.262 * 2.396
= 5.42
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (17.6-23.4) ± 5.42 ]
= [-11.22 , -0.38]
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DIRECT METHOD
given that,
mean(x)=17.6
standard deviation , s.d1=4.9933
sample size, n1=10
y(mean)=23.4
standard deviation, s.d2 =5.6999
sample size,n2 =10
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 = [( 17.6-23.4) ± t a/2 * sqrt((24.933/10)+(32.489/10)]
= [ (-5.8) ± t a/2 * 2.396]
= [-11.22 , -0.38]
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interpretations:
1. we are 95% sure that the interval [-11.22 , -0.38] 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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