The manufacturer of hardness testing equipment uses steel-ball indenters to pene
ID: 3316111 • Letter: T
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 be 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 method produce different hardness readings. The metal specimens to be tested are large enough so that two indentions can be made. Therefore, the manufacturer uses b indenters on each specimen and compares he hardness readings. Construct a 95% confidence interval o ge het er he wonder ers sul diff e measurements Note: A normal probability plot and boxplot of the data indicate that the differences are approximately normally distributed with no outliers. Click the icon to view the data table. o udge whether he t o indenters result in different measurements, where the diff sare computed as diamond minus Construct a 95% confidence interva ball er The 95% confidence interval to udge whether the two inden ers resu in different measurements is Round to the nearest tent s needed i Data Table Specimen 1 2 3 4 5 6 7 8 9 Steel ball 51 57 61 71 68 54 65 51 53 Diamond 53 55 63 74 69 5 68 51 56 Print Done Enter your answer in the edit fieldsExplanation / Answer
TRADITIONAL METHOD
given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
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((144/10)+(324/25))
= 5.231
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 * 5.231
= 11.832
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (296-315) ± 11.832 ]
= [-30.832 , -7.168]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=296
standard deviation , s.d1=12
sample size, n1=10
y(mean)=315
standard deviation, s.d2 =18
sample size,n2 =25
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 = [( 296-315) ± t a/2 * sqrt((144/10)+(324/25)]
= [ (-19) ± t a/2 * 5.231]
= [-30.832 , -7.168]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 95% sure that the interval [-30.832 , -7.168] 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
Related Questions
Navigate
Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.