The article “Analysis of the Modeling Methodologies for Predicting the Strength

Lipossig 2021-09-29 Answered
The article “Analysis of the Modeling Methodologies for Predicting the Strength of Air-Jet Spun Yarns” (Textile Res. J., 1997: 39–44) reported on a study carried out to relate yarn tenacity \(\displaystyle{\left({y},\ \in\ {\frac{{{g}}}{{{t}{e}{x}}}}\right)}\) to yarn count \(\displaystyle{\left({x}_{{{1}}},\ \in\ {t}{e}{x}\right)}\), percentage polyester \(\displaystyle{\left({x}_{{{2}}}\right)}\), first nozzle pessure \(\displaystyle{\left({x}_{{{3}}},\ \in\ {\frac{{{k}{g}}}{{{c}{m}^{{{2}}}}}}\right)}\), and second nozzle pressure \(\displaystyle{\left({x}_{{{4}}},\ \in\ {\frac{{{k}{g}}}{{{c}{m}^{{{2}}}}}}\right)}\) The estimate of the constant term in the corresponding multiple regression equation was 6.121. The estimated coefficients for the four predictors were -0.082, 0.113, 0.256, and -0.219, respectively, and the coefficient of multiple determination was 0.946
a) Assuming that the sample size was \(\displaystyle{n}={25}\), state and test the appropriate hypotheses to decide whether the fitted model specifies a useful linear relationship between the dependent variable and at least one of the four model predictors.
b) Again using \(\displaystyle{n}={25}\), calculate the value of adjusted \(\displaystyle{R}^{{{2}}}\).
c) Calculate a 99% confidence interval for true mean yarn tenacity when yarn count is 16.5, yarn contains 50% polyester, first nozzle pressure is 3, and second nozzle pressure is 5 if the estimated standard deviation of predicted tenacity under these circumstances is 0.350.

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Expert Answer

izboknil3
Answered 2021-09-30 Author has 6736 answers

Step 1
a) A model utility test for testing
\(\displaystyle{H}_{{{0}}}:\beta_{{{1}}}=\beta_{{{2}}}=\cdots=\beta_{{{k}}}={0}\)
versus alternative
\(\displaystyle{H}_{{{a}}}:\text{at least one,}\ \beta_{{{i}}}\ne{0},\ {\left({i}={1},\ {2},\ \cdots,\ {k}\right)}\)
use test statistic value
\(\displaystyle{f}={\frac{{{\frac{{{R}^{{{2}}}}}{{{k}}}}}}{{{\frac{{{\left({1}-{R}^{{{2}}}\right)}}}{{{\left[{n}-{\left({k}+{1}\right)}\right]}}}}}}}={\frac{{{\frac{{{S}{S}{R}}}{{{k}}}}}}{{{\frac{{{S}{S}{E}}}{{{\left[{n}-{\left({k}+{1}\right)}\right]}}}}}}}={\frac{{{M}{S}{R}}}{{{M}{S}{E}}}}\)
The test is upper-tailed - the P-value is the area under \(\displaystyle{F}_{{{k},\ {n}-{\left({k}+{1}\right)}}}\) to the right of the test statistic value.
A model utility test is of use. The coefficient of multiple determination, n, and k are given; thus, the value of f statistic is
\(\displaystyle{\frac{{{\frac{{{R}^{{{2}}}}}{{{k}}}}}}{{{\frac{{{\left({1}-{R}^{{{2}}}\right)}}}{{{\left[{n}-{\left({k}+{1}\right)}\right]}}}}}}}={\frac{{{\frac{{{0.946}}}{{{4}}}}}}{{{\frac{{{1}-{0.946}}}{{{20}}}}}}}={87.6}\)
As usual, one could use P-value under the mentioned curve, which is
\(\displaystyle{P}={0.000}\),
or the table in the appendix, for which the closest values is for \(\displaystyle\alpha={0.001}\)
\(\displaystyle{F}_{{{0.001},\ {4},\ {20}}}={7.1}\)
Since
\(\displaystyle{P}={0.00}{<}\alpha\)
reject null hypothesis at reasonable significance level. Similarly same conclusion can be made from the fact
\(\displaystyle{F}_{{{0.001},\ {4},\ {20}}}={7.1}\leq{87.6}={f}\)
Step 2
b) Adjusted \(\displaystyle{R}^{{{2}}}\) is given by
\(\displaystyle{1}-{\frac{{{n}-{1}}}{{{n}-{\left({k}+{1}\right)}}}}{\frac{{{S}{S}{E}}}{{{S}{S}{T}}}}\)
Using this, and the fact that
\(\displaystyle{\frac{{{S}{S}{E}}}{{{S}{S}{T}}}}={1}-{R}^{{{2}}}\)
the adjusted \(R^{2}\) is
\(\displaystyle{1}-{\frac{{{n}-{1}}}{{{n}-{\left({k}+{1}\right)}}}}{\left({1}-{R}^{{{2}}}\right)}={1}-{\frac{{{24}}}{{{20}}}}{\left({1}-{0.946}\right)}={0.935}\)
c) A \(\displaystyle{100}{\left({1}-\alpha\right)}\%\) confidence interval for mean of Y when \(\displaystyle{x}={x}^{{\cdot}},\ \mu_{{{Y}\cdot{x}^{{\cdot}}}}\), is
\(\displaystyle\hat{{{y}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}},\ {n}-{\left({k}+{1}\right)}\cdot{s}_{{{Y}}}}}\)
For \(\displaystyle{x}^{{\cdot}}={\left({16.5},\ {50},\ {3},\ {5}\right)}\), putting values into the given regression model (estimated coefficients are given)0 the following holds
\(\displaystyle\hat{{{y}}}_{{{16.5},\ {50},\ {3},\ {5}}}={6.121}-{0.082}{x}_{{{1}}}+{0.113}{x}_{{{2}}}+{0.256}{x}_{{{3}}}-{0.219}{x}_{{{4}}}\)
\(\displaystyle={6.121}-{0.082}\cdot{16.5}+{0.113}\cdot{50}+{0.256}\cdot{3}-{0.219}\cdot{5}\)
\(\displaystyle={10.091}\)
Using the value given in the exercise, a 95% confidence interval for the true average when \(\displaystyle{\left({x}_{{{1}}},\ {x}_{{{2}}},\ {x}_{{{3}}},\ {x}_{{{4}}}\right)}={\left({16.5},\ {50},\ {3},\ {5}\right)}\) is
\(\displaystyle\hat{{{y}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}},\ {n}-{\left({k}+{1}\right)}\cdot{s}_{{\hat{{{Y}}}}}={10.091}\pm{2.845}\cdot{0.350}}}\)
\(\displaystyle={\left({9.095},\ {11.087}\right)}.\)

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