Chapter 18, Problem 18.31E

(a)

To determine

To find:

The two sample $t-\text{statistic}$ and $p-\text{value}$ of the given sample.

Answer to Problem 18.31E

  $t=5.09$

Explanation of Solution

Given:

Group	$n$	$\overline{x}$	$s$
Tongue piercing	$46$	$4.0$	$3.5$
No Tongue piercing	$46$	$1.2$	$1.3$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

  $\begin{array}{l}{n}_1=46,\overline{x_1}=4.0,s_1=3.5\\ n_2=46,\overline{x_2}=1.2,s_2=1.3\end{array}$

Standard error

  $\begin{array}{l}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{\left(3.5\right)}^2}{46}+\frac{{\left(1.3\right)}^2}{46}}\\ =\sqrt{0.266+0.36}\\ =0.8757\end{array}$

Degree of freedom

  $\begin{array}{l}=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}\\ =57\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ t=\frac{4.0-1.2}{0.8757}\\ t=5.09\end{array}$

(b)

To determine

To explain:

The tongue piercing causes more enamel cracks of the given sample.

Explanation of Solution

Given:

Group	$n$	$\overline{x}$	$s$
Tongue piercing	$46$	$4.0$	$3.5$
No Tongue piercing	$46$	$1.2$	$1.3$

Concept used:

Formula

Standard error

  $\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}$

Degree of freedom

  $=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}$

Significance level

  $\alpha =1-\text{confidence}$

Calculation:

  $\begin{array}{l}{n}_1=46,\overline{x_1}=4.0,s_1=3.5\\ n_2=46,\overline{x_2}=1.2,s_2=1.3\end{array}$

Standard error

  $\begin{array}{l}=\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}\\ =\sqrt{\frac{{\left(3.5\right)}^2}{46}+\frac{{\left(1.3\right)}^2}{46}}\\ =\sqrt{0.266+0.36}\\ =0.8757\end{array}$

Degree of freedom

  $\begin{array}{l}=\frac{{\left[\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}\right]}^2}{\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_1-1}+\frac{{\left(\frac{s_1^2}{n_1}\right)}^2}{n_2-1}}\\ =57\end{array}$

The $95\%$ confidence interval for difference

  $\begin{array}{l}t=\frac{\overline{x_1}-\overline{x_2}}{\sqrt{\frac{s_1^2}{n_1}+\frac{s_2^2}{n_2}}}\\ t=\frac{4.0-1.2}{0.8757}\\ t=5.09\end{array}$

  $p-$ Value is less than $0.01$

So the null hypothesis are rejected

The claim that tongue piercing causes more enamel cracks.