Question

The manufacturer of hardness testing equipment uses​ steel-ball indenters to penetrate metal that is being tested.​ However, the manufacturer thinks it would be better to use a diamond indenter so that all types of metal can be tested. Because of differences between the two types of​ indenters, it is suspected that the two methods will produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made.​ Therefore, the manufacturer uses both indenters on each specimen and compares the hardness readings. Construct a​ 95% confidence interval to judge whether the two indenters result in different measurements.

​Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers.

Specimen   Steel ball   Diamond
1 50 52
2 57 55
3 61 63
4 71 74
5 68 69
6 54 56
7 65 68
8 51 51
9 53 56

Construct a​ 95% confidence interval to judge whether the two indenters result in different​ measurements, where the differences are computed as​ 'diamond minus steel​ ball'.

The lower bound is

The upper bound is

​(Round to the nearest tenth as​ needed.)

Answer 1

TRADITIONAL METHOD
given that,
mean(x)=60.4444
standard deviation , s.d1=8.2932
number(n1)=9
y(mean)=58.8889
standard deviation, s.d2 =7.7046
number(n2)=9
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((68.78/9)+(59.36/9))
= 3.77
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 8 d.f is 2.31
margin of error = 2.306 * 3.77
= 8.7
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (60.4444-58.8889) ± 8.7 ]
= [-7.15 , 10.26]
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DIRECT METHOD
given that,
mean(x)=60.4444
standard deviation , s.d1=8.2932
sample size, n1=9
y(mean)=58.8889
standard deviation, s.d2 =7.7046
sample size,n2 =9
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 = [( 60.4444-58.8889) ± t a/2 * sqrt((68.78/9)+(59.36/9)]
= [ (1.56) ± t a/2 * 3.77]
= [-7.15 , 10.26]
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interpretations:
1. we are 95% sure that the interval [-7.15 , 10.26] 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:
95% confidence interval to judge whether the two indenters result in different​ measurements, where the differences
are computed as​ 'diamond minus steel​ ball'. [-7.15 , 10.26]