Question

a.) In a study on low-back pain (LBP), a random sample of 31 workers who suffer from LBP were tested for lateral range of motion (in degrees). The average was found to be 88.3o and sample standard deviation 7.8o. A similar sample of 28 workers who did not suffer from LBP was found to have an average lateral range of motion of 91.5o and sample standard deviation 5.5o. Compute a 90% confidence interval for the difference in the true mean lateral range of motion for the two populations.

b.) A Brinell hardness test involves measuring the diameter of the indentation made when a hardened steel ball is pressed into material under a standard test load. Suppose that the Brinell hardness is determined for each specimen in a sample of size 51, resulting in a sample mean hardness of 64.3 and a sample standard deviation of 6.1. Calculate a 99% confidence interval for the true average Brinell hardness for material specimens of this type.

c.) In a study of pulverized fuel-ash concrete mix, it was found that the sample mean compressive strength was 27.0 with a standard deviation of 4.89 for a sample of 68 specimens which had been aged for 7 days. A sample of 74 specimens which had been aged for 28 days yielded a mean compressive strength of 35.8 with standard deviation 6.43. Is there strong evidence that mean compressive strength is higher after 28 days than after 7 days?   

d.) Is there a difference in the amount of airborne bacteria between carpeted and uncarpeted rooms? In an experiment, 15 rooms were carpeted (C) and 35 were left uncarpeted (U). The rooms are similar in size and function. After a suitable period of time, the concentration of bacteria in the air was measured (in units of bacteria per cubic foot) in all of these rooms. The mean concentration in the carpeted rooms was 184 with a standard deviation of 27.0 and the mean concentration in the uncarpeted rooms was 172 with a standard deviation of 17.9. If you were to conduct a hypothesis test your hypotheses would be

Answer 1

a.
TRADITIONAL METHOD
given that,
mean(x)=88.3
standard deviation , s.d1=7.8
number(n1)=31
y(mean)=91.5
standard deviation, s.d2 =5.5
number(n2)=28
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((60.84/31)+(30.25/28))
= 1.744
II.
margin of error = t a/2 * (standard error)
where,
t a/2 = t -table value
level of significance, α = 0.1
from standard normal table, two tailed and
value of |t α| with min (n1-1, n2-1) i.e 27 d.f is 1.703
margin of error = 1.703 * 1.744
= 2.971
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (88.3-91.5) ± 2.971 ]
= [-6.171 , -0.229]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
mean(x)=88.3
standard deviation , s.d1=7.8
sample size, n1=31
y(mean)=91.5
standard deviation, s.d2 =5.5
sample size,n2 =28
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 = [( 88.3-91.5) ± t a/2 * sqrt((60.84/31)+(30.25/28)]
= [ (-3.2) ± t a/2 * 1.744]
= [-6.171 , -0.229]
-----------------------------------------------------------------------------------------------
interpretations:
1. we are 90% sure that the interval [-6.171 , -0.229] contains the true population proportion
2. If a large number of samples are collected, and a confidence interval is created
for each sample, 90% of these intervals will contains the true population proportion
b.
TRADITIONAL METHOD
given that,
sample mean, x =64.3
standard deviation, s =6.1
sample size, n =51
I.
standard error = sd/ sqrt(n)
where,
sd = standard deviation
n = sample size
standard error = ( 6.1/ sqrt ( 51) )
= 0.854
II.
margin of error = t α/2 * (standard error)
where,
ta/2 = t-table value
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 50 d.f is 2.678
margin of error = 2.678 * 0.854
= 2.287
III.
CI = x ± margin of error
confidence interval = [ 64.3 ± 2.287 ]
= [ 62.013 , 66.587 ]
-----------------------------------------------------------------------------------------------
DIRECT METHOD
given that,
sample mean, x =64.3
standard deviation, s =6.1
sample size, n =51
level of significance, α = 0.01
from standard normal table, two tailed value of |t α/2| with n-1 = 50 d.f is 2.678
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 = [ 64.3 ± t a/2 ( 6.1/ Sqrt ( 51) ]
= [ 64.3-(2.678 * 0.854) , 64.3+(2.678 * 0.854) ]
= [ 62.013 , 66.587 ]
-----------------------------------------------------------------------------------------------
interpretations:
1) we are 99% sure that the interval [ 62.013 , 66.587 ] contains the true population mean
2) If a large number of samples are collected, and a confidence interval is created
for each sample, 99% of these intervals will contains the true population mean
c.
Given that,
mean(x)=35.8
standard deviation , s.d1=6.43
number(n1)=74
y(mean)=27
standard deviation, s.d2 =4.89
number(n2)=68
null, Ho: u1 = u2
alternate, H1: u1 > u2
level of significance, α = 0.05
from standard normal table,right tailed t α/2 =1.668
since our test is right-tailed
reject Ho, if to > 1.668
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =35.8-27/sqrt((41.3449/74)+(23.9121/68))
to =9.2231
| to | =9.2231
critical value
the value of |t α| with min (n1-1, n2-1) i.e 67 d.f is 1.668
we got |to| = 9.22307 & | t α | = 1.668
make decision
hence value of | to | > | t α| and here we reject Ho
p-value:right tail - Ha : ( p > 9.2231 ) = 0
hence value of p0.05 > 0,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 > u2
test statistic: 9.2231
critical value: 1.668
decision: reject Ho
p-value: 0
we have enough evidence to support the claim that mean compressive strength is higher after 28 days than after 7 days.
d.
Given that,
mean(x)=184
standard deviation , s.d1=27
number(n1)=15
y(mean)=172
standard deviation, s.d2 =17.9
number(n2)=35
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.145
since our test is two-tailed
reject Ho, if to < -2.145 OR if to > 2.145
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =184-172/sqrt((729/15)+(320.41/35))
to =1.579
| to | =1.579
critical value
the value of |t α| with min (n1-1, n2-1) i.e 14 d.f is 2.145
we got |to| = 1.57902 & | t α | = 2.145
make decision
hence value of |to | < | t α | and here we do not reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != 1.579 ) = 0.137
hence value of p0.05 < 0.137,here we do not reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: 1.579
critical value: -2.145 , 2.145
decision: do not reject Ho
p-value: 0.137
we do not have enough evidence to support the claim that difference in the amount of airborne bacteria between carpeted and uncarpeted rooms