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

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

\(\displaystyle={\left({9.095},\ {11.087}\right)}.\)