Chapter PVII, Problem 3RE

(a)

To determine

To explain is there association between the hardness of the water and the mortality rate and write the hypothesis.

Answer to Problem 3RE

Null hypothesis: There is no association between the hardness of the water and the mortality rate, i.e. $H_0:{\beta }_1=0$ .

Alternative hypothesis: There is an association between the hardness of the water and the mortality rate, i.e. $H_a:{\beta }_1\ne 0$ .

Explanation of Solution

In the question, the data were collected on the calcium concentration in the water and the mortality rates is given. Also the regression analysis and the scatterplot for the same is given. The regression analysis is as:

Variable	Coefficient	SE(Coeff)
Intercept	1676	29.3
Calcuim	-3.23	0.48

  $\begin{array}{l}{R}^2=43\%\\ s=143\\ df=59\end{array}$

Let us define the hypothesis as:

Null hypothesis: There is no association between the hardness of the water and the mortality rate, i.e. $H_0:{\beta }_1=0$ .

Alternative hypothesis: There is an association between the hardness of the water and the mortality rate, i.e. $H_a:{\beta }_1\ne 0$ .

(b)

To determine

To explain what do you conclude, assuming the assumptions for regression inference are met.

Answer to Problem 3RE

There is an association between the hardness of the water and the mortality rate.

Explanation of Solution

In the question, the data were collected on the calcium concentration in the water and the mortality rates is given. Also the regression analysis and the scatterplot for the same is given. The regression analysis is as:

Variable	Coefficient	SE(Coeff)
Intercept	1676	29.3
Calcuim	-3.23	0.48

  $\begin{array}{l}{R}^2=43\%\\ s=143\\ df=59\end{array}$

Let us define the hypothesis as:

Null hypothesis: There is no association between the hardness of the water and the mortality rate, i.e. $H_0:{\beta }_1=0$ .

Alternative hypothesis: There is an association between the hardness of the water and the mortality rate, i.e. $H_a:{\beta }_1\ne 0$ .

Now let us calculate the t -value and P-value from Excel as:

For P-value we will use the formula:

  $=\text{T}\text{.DIST}\text{.2T}(t,df)$

And for the t -value is calculated by dividing coefficient by SE(Coeff).

So, the calculation is as:

Variable	Coefficient	SE(Coeff)	t-ratio	P-value	P-value
Intercept	1676	29.3	=P14/Q14	=T.DIST.2T(R14,59)	<0.0001
Calcuim	-3.23	0.48	=P15/Q15	=T.DIST.2T(6.7292,59)	<0.0001

The result will be as:

Variable	Coefficient	SE(Coeff)	t-ratio	P-value	P-value
Intercept	1676	29.3	57.2014	2.20102E-53	<0.0001
Calcuim	-3.23	0.48	-6.7292	7.76098E-09	<0.0001

Thus, the test statistics and the P-value is as above.

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 that there is an association between the hardness of the water and the mortality rate.

(c)

To determine

To create a $95\%$ confidence interval for the slope of the true line relating calcium concentration and mortality.

Answer to Problem 3RE

  $(-4.19,-2.27)$ is the confidence interval.

Explanation of Solution

In the question, the data were collected on the calcium concentration in the water and the mortality rates is given. Also the regression analysis and the scatterplot for the same is given. The regression analysis is as:

  $\begin{array}{l}{R}^2=43\%\\ s=143\\ df=59\end{array}$

Variable	Coefficient	SE(Coeff)	t-ratio	P-value	P-value
Intercept	1676	29.3	57.2014	2.20102E-53	<0.0001
Calcuim	-3.23	0.48	-6.7292	7.76098E-09	<0.0001

Now, let us calculate the t -value for the confidence interval that i.e. from the student’s t -table from the Excel by the function:

  $\text{=TINV(Prob,df)}$

The calculation will be:

  $\text{=TINV(0}\text{.05,59)}$

So, the t -value comes out to be as:

  $t=2$

So, the confidence interval will be calculated as:

  $\begin{array}{l}{b}_1\pm t*\times SE\\ =-3.23\pm 2\times 0.48\\ =(-4.19,-2.27)\end{array}$

Thus, $(-4.19,-2.27)$ deaths per $100,000$ for each ppm calcium.

(d)

To determine

To interpret your interval.

Answer to Problem 3RE

We are $95\%$ confident that mortality (deaths per $100,000$ ) decreases on average, between $4.19\text{ and }2.27$ for each part per million of calcium in drinking water.

Explanation of Solution

In the question, the data were collected on the calcium concentration in the water and the mortality rates is given. Also the regression analysis and the scatterplot for the same is given. The regression analysis is as:

  $\begin{array}{l}{R}^2=43\%\\ s=143\\ df=59\end{array}$

Variable	Coefficient	SE(Coeff)	t-ratio	P-value	P-value
Intercept	1676	29.3	57.2014	2.20102E-53	<0.0001
Calcuim	-3.23	0.48	-6.7292	7.76098E-09	<0.0001

So, the confidence interval will be calculated as:

  $(-4.19,-2.27)$ .

Thus, based on the regression, we are $95\%$ confident that mortality (deaths per $100,000$ ) decreases on average, between $4.19\text{ and }2.27$ for each part per million of calcium in drinking water.