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

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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izboknil3

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
$$\displaystyle\hat{{{y}}}\pm{t}_{{{\frac{{\alpha}}{{{2}}}},\ {n}-{\left({k}+{1}\right)}\cdot{s}_{{\hat{{{Y}}}}}={10.091}\pm{2.845}\cdot{0.350}}}$$
$$\displaystyle={\left({9.095},\ {11.087}\right)}.$$