Chapter PVII, Problem 21RE

(a)

To determine

To describe what you see in this scatterplot.

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. By looking at the scatterplot, we can say that the points are moving upwards that implies that there is positive association between the interval and duration. Also the points are close together in the form of a straight line so, this means that it shows a generally linear trend and are moderately scattered.

(b)

To determine

To write an appropriate hypothesis.

Answer to Problem 21RE

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. Since we have to show the relationship between the interval and duration, so the null hypothesis states that there is no linear relationship between interval and duration. And the alternative hypothesis states that there is linear relationship between interval and duration. So, we have,

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

(c)

To determine

To explain do you think the assumptions for inference are met.

Answer to Problem 21RE

Yes, the assumptions for inference are met.

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. So, we have,

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

Since, the histogram given in the question shows a uni-modal graph and it is roughly symmetric in nature then it met the Normal condition. Also the points in the residual plot show a random scatter this implies they met the random condition. So, the assumptions for inference are met.

(d)

To determine

To state a conclusion based on the regression analysis.

Answer to Problem 21RE

There is evidence of a positive linear relationship between duration and time to next eruption of Old Faithful.

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. So, we have,

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

Thus, by the regression analysis we know that,

  $\begin{array}{l}{R}^2=77\%\\ s=6.159\\ df=220\end{array}$

Variable	Coefficient	SE(Coeff)	t-ratio	P-value
Intercept	34	1.428	23.8	<0.0001
Duration	10	0.3822	27.1	<0.0001

Thus, by using the calculator $\text{TI}-89$ , the test statistics and the P-value is as:

  $\begin{array}{l}P\le 0.001\\ t=27.1\end{array}$

As we know that if the P-value is less than or equal to the significance level then the null hypothesis is rejected, so we have,

  $P<0.05\Rightarrow \text{Reject }{H}_0$

Thus, we conclude thatthere is evidence of a positive linear relationship between duration and time to next eruption of Old Faithful.

(e)

To determine

To create a $95\%$ confidence interval for the mean length of time that will elapse following two-minute eruption.

Answer to Problem 21RE

The average time to next eruption after two-minute eruption is between $(53.24,56.12)$ minutes, with $95\%$ confidence.

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. So, we have,

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

Thus, by the regression analysis we know that,

  $\begin{array}{l}{R}^2=77\%\\ s=6.159\\ df=220\end{array}$

Variable	Coefficient	SE(Coeff)	t-ratio	P-value
Intercept	34	1.428	23.8	<0.0001
Duration	10	0.3822	27.1	<0.0001

Thus, we have,

  $\begin{array}{l}SE({\widehat{\mu }}_{\nu })=\sqrt{S{E}^2(b_1)\times {(x_{\nu }-\overline{x})}^2+\frac{s_e^2}{n}}\\ =\sqrt{{0.3822}^2\times {(2-3.57623)}^2+\frac{6.159}{222}}\\ =0.625\\ {\widehat{y}}_{\nu }=b_0+b_1{x}_v\\ =33.9668+10.3582\times 2\\ =54.683\\ t*=1.97\\ 54.683\pm 1.97(0.625)\\ =(53.24,56.12)\end{array}$

Thus, the average time to next eruption after two-minute eruption is between $(53.24,56.12)$ minutes, with $95\%$ confidence.

(f)

To determine

To create a $95\%$ prediction interval for the length of time that will wait to see the next eruption.

Answer to Problem 21RE

We will have to wait between $(63.23,87.57)$ minutes after a four-minute eruption, with $95\%$ confidence.

Explanation of Solution

It is given in the question eruptions do not occur at uniform intervals and may vary greatly. The scatterplot for the duration in minutes and interval in minutes is given. So, we have,

  $\begin{array}{l}{H}_0:{\beta }_1=0\\ H_0:{\beta }_1\ne 0\end{array}$

Thus, by the regression analysis we know that,

  $\begin{array}{l}{R}^2=77\%\\ s=6.159\\ df=220\end{array}$

Variable	Coefficient	SE(Coeff)	t-ratio	P-value
Intercept	34	1.428	23.8	<0.0001
Duration	10	0.3822	27.1	<0.0001

Thus, we have,

  $\begin{array}{l}SE({\widehat{y}}_{\nu })=\sqrt{S{E}^2(b_1)\times {(x_{\nu }-\overline{x})}^2+\frac{s_e^2}{n}+s_e^2}\\ =\sqrt{{0.3822}^2\times {(4-3.57623)}^2+\frac{6.159}{222}+6.159}\\ =2.493\\ {\widehat{y}}_{\nu }=b_0+b_1{x}_v\\ =33.9668+10.3582\times 4\\ =75.4\\ t*=1.97\\ 75.4\pm 1.97(2.493)\\ =(63.23,87.57)\end{array}$

Thus, based on this regression, we will have to wait between $(63.23,87.57)$ minutes after a four-minute eruption, with $95\%$ confidence.