Chapter 11, Problem 95SE

At temperatures approaching absolute zero (−273°C), helium exhibits traits that defy many laws of conventional physics. An experiment has been conducted with helium in solid form at various temperatures near absolute zero. The solid helium is placed in a dilution refrigerator along with a solid impure substance, and the fraction (in weight) of the impurity passing through the solid helium is recorded. (The phenomenon of solids passing directly through solids is known as quantum tunneling.) The data are given in the following table.

1. a Fit a least-squares line to the data.
2. b Test the null hypothesis H₀ : β₁ = 0 against the alternative hypothesis H_a : β₁ < 0, at the α = .01 level of significance.
3. c Find a 95% prediction interval for the percentage of the solid impurity passing through solid helium at −273◦C. (This value of x is outside the experimental region where use of the model for prediction may be dangerous.)

To determine

Fit a least-square line to the data.

Answer to Problem 95SE

The least-square line is $\widehat{y}=-13.49035-0.05283x$.

Explanation of Solution

The least-square line is as follows:

$\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x$

Where,

${\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}$, ${\beta }_1=\frac{S_{xy}}{S_{xx}}$, $S_{xy}={\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)$ and $S_{xx}={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$

$\overline{x}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{x}_i}$, $\overline{y}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{y}_i}$

From the table of data, the values of $\overline{x}$ and $\overline{y}$ can be computed as follows:

$\overline{x}=-268.28$, $\overline{y}=0.6826$

Now, the values of $S_{xy}$ and $S_{xx}$ can be obtained as follows:

$\begin{array}{c}{S}_{xy}={\displaystyle \sum _{i=1}^{10}\left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)\\ ={\displaystyle \sum _{i=1}^{10}\left(x_i-\left(-268.28\right)\right)}\left(y_i-0.6826\right)\\ =-15.728\end{array}$

Thus, the value of $S_{xy}$ is –15.728.

$\begin{array}{c}{S}_{xx}={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\\ ={\displaystyle \sum _{i=1}^n{\left(x_i-\left(-268.28\right)\right)}^2}\\ =297.716\end{array}$

The value of $S_{xx}$ is 297.716.

Now, the values of slope and intercept are calculated as follows:

$\begin{array}{c}{\widehat{\beta }}_1=\frac{S_{xy}}{S_{xx}}\\ =\frac{-15.728}{297.716}\\ =-0.05283\end{array}$

The value of slope ${\beta }_1$ is –0.05283.

$\begin{array}{c}{\widehat{\beta }}_0=\overline{y}-{\widehat{\beta }}_1\overline{x}\\ =0.6826-\left(-0.05283\right)\times \left(-268.28\right)\\ =-13.49035\end{array}$

The value of intercept ${\widehat{\beta }}_0$ is –13.49035.

Substitute the value of intercept and slope in the model; then the model becomes,

$\begin{array}{c}\widehat{y}={\widehat{\beta }}_0+{\widehat{\beta }}_{1}x\\ \widehat{y}=-13.49035-0.05283x\end{array}$

Thus, the least-square line is $\widehat{y}=-13.49035-0.05283x$.

To determine

Test the hypothesis $H_0:{\beta }_1=0$ against the alternative $H_a:{\beta }_1<0$ at the 0.01 significance level.

Answer to Problem 95SE

There is sufficient evidence to conclude that ${\beta }_1$ is less than zero.

Explanation of Solution

The null and alternative hypotheses are stated as follows:

$H_0:{\beta }_1=0$

$H_a:{\beta }_1<0$

The significance level is 0.01.

The test statistic is as follows:

$t=\frac{{\widehat{\beta }}_1}{\left(s\sqrt{c_{11}}\right)}$

Where,

${\widehat{\beta }}_1=\frac{S_{xy}}{S_{xx}}$, $S_{xy}={\displaystyle \sum _{i=1}^n\left(x_i-\overline{x}\right)}\left(y_i-\overline{y}\right)$, and $S_{xx}={\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$.

$s=\sqrt{\frac{SSE}{n-2}}$, $c_{11}=\frac{1}{S_{xx}}$

$SSE=S_{yy}-{\widehat{\beta }}_1{S}_{xy}$

From Part (a), it is seen that ${\widehat{\beta }}_1=-0.05283$, $S_{xx}=297.716$, $S_{xy}=-15.728$

Now, compute the value of $S_{yy}$ as follows:

$\begin{array}{c}{S}_{yy}={\displaystyle \sum _{i=1}^n{\left(y_i-\overline{y}\right)}^2}\\ ={\displaystyle \sum _{i=1}^{10}{\left(y_i-0.6826\right)}^2}\\ =0.97315\end{array}$

The value of $S_{yy}$ is 0.97315.

It is noticed that $SSE=S_{yy}-{\widehat{\beta }}_1{S}_{xy}$.

$\begin{array}{c}SSE=S_{yy}-{\widehat{\beta }}_1{S}_{xy}\\ =0.97315-\left(-0.05283\right)\left(-15.728\right)\\ =0.14225\end{array}$

Thus, the value of $SSE$ is 0.14225.

In addition, it is known that $s=\sqrt{\frac{SSE}{n-2}}$, $c_{11}=\frac{1}{S_{xx}}$.

$\begin{array}{c}s=\sqrt{\frac{SSE}{n-2}}\\ =\sqrt{\frac{0.14225}{10-2}}\\ =0.13335\end{array}$,

$\begin{array}{c}{c}_{11}=\frac{1}{S_{xx}}\\ =\frac{1}{297.716}\\ =0.00336\end{array}$

The test statistic can be obtained as follows:

$\begin{array}{c}t=\frac{{\widehat{\beta }}_1}{\left(s\sqrt{c_{11}}\right)}\\ =\frac{-0.05283}{\left(0.13335\right)\sqrt{0.00336}}\\ \approx -6.836\end{array}$

Thus, the test statistic value is –6.836.

Critical value:

Step-by-step procedure to obtain the t-critical value using Table 5 of Appendix 3:

• Locate the degrees of freedom as 8 in the column of df.
• Move left until column headed by $t_{0.01}$.
• Selecting the intersection of (8, 0.01) gives the critical value of t.

The critical value of t for the left-tailed test is $-2.896$.

Decision rule:

Reject the null hypothesis, if $\left|t_{cal}\right|>\left|t_{tab}\right|$, otherwise fail to reject the null hypothesis.

Conclusion:

The test statistic value is greater than the table value.

That is, $\left(\left|t_{cal}\right|=6.836\right)>\left(\left|t_{tab}\right|=2.896\right)$.

Hence, by the rejection rule, reject the null hypothesis $H_0$ at the 0.01 significance level.

Therefore, there is sufficient evidence to conclude that ${\beta }_1$ is less than zero.

To determine

Find a 95% prediction interval for the percentage of the solid impurity passing through solid helium at $–{273}^0\text{C}$.

Answer to Problem 95SE

A 95% prediction interval for the percentage of the solid impurity passing through solid helium is (0.599, 1.265).

Explanation of Solution

A 95% prediction interval for the percentage of the solid impurity passing through solid helium is as follows:

$\left({\widehat{\beta }}_0+{\widehat{\beta }}_1{x}^{\ast }-t_{0.05}.s\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{S_{xx}}},{\widehat{\beta }}_0+{\widehat{\beta }}_1{x}^{\ast }+t_{0.05}.s\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{S_{xx}}}\right)$

From Part (a), it is obtained that,

${\widehat{\beta }}_0=-13.49035$, ${\widehat{\beta }}_1=-0.05283$, $S_{xx}=297.716$, $\overline{x}=-268.28$, and $n=10$.

From Part (b), it is seen that the value of $s=0.13335$.

The t-critical value using Table 5 of Appendix 3 is $t_{0.025}=2.306$.

The lower limit is computed as follows:

$\begin{array}{c}{\widehat{\beta }}_0+{\widehat{\beta }}_1{x}^{\ast }-t_{0.025}.s\sqrt{1+\frac{1}{n}+\frac{{\left(x^{\ast }-\overline{x}\right)}^2}{S_{xx}}}\\ =-13.49035-0.05283\left(273\right)-2.306\left(0.13335\right)\sqrt{1+\frac{1}{10}+\frac{{\left(-273-\left(-268.28\right)\right)}^2}{297.716}}\\ =0.599\end{array}$

The upper limit is as follows:

$\begin{array}{c}{\widehat{\beta }}_0+{\widehat{\beta }}_1{x}^{\ast }+t_{0.025}.s\sqrt{1+\frac{1}{n}+\frac{\left(x^{\ast }-\overline{x}\right)}{S_{xx}}}\\ =-13.49035-0.05283\left(273\right)+2.306\left(0.13335\right)\sqrt{1+\frac{1}{10}+\frac{{\left(-273-\left(-268.28\right)\right)}^2}{297.716}}\\ =1.265\end{array}$

Thus, a 95% prediction interval for the percentage of the solid impurity passing through solid helium is (0.599, 1.265).