Question

M28 X B с 1 Right-hand Thread Left-hand Thread 2 86.2 123.3 3 106.5 97.8 4 74.5 104.0 5 83.8 101.6 6 154.3 140.7 7 127.1 99.0 8 106.9 131.7 9 99.8 91.2 10 111.8 115.6 11 118.8 123.2 12 120.6 127.8 13 142.3 111.7 14 76.4 130.7 15 145.5 121.0 16 124.4 138.7 17 119.6 133.1 18 122.8 107.8 19 85.5 99.4 20 118.3 149.1 21 102.0 126.5 22 102.3 83.8 23 118.1 145.5 24 132.0 110.4 25 101.1 136.0 26 118.2 131.6 27 105.4 122.0 28 112.1 113.6 29 109.4 95.5 30 109.5 128.5 31 111.6 115.8 32 106.2 95.0 33 103.4 112.5 34 115.7 102.6 35 105.7 97.5 36 120.9 133.9 37 143.9 161.7 38 145.5 124.1 39 134.1 147.8 40 128.1 115.0 41 120.4 133.0 42 145.3 125.8 Thread +

Answer 1

Given that,
mean(x)=115.024
standard deviation , s.d1=18.718
number(n1)=41
y(mean)=119.646
standard deviation, s.d2 =17.672
number(n2)=41
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, α = 0.05
from standard normal table, two tailed t α/2 =2.021
since our test is two-tailed
reject Ho, if to < -2.021 OR if to > 2.021
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =115.024-119.646/sqrt((350.36352/41)+(312.29958/41))
to =-1.15
| to | =1.15
critical value
the value of |t α| with min (n1-1, n2-1) i.e 40 d.f is 2.021
we got |to| = 1.14968 & | t α | = 2.021
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 != -1.1497 ) = 0.257
hence value of p0.05 < 0.257,here we do not reject Ho
ANSWERS
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null, Ho: u1 = u2
alternate, H1: u1 != u2
1.
test statistic: -1.15
critical value: -2.021 , 2.021
decision: do not reject Ho
p-value: 0.257
2.
we donot have enough evidence to support the claim that difference of means between right hand thread and left hand thread.
3.
TRADITIONAL METHOD
given that,
mean(x)=115.024
standard deviation , s.d1=18.718
number(n1)=41
y(mean)=119.646
standard deviation, s.d2 =17.672
number(n2)=41
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((350.364/41)+(312.3/41))
= 4.02
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 40 d.f is 2.021
margin of error = 2.021 * 4.02
= 8.125
III.
CI = (x1-x2) ± margin of error
confidence interval = [ (115.024-119.646) ± 8.125 ]
= [-12.747 , 3.503]
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DIRECT METHOD
given that,
mean(x)=115.024
standard deviation , s.d1=18.718
sample size, n1=41
y(mean)=119.646
standard deviation, s.d2 =17.672
sample size,n2 =41
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 = [( 115.024-119.646) ± t a/2 * sqrt((350.364/41)+(312.3/41)]
= [ (-4.622) ± t a/2 * 4.02]
= [-12.747 , 3.503]
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interpretations:
1. we are 95% sure that the interval [-12.747 , 3.503] 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
4.
margin of error = 8.125