= Solution
Let $s_i$ be speed and $j_i\in\{1,2,3,4\}$ the tool type. Each fit is a <normal linear model> with independent $N(0,\sigma^2)$ errors. The three mean specifications are
$$
\begin{aligned}
\text{model 1}:&\quad E(Y_i)=\alpha+\beta s_i,\\
\text{model 2}:&\quad E(Y_i)=\alpha+\tau_{j_i}+\beta s_i,\qquad\tau_1=0,\\
\text{model 3}:&\quad E(Y_i)=\alpha+\tau_{j_i}+(\beta+\delta_{j_i})s_i,
\qquad\tau_1=\delta_1=0.
\end{aligned}
$$
The constraints are treatment coding with type 1 as reference. The parameter counts are $2,5,8$, respectively. Model 2 gives parallel lines; model 3 allows <interaction terms> and distinct slopes.
In the <analysis of variance> interaction row, adding three independent slope differences costs three <degrees of freedom>. Its <extra sum of squares> is the reduction in <residual sum of squares>, $118.53-91.51=27.02$. Its mean square is $27.02/3\approx9.007$, and its <F-test> statistic is
$$
\boxed{F=\frac{27.02/3}{91.51/12}\approx1.181.}
$$
Thus the four missing entries are \b[$3$, $27.02$, $9.01$, and $1.18$], to the precision allowed by the rounded output. This tests $H_0:\delta_2=\delta_3=\delta_4=0$ against the alternative that at least one slope difference is nonzero. Under $H_0$ and the <normal linear model> assumptions, the statistic has distribution $F_{3,12}$. Its <p-value> $0.3579213$ gives no reason to reject at 5%.
The simpler common-line model is inadequate compared with the parallel-line model. Testing $H_0:\tau_2=\tau_3=\tau_4=0$ against at least one nonzero type effect, while retaining speed, gives the partial <F-test>
$$
F=\frac{(1282.08-118.53)/3}{118.53/15}\approx49.08,
\qquad F\sim F_{3,15}\quad\text{under }H_0.
$$
Its <p-value> is far below $0.001$, so reject the common-line restriction. This reduction differs from the sequential type sum of squares in the displayed table, because that table adds type before speed. \b[Recommend model 2: different intercepts and a common decreasing slope.]
The selected coefficient estimates give
$$
\boxed{\begin{aligned}
\widehat m_1(s)&=35.891690-0.024585s,\\
\widehat m_2(s)&=35.058282-0.024585s,\\
\widehat m_3(s)&=48.499350-0.024585s,\\
\widehat m_4(s)&=52.430811-0.024585s.
\end{aligned}}
$$
The intercept estimates the expected lifetime of type 1 at speed zero; if zero speed is outside the data range, it is only an extrapolated intercept. Its <standard error> is $4.039533$, and its <Student t-test> ratio is $8.885$, with two-sided <p-value> $2.31\times10^{-7}$. At any fixed speed, type 2 differs from type 1 by $-0.833408$ hours, with <standard error> $1.904707$, $t=-0.438$ and <p-value> $0.668$, giving little evidence of a difference. Types 3 and 4 exceed type 1 by $12.607660$ and $16.539121$ hours, with <standard errors> $1.777937$ and $1.876116$; their test ratios $7.091$ and $8.816$ and <p-values> $3.68\times10^{-6}$ and $2.55\times10^{-7}$ support positive differences. Each of these individual coefficient tests has null value zero, alternative nonzero, and null distribution $t_{15}$.
The speed coefficient means a reduction of $0.024585$ hours per extra revolution per minute, or $2.4585$ hours per additional 100 rpm, for every type. Its <standard error> is $0.004682$; the ratio $-5.251$ has null distribution $t_{15}$ when the common slope is zero, and two-sided <p-value> $9.77\times10^{-5}$.
The <residual standard error> $2.811$ estimates the common noise standard deviation in hours using 15 <residual degrees of freedom>, since $20-5=15$. The <coefficient of determination> $0.9247$ means that about $92.47\%$ of the corrected lifetime variation is explained by this fit. The overall <F-test> statistic $46.08$ tests the simultaneous restriction $\tau_2=\tau_3=\tau_4=\beta=0$ against at least one nonzero coefficient; its null distribution is $F_{4,15}$ and its <p-value> $2.974\times10^{-8}$ rejects an intercept-only mean. These conclusions remain conditional on suitable <regression diagnostics> for independent, homoscedastic, approximately normal errors.
The required sketch plots the four fitted lines. Its speed interval is illustrative because the observed speeds are not supplied; the ordering and vertical gaps are determined by the fitted coefficients.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2014/iii/paper-33-tool-lifetime.png]
{title=Parallel fitted tool-lifetime lines for the four tool types}
{height=552}
Back to article page
