sovle step by step and by your hand Note: Do not use programm yul tnIR that this

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sovle step by step and by your hand
Note:
Do not use programm

yul tnIR that this tion is valid? Explain. assump- 8.9 The quality engineer of a multinational company involved in manufacturing synthetic rubber wants to examine the hardness of the rubber. In the rubber industry, hardness is measured in degrees Shore. A random sample of 50 pieces of rubber was evaluated, with the following results: 62.4 66.0 67.6 63.2 66.3 61.8 61.8 67.5 61.3 65.0 64.6 61.3 64.7 63.5 66.4 60.0 65.8 64.3 62.3 61.4 63.6 69.5 64.9 62.2 67.7 67.7 66.6 64.5 64.0 61.6 68.5 65.2 66.9 66.5 68.3 64.3 63.3 63.6 65.8 60.1 62.6 66.4 67.2 62.5 63.6 63.4 63.1 62.5 68.7 62.2 4 33 045 RUBBER2 Set up a 95% confidence interval estimate of the average hardness (in degrees Shore) for all pieces of rubber of this type. (a) (b) What assumption must be made in (a)? On the basis of these data, do you think that this assump- tion is valid? Explain. ta evaluate the

Explanation / Answer

8.9
a.
TRADITIONAL METHOD
given that,
sample mean, x =64.484
standard deviation, s =2.411
sample size, n =50
I.
stanadard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 2.411/ sqrt ( 50) )
= 0.341
II.
margin of error = t /2 * (stanadard error)
where,
ta/2 = t-table value
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 49 d.f is 2.01
margin of error = 2.01 * 0.341
= 0.685
III.
CI = x ± margin of error
confidence interval = [ 64.484 ± 0.685 ]
= [ 63.799 , 65.169 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =64.484
standard deviation, s =2.411
sample size, n =50
level of significance, = 0.05
from standard normal table, two tailed value of |t /2| with n-1 = 49 d.f is 2.01
we use CI = x ± t a/2 * (sd/ Sqrt(n))
where,
x = mean
sd = standard deviation
a = 1 - (confidence level/100)
ta/2 = t-table value
CI = confidence interval
confidence interval = [ 64.484 ± t a/2 ( 2.411/ Sqrt ( 50) ]
= [ 64.484-(2.01 * 0.341) , 64.484+(2.01 * 0.341) ]
= [ 63.799 , 65.169 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 95% sure that the interval [ 63.799 , 65.169 ] contains the true population mean
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 mean

b.
above problem there is assumption
simply find out the sample mean and stanadard deviation for given samples
and done it confidence interval for t test for single mean

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