An article in the Journal of Aircraft (1986, Vol. 23, pp. 859-864) described a n
ID: 3315115 • Letter: A
Question
An article in the Journal of Aircraft (1986, Vol. 23, pp. 859-864) described a new equivalent plate analysis method formulation that is capable of modeling aircraft structures such as cranked wing boxes and that produces results similar to the more computationally intensive finite element analysis method. Natural vibration frequencies for the cranked wing box structure are calculated using both methods, and results for the first seven natural frequencies follow: SHOWN IN PICTURE. a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use alpha=0.05. Find the P-value. b) Find the 95% confidence interval on the mean difference between the two methods. Use alpha=0.05.Please be descriptive! And do it on paper, not with use of Excel or Minitab. Thank you very much!
An article in the Journal of Aircraft (1986, Vol. 23, pp. 859-864) described a new equivalent plate analysis method formulation that is capable of modeling aircraft structures such as cranked wing boxes and that produces results similar to the more computationally intensive finite element analysis method. Natural vibration frequencies for the cranked wing box structure are calculated using both methods, and results for the first seven natural frequencies follow: SHOWN IN PICTURE. a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use alpha=0.05. Find the P-value. b) Find the 95% confidence interval on the mean difference between the two methods. Use alpha=0.05.
Please be descriptive! And do it on paper, not with use of Excel or Minitab. Thank you very much!
a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use alpha=0.05. Find the P-value. b) Find the 95% confidence interval on the mean difference between the two methods. Use alpha=0.05.
Please be descriptive! And do it on paper, not with use of Excel or Minitab. Thank you very much!
3. (10-54) An article in the Journal of Aircraft (1986, Vol. 23, pp. 859-864) described a new equivalent plate analysis method formulation that is capable of modeling aircraft structures such as element analysis method. Natural vibration frequencies for the cranked wing box structure are calculated using both methods, and results for the first seven natural frequencies follow: FiniteEquivalent Element Plate, Freq. Cycle/s 4.58 48.52 97.22 113.99 74.73 212.72 14.76 49·10 99.99 117.53 181.22 220.14 294.80 7 277.38 (20 points total, 10 points each) (a) Do the data suggest that the two methods provide the same mean value for natural vibration frequency? Use = 0.05. Find the P-value. Use the 7-step procedure.
Explanation / Answer
Q1.
Given that,
mean(x)=296
standard deviation , s.d1=12
number(n1)=10
y(mean)=315
standard deviation, s.d2 =18
number(n2)=25
null, Ho: u1 = u2
alternate, H1: u1 != u2
level of significance, = 0.05
from standard normal table, two tailed t /2 =2.262
since our test is two-tailed
reject Ho, if to < -2.262 OR if to > 2.262
we use test statistic (t) = (x-y)/sqrt(s.d1^2/n1)+(s.d2^2/n2)
to =296-315/sqrt((144/10)+(324/25))
to =-3.6324
| to | =3.6324
critical value
the value of |t | with min (n1-1, n2-1) i.e 9 d.f is 2.262
we got |to| = 3.63242 & | t | = 2.262
make decision
hence value of | to | > | t | and here we reject Ho
p-value: two tailed ( double the one tail ) - Ha : ( p != -3.6324 ) = 0.005
hence value of p0.05 > 0.005,here we reject Ho
ANSWERS
---------------
null, Ho: u1 = u2
alternate, H1: u1 != u2
test statistic: -3.6324
critical value: -2.262 , 2.262
decision: reject Ho
p-value: 0.005
Q2.
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.
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((144/10)+(324/25))
= 5.231
II.
margin of error = t a/2 * (standard 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]
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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]
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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
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