Chapter 10, Problem 20CQ

a.

To determine

To construct: The scatterplot for the variablesthe age of a child and the number of cavities.

Answer to Problem 20CQ

Output using the MINITAB software is given below:

Explanation of Solution

Given info:

The data shows the age of a child (x) and the number of cavities (y) values.

Calculation:

Step by step procedure to obtain scatterplot using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose Simple and then click OK.
• Under Y variables, enter a column ofNo.of cavities.
• Under X variables, enter a column ofAge of child.
• Click OK.

b.

To determine

To compute: The value of the correlation coefficient.

Answer to Problem 20CQ

The value of the correlation coefficientis 0.842.

Explanation of Solution

Calculation:

Correlation coefficient r:

Software Procedure:

Step-by-step procedure to obtain the ‘correlation coefficient’ using the MINITAB software:

• Select Stat >Basic Statistics > Correlation.
• In Variables, select x and y from the box on the left.
• Click OK.

Output using the MINITAB software is given below:

From the MINITAB output, the value of the correlation is 0.842.

c.

To determine

To test: The significance of the correlation coefficient at $\alpha =0.05$ using Table I.

Answer to Problem 20CQ

The conclusion is that,there is a linear relation between the age of a child and the number of cavities.

Explanation of Solution

Given info:

The level of significance is $\alpha =0.05$ and the sample size is 6.

Calculation:

The hypotheses are given below:

Null hypothesis:

$H_0:\rho =0$

That is, there is no linear relation betweenthe age of a child and the number of cavities.

Alternative hypothesis:

$H_1:\rho \ne 0$

That is, there is a linear relation between the age of a child and the number of cavities.

The sample size is 6.

The formula to find the degrees of the freedom is $n-2$.

That is,

$\begin{array}{c}n-2=6-2\\ =4\end{array}$

From the “TABLE –I: Critical Values for the PPMC”, the critical value for 4 degrees of freedom and $\alpha =0.01$ level of significance is 0.811.

Rejection Rule:

If the absolute value of r is greater than the critical value then reject the null hypothesis.

Conclusion:

From part (b), the value of r is0.842that is the absolute value of r is 0.842.

Here, the absolute value of r is greater than the critical value $\left|r\right|>\text{critical value}$.

That is, $\left|r\right|\left(=0.842\right)<\text{critical value}\left(=0.811\right)$.

By the rejection rule,reject the null hypothesis.

There is asufficient evidence to support the claim that “there is alinear relation betweenthe age of a child and the number of cavities”.

d.

To determine

To find: The regression equation for the given data.

Answer to Problem 20CQ

The regression equation for the given datais $\widehat{y}=-1.918+0.551x$.

Explanation of Solution

Calculation:

Regression:

Software procedure:

Step by step procedure to obtain the regression equation using the MINITAB software:

• Choose Stat > Regression > Regression.
• In Responses, enter the column ofNo.of cavities.
• In Predictors, enter the column ofAge of child.
• Click OK.

Output using the MINITAB software is given below:

Thus, regression equation for the given datais $\widehat{y}=-1.918+0.551x$.

e.

To determine

To construct: The scatterplot for the variablesthe age of a child and the number of cavitieswith regression line.

Answer to Problem 20CQ

Output using the MINITAB software is given below:

Explanation of Solution

Calculation:

Step by step procedure to obtain scatterplot using the MINITAB software:

• Choose Graph > Scatterplot.
• Choose with line and then click OK.
• Under Y variables, enter a column of No.of cavities.
• Under X variables, enter a column ofAge of child.
• Click OK.

f.

To determine

To obtain: The predicted value of the number of cavities for a child of 11.

Answer to Problem 20CQ

Thepredicted value of the number of cavities is 4.143.

Explanation of Solution

Calculation:

Thus, regression equation for the given datais $\widehat{y}=-1.918+0.551x$.

Substitute x as 11 in the regression equation

$\begin{array}{c}\widehat{y}=-1.918+0.551x\\ =-1.918+\left(0.551\times 11\right)\\ =-1.918+6.061\\ =4.143\end{array}$

Thus, the predicted value of the number of cavities is 4.143.